AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions and Answers.

8th Class Maths 8th Lesson Exploring Geometrical Figures InText Questions and Answers

Do this

Question 1.
Identify which of the following pairs of figures are congruent.     [Page No. 184]
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 1
Answer:
The congruent figures are (1, 10), (2, 6, 8), (3, 7), (12, 14), (9, 11), (4, 13).

AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions

Question 2.
Look at the following pairs of figures and find whether they are congruent. Give reasons. Name them.    [Page No. 185]
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 2
Answer:
i) △ABC, △PQR
∠A = ∠Q Angle
There is no information about other angles (or) sides.
But if we overlap each other, they coincide.
∴ △ABC ≅ △PQR

ii) From △PLM, △QNM
PL = QN (S)
LM = MN (S)
PM = QM (S)
By S.S.S congruency, these two triangles are congruent.
∴ △PLM ≅ △QNM

iii) From △LMN, △PQR
NL ≠ PQ,LM ≠ QR, NM ≠ RP [∵ The corresponding angles are not given]
∴ △LMN ≆ △PQR

iv) From fig. ABCD is a parallelogram and LMNO is a rectangle.
In any case a rectangle and a parallelogram are not congruent.
∴ ▱ ABCD ≆ □ DLMNO

v) Both the circles are having same radii,
i.e., r1 = r2 = 2 units
∴ The given circles are congruent to each other.

AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions

Question 3.
Identify the out line figures which are similar to those given first.    [Page No. 186]
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 3
Answer:
The similar figures are (a) (ii), (b) (ii).

Question 4.
Draw a triangle on a graph sheet and draw its dilation with scale factor 3. Are those two figures are similar?      [Page No. 191]
Answer:
Step – 1: Draw a △ PQR and choose the center of dilation C which is not on the triangle. Join every vertex of the triangle from C and produce.
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 4
Step – 2: By using compasses, mark three points P’, Q’ and R’ on the projections
so that
CP’ = k(CP) = 3CP
CQ’ = 3 CQ
CR’ = 3 CR
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 5
Step- 3: Join P’Q’,Q’R’and R’P’.
Notice that △P’Q’R’ ~ △PQR
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 6

AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions

Question 5.
Try to extend the projection for any other diagram and draw squares with scale factor 4, 5. What do you observe? [Page No. 191]
Answer:
Sometimes we need to enlarge 10 the figures say for example while making cutouts, and sometimes we reduce the figures during designing. Here in every case the figures must be similar to the original. This means we need to draw enlarged or reduced similar figures in daily life. This method of drawing enlarged or reduced similar figure is called ‘Dilation’.
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 7
Observe the following dilation ABCD, it is a square drawn on a graph sheet.
Every vertex A, B, C, D are joined from the sign ‘O’ and produced to 4 times the length upto A, B, C and D respectively. Then A, B, C, Dare joined to form a square which 4 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and
\(\frac{OA’}{OA}\) = \(\frac{4}{1}\) = 4 is called scale factor.

Question 6.
Draw all possible lines of symmetry for the following figures.     [Page No. 193]
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 8
Answer:
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 9

Try these

AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions

Question 1.
Stretch your hand, holding a scale in your hand vertically and try to cover your school building by the scale (Adjust your distance from the building). Draw the figure and estimate height of the school building.      [Page No. 189]
Answer:
Illustration: A girl stretched her arm towards a school building, holding a scale vertically in her arm by standing at a certain distance from the school building. She found that the scale exactly covers the school building as in figure. If we compare this illustration with the previous example, we can say that
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 10
By measuring the length of the scale, length of her arm and distance of the school building, we can estimate the height of the school building.

Question 2.
Identify which of the following have point symmetry.     [Page No. 196]
1.
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 11
2. Which of the above figures are having symmetry ?
3. What can you say about the relation between line symmetry and point symmetry?
Answer:
1. The figures which have point symmetry are (i), (ii), (iii), (v).
2. (i), (iii), (v).
3. Number of lines of symmetry = Order of point symmetry.

Think, discuss and write

Question 1.
What is the relation between order of rotation and number of axes of symmetry of a geometrical figure?     [Page No. 195]
Answer:
The line which cuts symmetric figures exactly into two halves is called line of symmetry. The figure is rotated around its central point so that it appears two or more times as original. The number of times for which it appears the same is called the order of rotation.
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 12
From the above table number of lines of symmetry = Number of order of rotation.

AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions

Question 2.
How many axes of symmetry does a regular polygon has? Is there any relation between number of sides and order of rotation of a regular polygon?      [Page No. 195]
Answer:
Number of sides of a regular polygon are n. Then its lines of symmetry are also n.
AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions 13

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 10th Lesson Not For Drinking-Not For Breathing

8th Class Biology 10th Lesson Not For Drinking-Not For Breathing Textbook Questions and Answers

Improve Your Learning

Question 1.
How does air pollution lead to water pollution?
Answer:

  1. Air and water are not separate problem. There is a close link between the health of air and the health of water.
  2. Nitrogen and chemical contaminants are two types of pollutants that harm both the air and water.
  3. Up to l/3rd of the nitrogen that pollutes the bay and it’s rivers comes from the air.
  4. Air pollution from a very large geographic area can eventually wind up in the Bay.
  5. Use of fertilizers and pesticides in agriculture pollutes not only air but also land and water.
  6. Sources of air pollution includes vehicles, industries, power plants and farm operations which lead to water pollution.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 2.
What steps can be taken up to control air pollution and water pollution?
Answer:
Some of the methods for controlling air pollutions are:

  1. Tall chimneys should be installed in all factories to reduce air pollution at the ground level.
  2. Better designed fuel burning equipment should be used in homes and industries so that fuel burnt completely.
  3. Install electrostatic precipitators in the chimneys of industries.
  4. Reduce vehicular emissions by using non polluting fuels like CNG (compressed natural gas)
  5. Use LPG for domestic use (Liquify Petroleum Gas)
  6. Improve the quality of fuel in automobiles and use catalytic converters in them.
  7. Make use of renewable alternative source of energy like solar energy, wind energy and hydro energy.
  8. All motor vehicles should be maintained properly so that they comply with pollution norms.
  9. Use unleaded petrol.
  10. Plant and grow, more and more trees, we can protect plants and trees. Vanamahotsav should be continued every year in July month.

Prevention and controlling of water pollution

  1. Toxic industrial wastes should be treated chemically to neutralize the harmful substances present in it before discharging into rivers and lakes.
  2. The sewage should not be dumped in to the rivers directly.
  3. The use of excessive fertilizers and pesticides should be avoided.
  4. The use of synthetic detergent should be minimized or biodegradable detergents should be used.
  5. Dead bodies of human beings and animals should not be thrown into rivers.
  6. The excreta and other garbage should be treated in a biogas plant to get fuel as well as manure.
  7. The water of rivers, streams, ponds and lakes should be purified or cleaned. For example Ganga action plait launched by the Indian Government.
  8. Trees and shrubs should bp planted along the banks of the rivers.
  9. There should be general Rareness among the people regarding the harmful of water pollution and the ways of prevention.
  10. Waste papers, plastic, waste fbod materials and rotten food and vegetables should not be thrown into open drains.
  11. Go for the alternate energy resources that can replenish themselves without affecting our environment.
  12. Reuse the materials for secondary purpose. Recycling is the next stage of reuse.

Question 3.
Why does the increased level of nutrients in the water affect the survival of aquatic organisms?
Answer:
Plants nutrients:

  1. Phosphates and nitrates – chemical fertilizers from agriculture run – off due to rain and industrial wastes enter into water through sewage and pollute the water.
  2. It helps algae to bloom, weeds to grow and bacteria is spread. As a result water turn green and cloudy and smell bad.
  3. Decomposing plants use up the oxygen in water, disrupting aquatic life, reducing biodiversity and even killing aquatic life.
  4. Thus, this enrichment of water by nutrients leading to excessive plant growth and depletion of oxygen is known as ‘Eutrophication’. This affects aquatic life badly.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 4.
Road side plants cannot grow properly. Find your own reasons and explain with your argument.
Answer:

  1. The plantation along the roads mainly includes neem, peepal, banyan, almond etc.
  2. It was observed that vegetation at roadside with heavy traffic and markets was much affected by vehicular emissions.
  3. Significant decrease in total chlorophyll and protein content was observed with reduced leaf area.
  4. It is concluded that plants can be uses as indicators for urban air pollution and there is need to protect the road side plants from air pollution.
  5. Biomonitoring of plants is an important tool to evaluate the impact of air pollution on plants.
  6. A study suggests that plants have the potential to serve as excellent quantitative and qualitative indices of pollution levels.
  7. So plants should be grown and protected.

Question 5.
Sudheer is a traffic constable. What do you think about his health? Give some suggestions to protect his health during duty period.
Answer:

  1. Environmental pollution place a significance role in the development of various respiratory diseases. Different particles and gases from vehicular emissions like carbondioxide, carbon monoxide, sulphur, benzene, lead, nitrogen dioxide and black smoke are the root of the problem.
  2. Traffic police are increasingly becoming victims of a diabetes and allergies at a younger age. Irregular work schedule were posing a challenge to the health of the police. Besides physical strain, mental stress and asymmetrical food habits are also contributing the problem.
  3. The traffic police men who work at busy intersections are at the highest risk of developing asthma, bronchitis, shortness of breath, sore throat, chest pain, lung cancer, eye irritation, skin ailments. Impaired hearing, excessive carboxy haemoglobin and annoyance with noise also high blood pressure and cardiovascular problems.
  4. On the basis of the study, pollution masks should be used by the traffic police men on duty at higher polluted junctions.
  5. He should be recommended better and special medical care for his protection. Various duty places for them need to be scientifically evaluated for their exposure risk.
  6. To protect ears from deafening noises, they can put small cotton balls in the ear and also wear ear masks for this purpose.
  7. Ecofriendly solar traffic booths at traffic intersections will be provided for the traffic police men. These booths contains ionisers which supress the suspended particulate matter and provide a healthy environment for a police man seated inside.

Question 6.
Write a short note on the effects of water pollution in your village. Suggest precautions.
Answer:

  1. Patancheru is a suburban mandal headquarters in Medak district, located about 25 km from Hyderabad. It is a major industrial hub of the state.
  2. It is one of the most polluted areas in India. Nearly 14 villages were badly affected by pollution related diseases like cancer, respiratory diseases and heart diseases caused by number of poisons in air, water and on land.
  3. The presence of pharmaceutical and chemical industries, pesticide units, steel rolling industries, distilleries releasing the pollutants like chlorine, Hydrogen sulphide, which are entering into the atmosphere.
  4. Most of the agricultural lands became barren. The lives of people there depend on agriculture and animal husbandry. They became helpless. Most of the people converted themselves as workers in the factories.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 7.
Visit a pollution check centre. Observe the process of conducting a pollution check and record your findings. You may consider the following areas your record.
Average number of vehicles checked in a certain time period, Time taken to check each vehicle, Pollutants checked for, The process of testing, Permissible limits of emission of various pollutants, Measures taken if the emitted gases are above the permissible limits.

  • % Average number of vehicles checked in a certain time period.
    Answer:
    10-15
  • Time taken to check each vehicle.
    Answer:
    5-7 minutes
  • Pollutants checked for
    Answer:
    Carbon dioxide, carbon monoxide, nitrogen, nitrogen oxide, methane, hydro carbons, sulphur dioxide, particular matter, trace elements, water vapour etc.
  • The process of testing:
    1. The test shall be carried out with the engine mounted on a test bench and connected to a dynamometer.
    2. The gases emission from the exhaust of the engine include hydrocarbons, carbondioxide, carbon monoxide and oxides of nitrogen.
    3. During prescribed sequence of warmed up engine operating conditions, the amount of the above gases in the exhaust, shall be examined continuously.
    4. The prescribed sequence of operations consists of a number of speed and power modes which span the typical operating range of engine.
    5. During each mode, the concentration of each pollutant, exhaust flow and power out put shall be determined and the measured values, weighed and used to calculate the grams of each pollutant emitted for kilowatt hour.
  • Permissible limits of emission of various pollutants.
    Pefrol engine
    Carbondioxide                                  14%
    Carbon monoxide                            1 to 2%
    Nitrogen oxide less than                  0.5%
    Hydrocarbons                                   0.5%
    Sulphur dioxide                                possible traces
    Particular matter less than                0.5%
    Trace elements less than                   0.5%
    Nitrogen                                            71%
    Water vapour                                    12%
    Diesel Engine
    Nitrogen                                                   67%
    Carbondioxide                                          13%
    Water vapour                                            11%
    Carbon monoxide                             less than 0.045%
    Nitrogen oxide                                 less than 0 to 1.5%
    Sulphur dioxide                                less than 0.03%
    Hydro carbons                                  less than 0.43%
    Particulate Pollutants                       less than 0.045%
    Trace elements                                           0.3%
  • Measures taken if the emitted gases are above the permissible limits.
    Answer:
    The carbon particles (soot) deposited in the engine head will be checked and cleaned or the vehicle will be ceased by .R.T.O.

Question 8.
Organize a field visit to a pond/lake/river present in or near to your village with the help of your teachers.
Observations followed by discussion could focus on…. The history of the pond or lake or river, Water resources available other than that river/pond/or lake, Cultural traditions, Pollution concerns, Source of pollution, Effects of pollution on the people living by the river side as well as those living far away.
Answer:
Kolleru lake: It is a 2nd largest fresh water lake located in Andhra Pradesh located between Krishna and Godavari delta.
History: Two copper plates of the early Pallava dynasty have been found in the lake, tracing it’s history to Langula Narasimha Deva an Ganga Vanshi Odisha King. According to legend, the Gajapathi fort was located at Kolleti kota on one of the eastern islands of the lake. The enemy general “Muhammadan” general escamped at “Chiguru kota” located on the shores. In some ways the lake protected the Oriya forces. The enemy finally tried to excavate a channel, the modern day Upputeru. So that the water of the lake would empty into the sea and the level would fall so that they could attack the Gajapathi fort. The royal oriya army general sacrificed his own daughter to propitiate Gods and ensure his success against Muhammadan and her name was “Perantala Kanama”. Therefore the channel Was called perantala Kanama. “Sri Peddinti Ammavari Temple” is one of the oldest and famous temples found in Kolleru.
Water sources available other than Kolleru: Wells, taps. The lake is fed directly water from seasonal Budameru and Tammileru streams and connected to the Krishna and Godavari systems by channels.
Cultural traditions: The vast majority of the district is rural in nature. Thus the culture of the kolleru lake people is mostly conservative and traditional. The joint family system, the arranged marriages are the norms. Telugu language is spoken in this place. Pollution concerns: Kolleru lake is suffering from the unsatisfied greed of people and selfish interests of mankind who exploit the lake’s integrity. Thousands of fish tanks were dug up effectively converting the lake into a mere drain. This had great impact in terms of pollution, leading to difficulty in getting drinking water for the local people.
Source of pollution: Satellite images taken on February 9, 2001 by the Indian remote sensing satellite found that approximately 42% of the 245 km2 lake was occupied by aquaculture. While agriculture had encroached another 8.5% they were mostly rice paddies. Surprisingly no clear water could be found in the satellite image. The rest of the lake is being diminished by weeds like elephant grass and water hyacinth.
Effects of pollution on people: Thousands of fish tanks were dug up effectly leading to difficulty in getting drinking water for the local people. An adverse effect on the thousands of acres of crop in the upper reaches of water flow into the sea because of obstruction by bunds of the fish tanks that appeared illegally.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 9.
What is air pollution? Make a flow-chart to describe its causes and effects.
Answer:
If solid, liquid and gaseous substances are present in higher volumes than required in the air, it is harmful to air. It is called air pollution.
Table shows causes and effects of air pollution:
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 1
COMMON POLLUTANTS AND THEIR SOURCES
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 2

Question 10.
Clear and transparent water is always suitable for drinking. Comment.
Answer:

  1. No. Clear and transparent water is not always suitable for drinking.
  2. Water might appears clean but it may contain some disease causing micro-organisms and other dissolved impurities.
  3. Hence it is advised to purify water before drinking.
  4. Purifying can be done by water purifying systems or by boiling the water.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 11.
If our monument like Tajmahal is effected by air pollution, what is your advise to protect it?
Answer:
The Taj Mahal one of the seven wonders of the world is located in Agra. It is made of white marbles. The effect of pollutants like SO2, NO2, smoke, dust, soot, etc. on it has turned the marble from white to yellow.
Precautions to protect Taj Mahal:

  1. Switch over to cleaner fuels like CNG (Compressed Natural Gas) and LPG (Liquify Petroleum Gas)
  2. Use unleaded petrol in vicinity of Taj Mahal.
  3. Shift polluting industries to the outside of Agra city.
  4. Industrial pollution should be banned around the Taj Mahal.
  5. Limited vehicular use in a specific radius.
  6. An electronic board should be installed in the Taj Mahal premises, that compares current pollution levels, to the levels that deemed safe.

Question 12.
Reshma going to talk about controlling measures of soil pollution. Prepare write up for her.
Answer:
Controlling measures of soil pollution:

  1. Limit the use of fertilizers and pesticides.
  2. Awareness about biological control methods and their implementation.
  3. The grazing of cattle must be controlled and forest management should be done properly.
  4. The afforestation and reforestation must take place.
  5. Proper preventing methods like shields should be used in areas of wind erosion .and wind breakes.
  6. Remember to carry paper bags and minimising plastic bags.
  7. The soil binding grass must be planted and the large trees must be plant along the banks.
  8. Industrial waste must be dumped in the low lying areas.
  9. There should be a definite technic of cropping which does not allow weeds to settle on the fields.
  10. The mining waste must be improved along with the transportation.
  11. The are must not be left barren and dry.

Question 13.
To conduct a quiz program on air and water pollution, prepare five thought provoking questions.
Answer:

  1. What is environmental pollution?
  2. What are the natural disasters of pollution?
  3. What are the human activities that leads to pollution?
  4. What is the unforgettable industrial tragedy took place in Bhopal on second December
  5. Who is responsible for thepollution in Patancheru ? What are the interim orders, released by supreme court of India for the sake of people and environment?
  6. What is the sad story of River Musi?
  7. Can we save Taj Mahal from pollution?
    What is the present situation of Taj Mahal?
  8. Natural resources are the devine gift for us by nature. Can we keep these resources clean and healthy for the future generations? What is your role and response towards this?

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 14.
“Use Bicycle – Avoid motor bikes and cars”. This slogan is prepared by Sravani. Prepare some more slogans on pollution.
Answer:

  1. Slogans towards biodiversity:
  2. Segregate the reusable waste.
  3. Don’t produce lot of waste.
  4. Don’t use plastic covers.
  5. Compost the wet waste.
  6. Turn the waste into compost.
  7. Make Compost out of fallen leaves.
  8. Never burn the fallen leaves of trees.
  9. Use bicycles if the destination is manageable.
  10. Plant trees in vacant places and take care to ensure their growth.
  11. Grow plants and protect them for fresh air and ventilation.
  12. Use cattle dung, organic fertilizers.
  13. Practice eco-friendly methods.
  14. Strictly follow environmental policies and laws.
  15. Limit the use of fire wood and use bio fuels for cooking.
  16. Plant a sapling on your birthday water it every day.
  17. Wash your hands, feet close to the trees and plants.

Question 15.
If you are a general manager of a chemical industry, what precautions would you take to control air and water pollution?
Answer:
Precautions to control air and water pollution :

  1. Stoppage of effluent flowing into air and water bodies immediately.
  2. Installing tall chimneys in factory to reduce air pollution at the ground level.
  3. Installing electrostatic precipitators in the chimney of industry to reduce.
  4. Plant and grow, more and more. Trees in the industry surroundings.
  5. Protecting plants and trees.
  6. Toxic industrial wastes should be treated chemically to neutralize the harmful substances present in it before discharging into water bodies.
  7. Waste water from the industry is to be filtered first to remove particulate material and stored in shallow tanks where bacterial degradation of organic compounds takes place.

8th Class Biology 10th Lesson Not For Drinking-Not For Breathing InText Questions and Answers

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 1.
Observe this certificate try to find out answers for the following questions.
Answer:
The pollution under control certificate
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 3
a) Which department issues the pollution under control certificate?
Answer:
The pollution check up centre issues the certificate.

b) For how much time is it valid?
Answer:
It is valid for six months.

c) For which type of vehicle has it been issued?
Answer:
Motor bike, scooters, cars, bus, lorry all type of vehicles.

d) What is emission test ? What components are tested ini the pollution check up centre?
Answer:
The test conducting of the gases releasing from the vehicle is called emission test. Components tested in the pollution check up center are carbondioxide, carbon monoxide, nitrogen, hydrocarbons, sulphur dioxide etc.

e) What will happen if carbon monoxide (CO) and hydro carbons (HQ readings are higher that the permissible limits reading ?
Answer:
If the above said gases are higher than the permissible limits reading it leads to pollution is harmful to the living organisms and hurt the health and well of living organisms.

f) Think of why there is a peed of “pollution under control certificate”?
Answer:
With a rapid increase in Ifie number of vehicles the problem of automobile pollution has assumed greater significance. The emission of smoke from motor vehicles is a major source of air pollution.

Question 2.
What will happen if harmful organisms or substances enter your body? How do you feel?
Answer:
If harmful organisms enter the body, the normal functioning of the body will be disrupted or disturbed. We feel sick.

What is air pollution ?
Question 3.
List out the gases that you know present in the air.
Answer:
The gases present in the air are nitrogen, oxygen, carbondioxide, inert gases mainly argon, and water vapour.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 4.
What are the four major gases in the air?
Answer:
The four major gases are nitrogen oxygen, argon and carbondioxide.

Question 5.
Draw a neat diagram showing the composition of air in the atmosphere arid write the percentage of gases.
Answer:
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 4

Question 6.
Collect the pictures of natural activities and human activities which leads to pollution and paste them in your record book.
Answer:
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 5AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 6

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 7AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 8

Question 7.
If a person burnt out types or dried leaves at a particular place. Where shall go the smoke and ash goes?
Answer:
The smoke and ash raise up, mixes with gases in the atmosphere which leads to pollution.

Water Pollution
Question 8.
Let us read the following news paper clipping.
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 9
Answer the following questions based on your understanding of the paper clipping.
a. What do you understand after reading the news paper clippings?
Answer:
Some areas in Nalgonda district, the water got polluted because of the chemicals dissolved in it. These chemicals are released Into the water by the oil factory, textiles, pesticide factory and steel factory.

b. What are the issues discussed in this news paper clipping?
Answer:
Total dissolved solids (T.D.S.) in water should be maximum 500. Due to pollution it is estimated that T.D.S. has reached to 10,000. The pollution in water is higher which is dangerous for use.

c. What are its causes and effects?
Answer:
The causes for the TDS: The chemical pollutants released by the factories into the water. Effects: The water becomes unfit for drinking. By using the water the crops are beeing dead.

d. How does the problem arise?
Answer:
Large amounts of chemicals discharged into the ground water leads to water pollution. Due to water pollution the problem arise in those areas.

e. Are you also facing this type of problems in your area? Can you explain reasons behind?
Answer:
Yes, I am facing this type problems in our area. Because industries are developed in our area. So population is increasing. That’s why water is polluted in our area.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

8th Class Biology 10th Lesson Not For Drinking-Not For Breathing Activities

Activity – 1

Question 1.
NATURAL DISASTERS POLLUTION
Collect information from your school library for the following natural disasters in the world.
Answer:
Volcanic eruptions:
Deep under the earth’s surface, it’s so hot that even rock melts. Sometimes this molten rock, called ‘magma’, is pushed up to the surface. At this point it is referred to as lava. And the opening or vent that lets the lava out is a volcano. ‘
A volcano may explode violently throwing out rocks for miles around. It releases various gases and ash into the atmosphere.

Forest fires:
Forest fire is a moving combustion reaction spreading outwards in a band from it’s ignition point leaving burnt forest behind it and also called as wild fire.
It can be large uncontrolled disasters that burnt through hundred to hundred thousand acres.
Causes : Natural cause, lightening, volcanic eruptions, sparks from rocks falls and spontaneous combustion.
Pollution :
Forest fires release carbon particles (ash) into the air and pollute the air.

Sand storms and Tsunamis:
A sand storm or a dust storm is a meteorological phenomenon common in arid and semi arid regions. They arise when a gust front or other strong wind blows, loose sand and dirt from a dry surface. Particles are transported by saltation and suspension. The Sahara where sand is more prevalent, soil type than dirt or rock.
Causes:
As a force of wind passing over loosly held particles increased, the particles of sand first start to vibrate, then to saltate. As they repeatedly strike the ground, they loosen and break off smaller particles of dust, which then begin to travel in suspension.

Tsunamis:

  1. Tsunami is also called as a ‘seismic sea wave’ is a series of waves in a water body caused by the displacement of a large volume of water.
  2. Tsunami impact is limited to coastal areas. But their destructive power can be enormous, and they can effect entire ocean basin.
  3. The 2004 Indian Ocean tsunami was among the deadliest natural disasters in human history with, atleast 2,30,000 people killed or missing in 14 countries bordering the Indian Ocean.
  4. Tsunami causes much damage by two mechanisms. The smashing force of water travelling at a high speed and destructive power of a large volume of water draining of the land and carrying away large amount of debris with it.
  5. About 80% of tsunamis occur in the Pacific ocean, and there is possibility for tsunamis wherever there are large bodies of water including lakes.

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Activity – 2

Question 2.
A) OIL PAPER EXPERIMENT
Answer:
Take three square pieces of white paper of 5 x 5 cm size dipped in oil. Hang these oil dipped paper at three different locations, say your backyard, your school, near a park or parking lot etc. Let it be there for 30 minutes. Observe and compare all three papers.
a) What did you found on those papers dipped in oil ?
Answer:
Some dust particles sticking to the oil paper.

b) Is there any difference observed for all the three locations ?
Answer:
Some differences are observed. The soot and ash dust particles are seen on the oil paper hung at back yard which comes from burning of wood.
More dust particles seen on the oil paper at the school and less dust particles on the oil paper which was hung near a park.

c) Try to find out the answer why this difference occurred.
Answer:
When different types of fuels are burnt they release different types of particles into air. These particles are different according to different locations.

d) Do you know where the dust particles could have come from ?
Answer:
The dust particles could have come from the effect of air pollution caused by man made resources largely affects nature.

B) HUMAN ACTIVITIES:
Name of the fuels we burnt in our daily activities including rural and urban areas.
Answer:
Petrol, diesel, wood, charcoal, tyres etc.

Activity – 3

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 3.
POWER GENERATION PLANTS.
Go to your school library and collect information to make a list of these power generation plants and where they are located.
Answer:
THERMAL POWER PLANTS
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 10 AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 11 AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 12 AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 13AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 14

NUCLEAR POWER PLANTS
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 15

Discuss about the adverse effects of Global warming.
Answer:
Effects of Global Warming: Global warming is having measurable effects on the planet right now. They are

  1. Ice is melting in both polar ice caps and mountain glaciers.
  2. Lakes around the world, including lake superior are warming rapidly.
  3. Animals are changing migration patterns.
  4. Plants are changing the dates of activity. Ex : Leaf flash.
  5. Global warming increase in temperature around the world.
  6. The average global temperature increased 0.8°C over the post 100 years.
  7. Weather is changed by globed warming, some places become more hot and some more cool.
  8. Global warming increases the sea level. So the costal areas will sink into the sea.
  9. Ice is melted which increases the sea level that leeds to ocean acidification.

Ask your teacher about secondary pollutants why they are called so?
Answer:
Oxides of nitrogen NO, NO2 (NOx), peroxy acetyl nitrate, formaldehyde, ozone, etc., are the secondary pollutants.
Pollutants are defined as primary pollutants resulting from combustion of fuels and industrial operations and secondary pollutants, those which are produced due to reaction of primary pollutants in the atmosphere.

Activity – 4

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 4.
FIELD VISIT:
Visit nearby factory, industry (boiled rice mill, brick making kiln, oil mill food processing mill etc.,) present in your area and observe.

  • How are they polluting air and water ?
    Answer:
    They are releasing smoke, smooth dust particles into air and releasing waste material into water sources and polluting air and water.
  • Is there any green belt around the factory ? Name the trees they are growing.
    Answer:
    Ashoka, Gulmohar, Neem, Eucalyptus are growing around the factory.
  •  What precautions are they taking to prevent pollution ?
    Answer:
    Suction devices known as vacuum pans are used to collect the pollutants from the water.
    To control air pollution, ventury type wet scrubber and meeting the norms described by A.P. Pollution board -have arranged.

Lab Activity 

Question 5.
POLLUTANTS:
Aim: Observation of pollutants in local available water samples.
Material: Glass tumblers, water samples from tap, pond, river, well, lake, red, blue litmus papers, soap.
Procedure: Collect water samples from a tap, pond, river, well and lake. Pour each into separate glass containers. Compare these for smell, colour pH and hardness.

  • pH of water samples can be determined by using litmus paper. If blue litmus paper turns to the red colour, that water sample is acidic in nature and if red litmus turns to blue, water sample is basic in nature.
  • Hardness of water can be determined using soap. If water produces lesser foam it is referred as hard water.

Observations and findings:
Record your observation in the following table.
AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing 16

Activity – 5

Question 6.
Visit your nearby pond/lake or river and find out the material being discharged in it. Prepare a biography on it.
Answer:
As Hyderabad has grown in size and is emerging as a global mega city, its growing water requirements have been met by under taking long distance water projects over the years. These projects are dependent on Musi River. Thousands of people depend on it for their daily needs and livelihood. The Moosi has been polluted for many years.
The people living near the Musi river, throw large quantities of garbage, untreated sewage, industrial waste, dead bodies, polythene bags, hot water and statues of deities and many other materials directly in to the river. –
The ‘Musi reservoir action plan project’ was undertaken to reduce the pollution level in the river. Pollution control activities include under the project are:

  • Solid waste management.
  • Installation of sewage treatment plant.
  • Provision of low cost sanitary facilities.
  • Development of river front.
  • Efforts to develop public awareness.

Ask your teacher about aerobic bacteria and write a note on it with some examples.
Answer:
An aerobic bacteria is the one that can survive and grow in an oxygenated environment. Bacillus, nocardic and pseudomonas space aeruginosa, myobacterium tuberculosis are some of the examples of aerobic bacteria.

Do you know oil slick on sea water ?In what way it is dangerous to aquatic life?
Answer:
Several kinds of plants and animals live in oceans. Oceans maintain equilibrium in nature. We transport several kinds of oils and fuels over seas. The spillage of these oils and fuels by accident creates a layer of oil over the surface of the sea water for hundreds of kilometers. This is called oil spill or oil slog. When it happens, air and light cannot enter the sea water and several marine creatures like fish, tortoise and other marine life forms die of asphyxiation.

Think and Discuss

AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing

Question 1.
When we go on a busy road in the evening a lot of smoke is spread in the surroundings. We get cough and feel uneasy even when we close the nose with napkins.
a) Why this type of symptoms we observe? Think about it.
Answer:
Carbon monoxide is poisonous gas combines with haemoglobin of our blood and forms carboxy haemoglobin. Due to this haemoglobin is unable to carry oxygen to various parts. This leads to respiratory problems. It causes suffd&Stion and may cause even death.

b) If these symptoms will continue, what happens?
Answer:
Air pollution is like a slow poison. The effect of air pollution are not seen immediately. But over a long period of time the pollutants present in air damage our health and property.

Question 2.
Do you find any relation between pH and hardness of water?
Answer:
No relation between pH and hardness of water. pH is scale to measure acid, base or neutral hardness is the percentage of salts dissolved in the water.

Question 3.
Which water sample is colourless?
Answer:
Tap water.

Question 4.
Which water sample is suitable for drinking and why?
Answer:
Tap water is suitable for drinking because the tap water is cleaned and chlorinated and safe for drinking.

Question 5.
Do you find any change in colour and smell of water in some water samples ? What are your reasons ?
Answer:
The colour and smell of water is due to the nature of the soil and the plants grown in the water.

AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions and Answers.

8th Class Maths 13th Lesson Visualizing 3-D in 2-D InText Questions and Answers

Do This

Question 1.
Name some 3 – Dimensional objects.   [Page No. 282]
Answer:

  1. Cube
  2. Cylinder
  3. Sphere
  4. Cuboid
  5. Cone

AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions

Question 2.
Give some examples of 2 – Dimensional objects.     [Page No. 282]
Answer:

  1. Square
  2. Rectangle
  3. Line segment
  4. Circle
  5. Triangle

Question 3.
Draw a kite in your notebook. Is it 2 – D or 3 – D object?      [Page No. 282]
Answer:
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 1
Kite is a 2 – D object.

Question 4.
Identify some objects which are in cube or cuboid shape.      [Page No. 282]
Answer:
Shapes of Cube                           Shapes of Cuboid
a) Chalk piece box                       a) Duster
b) Dice                                         b) Cell phone (Cuboidal in shape)
c) Cube shaped cake                   c) Plasma T.V.

Question 5.
How many dimensions that a circle and sphere have?      [Page No. 282]
Answer:
Circle has 2 dimensions.
Sphere has 3 dimensions.

AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions

Question 6.
Identify the faces, edges and vertices of given figures.      [Page No. 288]
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 2
Answer:
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 3

Question 7.
Write the names of the prisms given below:     [Page No. 290]
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 4
Answer:
(i) Cube
(ii) Triangular prism
(iii) Pentagonal prism
(iv) Hexagonal prism
(v) Rectangular prism

Question 8.
Write the names of the pyramids given below:       [Page No. 290]
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 5
Answer:
(i) Square pyramid
(ii) Pentagonal pyramid
(iii) Hexagonal pyramid

AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions

Question 9.
Fill the table:      [Page No. 290]
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 6
Answer:
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 7

Question 10.
Explain the difference between prism and pyramid.      [Page No. 290]
Answer:
Upper and lower sides of a prism are equal in number. But, in a pyramid the base is a plane and all the edges are coincide in a single point on the top.

Try These

Question 1.
Name three things which are the examples of polyhedron. [Page No. 287]
Answer:
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 8

AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions

Question 2.
Name three things which are the examples of non-polyhedron. [Page No. 287]
Answer:
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 9

Think, Discuss and Write

Question 1.
How to find area and perimeter of top view and bottom view of the given figure.       [Page No. 283]
AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions 10
Answer:
Let the side of each face be ‘1’ unit say.
Shapes of different positions (I)                      Their areas (II)
1. Front view                                                    A = (1 × 1) + (1 × 1) + (1 × 1) = 3 Sq. Units
2. Top view                                                      A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
3. Bottom view                                                A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
Perimeters (III)
1. —————>                                             1 + 1 + 1 = 3 Units
2. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units
3. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions and Answers.

8th Class Maths 7th Lesson Frequency Distribution Tables and Graphs InText Questions and Answers

Do this

Question 1.
Here are the heights of some of Indian cricketers. Find the median height of the team.   [Page No. 154]
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 1
Answer:
The ascending order of heights is 5’3″, 5’5″, 57″, 5’8″, 5’9″, 571″, 571″, 6’0″, 6’0″, 6’7″
Number of players = 10 (is an even)
Median = Mean of \(\left(\frac{\mathrm{n}}{2}\right)\) and \(\left(\frac{n}{2}+1\right)\) terms = Mean of \(\left(\frac{10}{2}\right)\) and \(\left(\frac{10}{2}+1\right)\) terms = Mean of 5, 6 terms = AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 23

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 2.
Ages of 90 people in an apartment are given in the adjacent grouped frequency distribution.
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 2
i) How many Class Intervals are there in the table?
ii) How many people are there in the Class Interval 21 – 30?
iii) Which age group people are more in that apartment?
iv) Can we say that both people the last age group (61-70) are of 61, 70 or any other age?    [Page No. 158]
Answer:
i) 7    ii) 17     iii) 31 – 40     iv) Yes, they are 62, 63, ……, 69

Question 3.
Long jump made by 30 students of a class are tabulated as
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 3
I. Are the given class intervals inclusive or exclusive?
II. How many students are in second class interval?
III. How many students jumped a distance of 3.01 m or more?
IV. To which class interval does the student who jumped a distance of 4.005 m belongs?    [Page No. 160]
Answer:
I. Inclusive
II. 7
III. 15 + 3 + 1 = 19
IV. 401 – 500

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 4.
Calculate the boundaries of the class intervals in the above table.     [Page No. 160]
Answer:
Boundaries:
100.5 – 200.5
2005 – 300.5
300.5 – 400.5
400.5 – 500.5
500.5 – 600.5

Question 5.
What is the length of each class interval in the above table?     [Page No. 160]
Answer:
100

Question 6.
Construct the frequency polygons of the following frequency distributions.       [Page No. 174]
i) Runs scored by students of a class in a cricket friendly match.
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 4
ii) Sale of tickets for drama in an auditorium.
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 5
Answer:
i)
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 6AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 7
Steps of construction: Runs scored (Mid values of C.I.)
Step – 1: Calculate the mid points of every class interval given in the data.
Step – 2: Draw a histogram for this data and mark the mid points of the tops of the rectangles, (like B, C, D, E, F respectively).
Step – 3: Join the mid points successively.
Step – 4: Assume a class interval before the first class and another after the last class. Also calculate their mid values (A and G) and mark on the axis. (Here, the first class is 10 – 20. So, to find the class preceding 10 – 20, we extend the horizontal axis in the negative direction and find the mid point of the imaginary class interval 60 – 70.
Step – 5: Join the first end point B to A and last end point F to G which completes the frequency polygon.
Frequency polygon can also be drawn independently without drawing histogram. For this, we require the midpoints of the class interval of the data.

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

ii)
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 8AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 9
Steps of construction:
Step -1: Calculate the mid points of every class interval given in the data.
Step – 2: Draw a histogram for this data and mark the mid points of the tops of the rectangles (like B, C, D, E, F respectively).
Step – 3: Join the mid points successively.
Step – 4: Assume a class interval before the first class and another after the last class.
Step – 5: To find the class preceding 2.5 – 7.5, we extend the horizontal axis in the negative direction and find the mid point of the imaginary class interval 32.5 – 37.5 like A, G.
Step – 6: Join A to B and G to F.
∴ The required ABCDEFG polygon is formed.

Try these

Question 1.
Give any three examples of data which are in situations or in numbers.      [Page No. 148]
Answer:
1) The data of 35 students who like different games:
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 10
2) The data of 35 students who like different colours:
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 11
3) The data of 35 students who like different fruits:
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 12

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 2.
Prepare a table of estimated mean, deviations of the above cases. Observe the average of deviations with the difference of estimated mean and actual mean. What do you infer?
[Hint: Compare with average deviations]      [Page No. 151]
Answer:
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 13
Mean = \(\frac{\Sigma x_{i}}{N}\) = \(\frac{80}{5}\) = 16
Mean of the deviations = \(\frac{-5}{5}\) = -1
Mean = Assumed mean + Mean of deviations = 17 + (-1) = 16
∴ Assumed mean, original arithmetic mean are equal.

Question 3.
Estimate the arithmetic mean of the following data.      [Page No. 153]
i) 17, 25, 28, 35, 40
ii) 5, 6, 7, 8, 8, 10 10, 10, 12, 12, 13, 19, 19, 19, 20
Verify your answers by actual calculations.
Answer:
i) 17, 25, 28, 35, 40
Assumed Mean = 35
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 14
A.M in general method = \(\frac{\text { Sum of the observations }}{\text { No. of the observations }}\)

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 15

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

ii) 5, 6, 7, 8, 8, 10, 10, 10, 12, 12, 13, 19, 19, 19, 20
Assumed Mean = 10
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 16

Question 4.
Find the median of the data 24, 65, 85, 12, 45, 35, 15.       [Page No. 155]
Answer:
The ascending order of the data is 12, 15, 24, 35, 45, 65, 85
Number of observations (n) = 7 (odd)
∴ Median = \(\frac{n+1}{2}\) = \(\frac{7+1}{2}\) = 4th term
∴ Median = 35

Question 5.
If flie median of x, 2x, 4x is 12, then find mean of the data.       [Page No. 155]
Answer:
Given observations are x, 2x, 4x
∴ Median = 2x
According to the sum
2x = 12 ⇒ x = 6
2x = 2 × 6 = 12
4x = 4 × 6 = 24
∴ The mean of 6, 12, 24 = \(\frac{6+12+24}{3}\) = \(\frac{42}{3}\) = 14

Question 6.
If the median of the data 24, 29, 34, 38, x is 29 then the value of ‘x’ is
i) x > 38   ii) x < 29   iii) x lies in between 29 and 34   iv) none       [Page No. 155]
Answer:
Median of 24, 29, 34, 38, x is 29.
n = 5 is an odd.,
∴ Median = \(\frac{n+1}{2}\) = \(\frac{5+1}{2}\) = 3rd term
If x is less than 29, then only 29 should be a 3rd term.
∴ x < 29

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 7.
Less than cumulative frequency is related to …….     [Page No. 165]
Answer:
Upper boundaries

Question 8.
Greater than cumulative frequency is related to ……..      [Page No. 165]
Answer:
Lower boundaries

Question 9.
Write the Less than and Greater than cumulative frequencies for the following data.       [Page No. 165]
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 17
Answer:
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 18
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 19

Question 10.
What is total frequency and less than cumulative frequency of the last class above problem? What do you infer?    [Page No. 165]
Answer:
The sum of the frequencies in the above distribution table = 30
Less than C.F of the last C.I = 30
∴ Sum of the observations = Less than C.F of last C.I.

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 11.
Observe the adjacent histogram and answer the following questions.     [Page No. 169]
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 20
i) What information is being represented in the histogram?
ii) Which group contains maximum number of students?
iii) How many students watch TV for 5 hours or more?
iv) How many students are surveyed in total?
Answer:
i) The histogram represents students who watch the T.V.’s .(Duration of watching T.V).
ii) 4th class interval contains maximum number of students.
iii) 35 + 15 + 5 = 55
iv) Number of students are surveyed = 10 + 15 + 20 + 35 + 15 + 5 = 100

Think, discuss and write

Question 1.
Is there any change in mode, if one or two or more observations, equal to mode are included in the data?    [Page No. 155]
Answer:
If one or two or more observations equal to mode are included there will be no change in the mode.
Ex: The mode of 5, 6, 7, 8, 7, 9 is 7.
If 3, 7’s are added to above observations there will be no change in the mode.

Question 2.
Make a frequency distribution of the following series.
1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7.    [Page No. 161]
Answer:
The range of the observations = Highest value – Least value
∴ Range = 7 – 1 = 6
If number of classes = 7 then
Class Interval = \(\frac{\text { Range }}{\text { No. of classes }}\) = \(\frac{6}{7}\) = 0.8 (approx.)
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 21

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 3.
Construct a frequency distribution for the following series of numbers.
2, 3, 4, 6, 7, 8, 9, 9, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 16, 17, 18, 18, 19, 20, 20, 21, 22, 24, 24, 25. (Hint: Use inclusive classes)      [Page No. 161]
Answer:
Range = Maximum value – Minimum value = 25 – 2 = 23
Class Interval = \(\frac{\text { Range }}{\text { No. of classes }}\) = \(\frac{23}{5}\) = 4.6 = 5 (approx.) [∵ No. of classes = 5]
AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions 22

Question 4.
What are the differences between the above two frequency distribution tables?      [Page No. 161]
Answer:
The class intervals of first frequency distribution table are exclusive class intervals. The C.I’s of 2nd frequency distribution table are inclusive class intervals.

Question 5.
From which of the frequency distributions we can write the raw data again?      [Page No. 161]
Answer:
Classes

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 6.
All the bars (or rectangles) in a bar graph have     [Page No. 168]
a) same length b) same width c) same area d) equal value
Answer:
b) same width

Question 7.
Does the length of each bar depend on the lengths of other bars in the graphs?     [Page No. 168]
Answer:
No

Question 8.
Does the variation in the value of a bar affect the values of other bars in the same graph?      [Page No. 168]
Answer:
No

Question 9.
Where do we use vertical bar graphs and horizontal bar graphs?     [Page No. 168]
Answer:
Vertical and horizontal bar graphs are used to present the equal widths corresponding to the given frequencies.

Question 10.
Class boundaries are taken on the X-axis. Why not class limits?      [Page No. 172]
Answer:
The difference between upper and lower boundaries gives the class interval i.e., we take class boundaries on X-axis.

Question 11.
Which value decides the width of each rectangle in the histogram?      [Page No. 172]
Answer:
Class Interval

Question 12.
What does the sum of heights of all rectangles represent?     [Page No. 172]
Answer:
Sum of the frequencies.

Question 13.
How do we complete the polygon when there is no class preceding the first class?       [Page No. 173]
Answer:
The frequency of preceding class should be taken as ‘0’ (zero) then it should be joined.

Question 14.
The area of histogram of a data and its frequency polygon are same. Reason how?       [Page No. 173]
Answer:
Because both the figures are constructed on the basis of mid values of class intervals.

AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs InText Questions

Question 15.
Is it necessary to draw histogram for drawing a frequency polygon?       [Page No. 173]
Answer:
No need.

Question 16.
Shall we draw a frequency polygon for frequency distribution of discrete series?       [Page No. 173]
Answer:
No, we can’t.

Question 17.
Histogram represents frequency over a class interval. Can it represent the frequency at a particular point value?            [Page No. 175]
Answer:
Yes, the histogram represents the frequency at a particular point value. Since the length of a histogram represents the value of its corresponding frequency (length of the frequency).

Question 18.
Can a frequency polygon give an idea of frequency of observations at a particular point?       [Page No. 175]
Answer:
Yes, we can identify the frequency of observation with a frequency polygon at a particular point. Since the height of the polygon is equal to frequency of polygon.

AP Board 8th Class Biology Solutions Chapter 1 What is Science?

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 1 What is Science Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 1st Lesson What is Science

8th Class Biology 1st Lesson What is Science Textbook Questions and Answers

Improve Your Learning

Question 1.
What is Science?
Answer:
Science:

  1. Science is the concerted human effort to understand the history of the natural world how the natural world works, with observable physical evidence.
  2. Science is an organized study of knowledge which is based on experimentation.
  3. Science is a tool for searching truths of nature.
  4. Science is the way of exploring the world.

AP Board 8th Class Biology Solutions Chapter 1 What is Science

Question 2.
What are scientific methods ? Write the steps involved in a scientist’s work.
Answer:
Scientific Methods:

  1. Scientists solve a problem or answer a question by using organized ways. They are called “Scientific Methods”.
  2. The following are the steps involved in a scientific work.
    a) Step -1 : Ask questions.
    b) Step – 2 : Form hypothesis.
    c) Step – 3 : Plan experiments.
    d) Step – 4 : Conduct experiments.
    e) Step – 5 : Draw conclusions.

Question 3.
What is the way to find out solutions for the problems in scientific way?
Answer:
To find out solutions for the problems in scientific way we need to follow a sequential order. So we go through the following.

  1. Identifying problem : Let us identify any problems from your surroundings.
    Ex: The bulb did not lit in the room.
  2. Making hypothesis: List out different solutions which your think for the identifying problem.
    Ex: De filament, fuse failure, switch problem, wire problem.
  3. Collecting information: To solve the identifying problem collect material, apparatus, Information, persons.
    Ex: Collect material like tester, screwdriver, wooden scale, wires, insulation tape, table and blade.
  4. Data analysis: Arrange the collected data or information to conduct experiment.
  5. Experimentation: To prove selecting hypothesis conduct experiment.
    Ex: Observe filament of the bulb.
  6. Result analysis: Analyzing the results to find out the solution for the problem based on the results you need to select another hypothesis to prove.
    Ex: Filament of the bulb is good in condition so we need to observe fuse.
  7. Generalisation: Based on the experiment and its results explain the solution for the problem.
    Ex: Fuse is damaged so the bulb not glow, so we need to replace the fuse.

AP Board 8th Class Biology Solutions Chapter 1 What is Science

Question 4.
During investigation in science, name some rules to be followed for the safety.
Answer:

  1. Think ahead: Study the steps of the investigation, so you know what to expect.
  2. Be neat: Keep your work area clean.
  3. Oops!: If you should spill or break something or get cut, tell your teacher right away.
  4. Watch your eyes: Wear safety goggles anytime you are directed to do so.
  5. Yuck!: Never eat or drink anything during a scientific activity.
  6. Protect yourself from shocks: Be especially careful if an electric appliance is used. Be sure that electric cords are in a safe place where you can’t trip over them. Don’t ever pull a plug out an outlet by pulling on the cord.
  7. Keep it clean: Always clean up when you have finished.

Question 5.
If you plan a test to find out how much water different brands of paper towels absorb, write down the steps of the experiment.
Answer:

  1. Pour 1 liter of water into each of three beakers.
  2. Put a towel from each of the three brands into a different beaker for 10 seconds.
  3. Pull the towel out of the water, and let it drain back into the beaker for 5 seconds.
  4. Measure the amount of water left in each beaker.
  5. The towel put in the beaker in which less amount of water is left absorbs more water.

Question 6.
Collect information about scientists and their works and prepare a chart and paste it in your classroom.
Answer:
AP Board 8th Class Biology Solutions Chapter 1 What is Science 1

AP Board 8th Class Biology Solutions Chapter 1 What is Science

Question 7.
What are process skills ?
Answer:
Observe, compare, classify are the process skills.

  1. Observe – Use the senses to learn about objects and events.
  2. Compare – Identify characteristics of things or events to find out how they are alike and different.
  3. Classify – Group or organize objects or events in categories based on specific characteristics.

Question 8.
Read the following.
Answer:
Endemic Species:
AP Board 8th Class Biology Solutions Chapter 1 What is Science 2
You may find that these animals are specifically found in certain regions of the world.
You are also aware of the fact that many plants and animals are widely distributed throughout the world. But some species of plants and animals are found restricted to some areas only. Plants or animal species found restricted to a particular area of a country are called Endemic Species.
Now answer the following questions.
a) Name an Endemic Species of our State.
Answer:
Tiger, Peacock and Crane.

b) You may notice that Kangaroo is endemic to Australia and Kiwi to New Zealand. Can you tell which among the above picures represent an endemic species of India ?
Answer:
Peacock and Tiger.

c) Name some other endemic species of India.
Answer:
Indian Lion, Indian Wolf, Great Indian bustard bill.

AP Board 8th Class Biology Solutions Chapter 1 What is Science

Question 9.
Reading the following poster.
AP Board 8th Class Biology Solutions Chapter 1 What is Science 3
Answer:
Biotic Components:
Producers – Mangrove, spirogyra, euglena, oscillatoria, blue green algae, ulothrix, etc.
Consumers – Shrimp, crab, hydra, protozoans, mussel, snails, turtle, daphnia, brittle word, tube worm, etc.
Decomposers – Detritus feeding bacteria, etc.
Abiotic components – Salt and fresh water, air, sunlight, soil, etc.

Question 10.
What is Generalisation? Give an example.
Answer:
Based on the experiment and its results explaining the solution for the problem is called Generalisation.
Example:

a) The bulb did not light in the room.
b) Identify the problem that may be defilament, fuse failure, switch problem, wire problem.
c) Then take tester, screwdriver, wooden scale, wires, insulation tape, table and blade.
d) Observe the filament of the bulb.
e) If the filament of the bulb is good, then observe fuse.
f) As the fuse is damaged, we need to replace the fuse.

Based on this, the bulb did not lit in the room because the fuse is damaged. This is the generalisation in the above experiment.

AP Board 8th Class Biology Solutions Chapter 1 What is Science

Question 11.
What are the different types of writings used by scientists to describe what they are doing or learning?
Answer:

  1. Informative writing, Narrative writing, Expressive writing, Persuasive writing are used by the Scientists.
  2. In informative writing, scientists describe the observation, inferences and their conclusion.
  3. In narrative writing the scientists describe about something, give examples or tell a story.
  4. In expressive writing, they may write letters, poems, or songs.
  5. In persuasive writing they write letters about important issues in science and also write about what they have learned about science which helps others to understand about their thinking.

Question 12.
How do scientists use numbers in their investigations?
Answer:

  1. Measuring, interpreting data, using number sense are few methods used by scientists in their investigations.
  2. Scientists make accurate measurements by using different measuring instruments like thermometer clocks, timers, rules, a spring scale, beakers, balance and other containers to measure liquids.
  3. Scientists collect, organize, display and interpret data by using tables, charts and graphs.
  4. Scientists compare and order numbers, compute with numbers shown on graphs and read the scales on thermometers, measuring cups, beakers and other tools.
  5. Good scientists apply their maths skills to help them display and interpret the data they collect.

Question 13.
What is Hypothesis? What are variables?
Answer:

  1. Making a statement about an expected outcome is called Hypothesis.
  2. Variables are factors that can affect the outcome of the investigation.

AP Board 8th Class Biology Solutions Chapter 1 What is Science

Question 14.
What do the following persons do?
a) A geologist
b) A chemist
c) A biologist
d) An ecologist
e) A climatologist
Answer:
a) A geologist examines the distribution of fossils and makes observations to find patterns in natural phenomena.
b) A chemist observes the rate of a chemical reaction at a variety of temperatures.
c) A biologist observes the reaction of a particular tissue to various stimulants.
d) An ecologist observes the territorial behaviours of different animals and birds.
e) A climatologist collects data from weather balloons and makes observations basing on it.

AP SSC 10th Class English Solutions Chapter 2C The Brave Potter

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 2C The Brave Potter Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 2C The Brave Potter

10th Class English Chapter 2C The Brave Potter Textbook Questions and Answers

Comprehension

Answer the following questions.

Question 1.
What did the tiger think the mysterious creature was? Why did he allow himself bound around the neck with a thick rope?
Answer:
The tiger thought that the ‘leak’ was a mysterious creature. He also thought that it was terrible, dangerous and strong. While the tiger was sleeping, he was suddenly awakened by an angry voice shouting in his ear and felt heavy blows fall upon his head and shoulders. The voice warned him that he would kill him as he had run away. The tiger shivered with fright and thought that it must be the leak’ who had come out of the hut. So the tiger allowed himself found around the neck with a thick rope.

AP SSC 10th Class English Solutions Chapter 2C The Brave Potter

Question 2.
What made the potter angry? What made him angrier?
Answer:
When the storm began, the drunken potter suddenly remembered that he had left his donkey tied under a tree. He rushed out of his hut to take the animal into the stable but the donkey was not there. This made the potter angry. The potter walked through the wet forest searching for the animal. It became dark and he often stumbled over roots and fallen branches. Each step of the potter made him angrier.

Question 3.
Why did the king make the potter the General of the army?
Answer:
The king of the potter’s country gathered a large army when the war broke out between their country and a much stronger neighbour. But the king realized that it was not strong enough to save his country from defeat. So, he searched for a hero to lead his army. When he asked his ministers’ advice, one of them told him about the brave potter who had captured a tiger with his bare hands. The king sent for him and the potter went with his wife to the capital. The king was pleased to see him and ordered him to lead the army into battle the next day.

Question 4.
Why do you think the sentry feels that the potter is a giant?
Answer:
The sentry saw the potter galloping towards the camp with a tree in one hand and his reins in the other. The sentry thought that he must be the General who had captured a tiger with his bare hands. Hence the sentry felt that the potter was a giant.

Question 5.
Do you think that the potter is really brave or lucky? Give your reasons.
Answer:
I think that the potter is really lucky. The incidents of his catching the tiger and the enemy’s fleeing proved this.

AP SSC 10th Class English Solutions Chapter 2C The Brave Potter

Question 6.
What is the most humorous and thrilling incident in the story? Write the incident and say why it is humorous and thrilling.
Answer:
The most humorous and thrilling incident in the story is the potter’s catching the tiger. One day he drank more wine after a hard day’s work. Then the storm began and he remembered that he had left his donkey tied under a tree. He searched for it and finally found a sleeping tiger under thatched roof of a hut and thought that it was his donkey. As he was drunk, he couldn’t find the difference between a tiger and a donkey. The incident of his riding the tiger is the most humorous and thrilling incident. The writer created humour by creating situations where the tiger took the word ‘leak’ to be ‘a more powerful and dangerous thing’, the potter didn’t notice the difference between his donkey and the tiger, the potter’s riding the tiger and people mistook the potter to be a brave man. All these things made the incident humorous and thrilling.

Project work

I. You have read the story ‘The Brave Potter’. It is a humorous story. The writer of the story created humour by creating situations where the tiger took the word ‘leak’ to be ‘a more powerful thing’ than him and people mistook the potter to be a brave man.
Work in groups and collect a humorous story. Analyse how the writer created humour in it.
Answer:
Guru Govind had four disciples. One day he told them not to do anything without his permission. One day while they were on their way to a distant city Guru Govind fell asleep in the bullock cart they were travelling in. The Guru’s head rolled from side to side and suddenly his turban slipped from his head and fell onto the road. But the disciples did not make a move to get down and pick it up as their Guru had instructed them not to do anything on their own. After some time, the Guru woke up and his disciples told him about the loss of his turban. He was angry with them. He roared, “If anything falls of next time, pick it up at once !” After some time, the bullock dropped its dung and the four foolish disciples leaped down and picked the dung up. Guru was annoyed with them. Then he made a list of things that could fall off from a moving cart and said to them, “Pick up any of these things if they fall”. He also said, “Don’t pick up anything that is not in the list.” Just then the cart lurched violently and Guru Govind was thrown into a ditch. He yelled, “Pull me out; pull me out”. “We can’t, guruji,” said his disciples, “Your name is not in the list you have given to us.” Guru Govind pleaded with them to pull him but in vain. “We know you are testing us, guruji,” they said to him, “but we are not going to disobey your words. You have told us not to pick up anything that is not mentioned in the list and so we won’t do it.” “Give me the list!” yelled Guru. When they threw the Isit, he included his name among the other things. Then only the obedient disciples pulled out their beloved Guru out of the ditch. Here, the writer creates humour through the innocence and foolishness of the obedient disciples. He tries to produce humour by creating the situations where the obedient disciples misunderstood their guru’s words.

AP SSC 10th Class English Solutions Chapter 2C The Brave Potter

II. Writing anything funny or humorous is one of the hardest forms of the craft. You may have a great sense of humour, but capturing that in your writing takes skill and practice. Work in groups and recall incidents that made you laugh. Analyse the incidents to find out what made you laugh. It could be the use of some inappropriate word, the way a person is dressed up, an inappropriate timing of an action, etc. Also look at some cartoons and analyse what makes you laugh.
Answer:
There are a lot of humorous incidents we may come across in the stories we read, the movies and cartoons we watch and in our day-to-day lives. If we observe the movements of the famous comedian Charlie Chaplin, we can understand how he created such a humour. His style of walking, using the slapstick, his dress, his hat, his moustache, his face all produce the humour. He is a gifted artiste.

‘Tom and Jerry’ is a series of animated cartoon films. We find humour with the rivalry between a cat (Tom) and a mouse (Jerry), Tom’s chasing Jerry and slapstick scenes. “Tom’s making numerous attempts to capture Jerry which leads to destruction” – it creates fun. The scenes such as slicing Tom in half, shutting his head in a window or a door, stuffing Tom’s tail in a mangle, kicking him into a refrigerator, plugging his tail into an electric socket, sticking matches into his feet and lighting them, etc. amuse all the viewers.

The Brave Potter Summary in English

‘The Brave Potter’ is a very popular Telugu folktale collected by Marguerite Siek. It is a humorous story. A potter is the hero of this story.
It was a dark evening and the sky was full of clouds. The rain was about to start. It was starting to rain and an old tiger ran through the rain for shelter. The tiger crawled under the thatched roof of an old hut and lay down by the door. It was an old woman’s hut which had a leak. When the tiger began to fall asleep, he heard a woman’s voice complaining the leak in her hut was very terrible. She also complained aloud that she would rather meet a tiger than have the leak in her house. When the tiger heard her words, he thought that the ‘leak’ was a very dangerous and strong animal. He was doubtful whether they were all not afraid of him. While these thoughts were lingering in his mind, he fell asleep.

On the afternoon of that day, a potter had drunk more wine and no longer felt tired. When the rain began, he suddenly remembered that he had left his donkey tied under a tree. He rushed to the spot but couldn’t find it. He started searching for it. While he was searching, he often stumbled over roots and fallen branches. He felt angrier and wanted to give the donkey a good beating when he caught it. He reached the old woman’s hut and mistook the tiger for his donkey. He couldn’t notice the difference between a donkey and a tiger as he had drunk. He kicked and beat the sleeping tiger. The tiger thought that it must be the ‘leak’ who came out of the hut. The tiger shivered with fright and wanted to do as the ‘leak’ said. The potter jumped onto the frightened tiger’s back, rode it home and tied it up with the iron chain.

AP SSC 10th Class English Solutions Chapter 2C The Brave Potter

The next morning, the news spread throughout the village that the potter had caught a tiger and tied it to a tree in his yard. All of them praised his courage. But the potter couldn’t understand how it all had happened! They didn’t believe him saying that he had only brought his donkey home. The villagers even praised him for his modesty. A few years later a war broke out. The enemy army was much stronger. The king was worrying how he could save the country. He wanted a brave man to lead the army. One of his ministers told the king about the brave potter. He called for the potter, made him the General of the army and ordered him to lead the army into battle the next day. But the potter was worrying very much as he never carried a sword, nor had he ever ridden a horse. Hence, he wanted to practise riding the horse. He woke up early the next morning and climbed onto the horse’s back with great difficulty. Then he asked his wife to tie his feet to the stirrups. His wife bound his feet tightly to the stirrups and tied the two stirrups together. She also tied him to the saddle. Suddenly, the horse jumped free and galloped out of the stable. The potter held to the horse’s neck tightly and prayed to all the gods to save his life.

The horse galloped through the streets, the city gates and began to head for the enemy’s camp. Though he tried to pull on the reins and control the horse, the horse didn’t stop. The potter grabbed a branch when they passed a young tree. When a sentry from the enemy camp saw him with a tree in one hand and the reins in the other, he thought that the horse-rider must be the General who captured a tiger with his bare hands. Immediately he made loud shouts warning his men about the famous Tiger-General who was rushing towards them to attack. The frightened soldiers fled. Even their king followed his soldiers leaving a letter in the tent. The potter was surprised to find the camp empty. He brought the letter and gave it to his wife. He requested her to take the letter to their king and tell him the enemy had run away. His wife gave the letter to the king and he read it. He praised the potter and rewarded the potter so well that he didn’t need to work again.

The Brave Potter About the Author

Marguerite Siek was a great story teller. He was very much interested in telling folk and mythological stories of Asia. He travelled across many Asian countries and collected interesting short stories from various countries and published them in English. He translated many famous Indian folk stories into English. The present short story The Brave Potter’ is a very popular Telugu one collected by him from India.

The Brave Potter Glossary

blinding (adj): very bright

thatched (adj): covered with dried straw

nod off (phr.v): fall asleep

leak (n): a small hole that lets liquid or gas flow into or out of something

bound (v): tied someone so that they couldn’t move or escape

scream (v): to make a high loud noise with one’s voice because one is hurt, frightened, excited etc.

head (v): go to or travel towards a particular place

palm-wine (n): toddy/fermented palm juice drunk by village folk (kallu In Telugu)

AP SSC 10th Class English Solutions Chapter 2C The Brave Potter

stable (n): a building where horses are kept

grumble: to keep complaining in an unhappy way

stumbled: walked In an unsteady way and often almost Fell

muttered (v): spoke something that cannot be heard

groaned (v): made a long deep sound because one is in pain. upset or disappointed

Your Majesty (phr): way of addressing a king or a queen

saddle (n): a leather seat for a rider on a horse

stirrups (n): metal rings that hang down on each side of a horse’s saddle, used to support the riders foot

pawing (v): touching something repeatedly with a paw

hooves (n): the hard parts of the feet of some animals like horses (‘Hooves’ is the plural form of ‘hool.)

crashing: falling

reins (n): long leather bands held by a horse rider to control it

sentry (n): guard/a soldier whose job is to guard something

uprooted (v): pulled a tree or a plant out of the ground

deserted (adj): empty and quiet because flO people are there

cheering crowds (phr): a large gathering of people shouting in joy

AP Board 1st Class English Textbook Solutions Study Material Guide State Syllabus

Andhra Pradesh SCERT AP Board 1st Class English Textbook Solutions State Syllabus Pdf, AP 1st Class English Solutions Study Material Guide Pdf Free Download are part of AP Board 1st Class Textbook Solutions.

AP State Syllabus 1st Class English Textbook Solutions Study Material Pdf Free Download

AP 1st Class English Textbook Pdf | AP Board 1st Class English Solutions

Unit 1 Me & Myself

Unit 2 My Family

Unit 3 Actions & Fun

Unit 4 Numbers & Colours

Unit 5 Healthy Eating

Unit 6 My Surroundings

Unit 7 Play & Dance

Unit 8 Calendar

We hope these detailed Andhra Pradesh SCERT AP Board 1st Class English Textbook Solutions State Syllabus Pdf will be useful for students to understand all the basic concepts in a much better way. If you have any doubts related to AP 1st Class English Textbook Solutions, then you can ask us and we will be happy to assist you.

Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(a)

I.

Question 1.
Find the acute angle between the pair of lines represented by the following equations.
(i) x² – 7xy + 12y² = 0
(ii) y² – xy – 6x² = 0
(iii) (x cos α – y sin α)² = (x² + y²) sin² α
(iv) x² + 2xy cot α – y² = 0
Solution:
(i) x² – 7xy + 12y² = 0
a = 1, b = 12, h = –\(\frac{7}{2}\)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 1

(ii) y² – xy – x² = 0
a = -6, b = 1, h = –\(\frac{1}{2}\)
\(\cos \theta=\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 2

(iii) (x cos α – y sin α)² = (x² + y²) sin² α
x2 cos² α + y² sin² a – 2xy cos α sin α = x² sin² α + y² sin² α
∴ x² (cos² α – sin² α) – 2xy cos α sin α = 0
x².cos 2α – xy sin 2α = 0
a = cos 2α, b = 0, 2h = -sin 2α
\(\cos \theta=\frac{\|\cos 2 \alpha+0\|}{\sqrt{(\cos 2 \alpha-0)^{2}+\sin ^{2} 2 \alpha}}\)
= cos 2α
∴ θ = 2α

(iv) x² + 2xy cot a – y² = 0
a + b = 1 – 1 = 0
∴ θ = \(\frac{\pi}{2}\)

II.

Question 1.
Show that the following pairs of straight lines have the same set of angular bisectors (that is they are equally inclined to each other).
i) 2x² + 6xy + y² = 0,
4x² + 18xy + y² = 0.
ii) a²x² + 2h(a + b) xy + b²y² = 0,
ax² + 2hxy + by² = 0, a + b ≠ 0.
iii) ax² + 2hxy + by² + λ(x² + y²) = 0; (λ ∈ R),
ax² + 2hxy + by² = 0.
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 3
Solution:
(i) Combined equation of OA, OB is
2x² + 6xy + y² = 0
Equation of the pair of bisectors is
3(x² – y²) = (2 – 1) xy
3(x² – y2² = xy ………… (1)
Combined equation of OP, OQ is
4x² + 18xy + y² = 0
Equation of the pair of bisectors is
9(x² – y²) = (4 – 1) xy 9(x² – y²) = 3xy
3(x² – y²) = xy ………….. (2)
(1), (2) are same.
∴ OA, OB and OP, OQ are inclined to each other.

(ii) Combined equation of OA, OB is
a²x² + 2h(a + b) xy + b²y² = 0
Equation of the pair of bisectors is
h (a + b) (x² – y²) = (a² – b²) xy
h (a + b) (x² – y²) = (a + b)(a – b) xy
i.e., h(x² – y²) = (a – b) xy ………… (1)
Combined equation of OP, OQ is
ax² + 2hxy + by² = 0
Equation of the pair of bisectors is
h (x² – y²) = (a – b) xy …………. (2)
(1), (2) are same.
∴ OA, OB and OP, OQ are equally inclined to each other.

(iii) Combined equation of OA, OB is
ax² + 2hxy + by² + λ(x² + y²) = 0
(a + λ) x² + 2hxy + (b + λ) y² = 0
Equation of the pair of bisectors of OA, OB is
h (x2 – y2) = (a + λ – b – λ)xy
= (a – b)xy ……….. (1)
Combined equation of OP, OQ is
ax² + 2hxy + by² = 0
Equation of the pair of bisectors of OP, OQ is
h(x² – y²) = (a – b) xy …………. (2)
(1), (2) are same.
∴ OA, OB and OP, OQ are equally inclined to each other.

Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a)

Question 2.
Find the value of h, if the slopes of the lines represented by 6x² + 2hxy + y² = 0 are in the ratio 1: 2.
Solution:
Combined equation of the lines is
6x² + 2hxy + y² = 0
Suppose individual equations of the given lines are y = m1x and y = m2x
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 4

Question 3.
If ax² + 2hxy + by² = 0 represents two straight lines such that the slope of one line is twice the slope of the other, prove that 8h² = 9ab.
Solution:
Combined equation of the lines is
ax² + 2hxy + by² =0
Suppose, y = m1x and y = m2x are the individual equations of the lines.
∴ m1 + m2 = –\(\frac{2h}{b}\), m1m2 = \(\frac{a}{b}\)
Given m2 = 2m1
∴ 3m1 = –\(\frac{2h}{b}\) ; 2m1² = \(\frac{a}{b}\)
m1 = –\(\frac{2h}{3}\) ; m1² = \(\frac{a}{2b}\)
∴ (-\(\frac{2h}{3b}\))² = \(\frac{a}{2b}\)
\(\frac{4 h^{2}}{9 b^{2}}=\frac{a}{2 b}\)
8h² = 9ab.

Question 4.
Show that the equation of the pair of straight lines passing through the origin and making an angle of 30° with the line 3x – y – 1 = 0 is 13x² + 12xy – 3y² = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 5
Solution:
Equation of AB is 3x – y – 1 = 0
OA, OB make an angle of 30° with AB and pass through the origin.
Suppose slope of OA is m
∴ Equation of OA is
y – 0 = m (x – 0) = mx or mx – y = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 6
∴ \(\frac{\sqrt{3}}{2}=\frac{|3 m+1|}{\sqrt{10} \sqrt{m^{2}+1}}\)
Squaring and cross multiplying
\(\frac{3\left(m^{2}+1\right)}{4}=\frac{(3 m+1)^{2}}{10}\)
15(m² + 1) = 2 (3m + 1)²
15m² + 15 = 2 (9m² + 6m + 1)
= 18m² + 12m + 2
3m² + 12m -13 = 0
Suppose m1, m2 are two roots of the equation
m1 + m2 = -4, m1 m2 = \(\frac{-13}{3}\)
Combined equation of OA and OB is
(m1x – y) (m2x – y) = 0
m1m2x² – (m1 + m2) xy + y² = 0
\(\frac{-13}{3}\)x² + 4xy + y² = 0
-13 x² + 12xy + 3y² = 0 or
13x² – 12xy – 3y² = 0

Question 5.
Find the equation to the pair of straight lines passing through the origin and making an acute angle a with the straight line x + y + 5 = 0.
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 7
Solution:
Equation of AB is x + y + 5 = 0 ……….. (1)
Slope of AB = – λ
Suppose OA and OB are the required lines
Suppose equation of OA is
y = mx ⇒ mx – y = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 8
2(m² + 1) cos² α = (m – 1)²
2(m² + 1) = \(\frac{(m-1)^{2}}{\cos ^{2} \alpha}\) = (m – 1)² sec² α.
2m² + 2 = m² sec² α – 2m sec² α + sec² α.
m² (sec² α – 2) – 2m sec² α + (sec² α – 2) = 0
m1 + m2 = \(\frac{2 \sec ^{2} \alpha}{\sec ^{2} \alpha-2}\), m1m2 = 1
Combined equation of OA and OB is
(y – m1x) (y – m2x) = 0
y² (m1 + m2) xy + m1m2 x² = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 9
Combined equation of OA and OB is
x² + 2xy sec 2a + y² = 0

Question 6.
Show that the straight lines represented by (x + 2a)² – 3y² = 0 and x = a form an equilateral triangle.
Solution:
Combined equation of OA, OB is
(x + 2a)² – 3y² = 0
(x + 2a)² – (√3y)² =0
(x + 2a + √3 y) (x + 2a – √3 y) = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 10
Equation of OA is
x + √3y + 2a = 0 ………….. (1)
Equation of OB is
x – √3y + 2a = 0 ………. (2)
Equation of AB is x – a = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 11
∴ ∠OBA =60°
∴ ∠AOB = 180°- (∠OAB + ∠OBA)
= 180° – (60° + 60°)
= 180° – 120°
= 60°
∴ ∆OAB is an equilateral triangle.

Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a)

Question 7.
Show that the pair of bisectors of the angles between the straight lines (ax + by)² = c (bx – ay)², c > 0 are parallel and perpen-dicular to the line ax+ by + k= 0.
Solution:
Combined equation of the given lines is (ax + by)² = c (bx – ay)²
a²x² + b²y² + 2ab xy = c (b²x² + a²y² – 2abxy)
= cb²x² +ca²y² – 2cabxy
(a² – cb²)x² + 2ab (1 + c²) xy + (b² – ca²)y² = 0
Equation of the pair of bisectors is
h (x – y²) = (a – h) xy ^
ab (1 + c) (x² – y²)
= (a² – cb² – b² + ca²) (x² – y²) = 0
= (a² – b²)(1 + c) xy.
i.e., ab (x² – y²) – (a² – b²) xy = 0
(ax + by) (bx – ay) = abx² – a2xy +b²xy – aby²
= ab(x² – y²) – (a² – b²)xy
∴ The equation of the pair of bisectors are (ax + by) (bx – ay) = 0
The bisectors are ax + by = 0 and bx – ay = 0
ax + by = 0 is parallel to ax + by + k = 0
bx – ay = 0 is perpendicular to ax + by +k=0.

Question 8.
The adjacent sides of a parallelogram are 2x² – 5xy + 3y² = 0 and one diagonal is x + y + 2 = 0. Find the vertices and the other diagonal.
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 12
Solution:
Combined equation of OA and OB is 2x² – 5xy + 3y² = 0
Equation of AB is x + y+ 2 = 0
y = -(x + 2)
Substituting in (1)
2x² + 5x (x + 2) + 3(x +2)² = 0
2x² + 5x² + 10x + 3(x² + 4x + 4) = 0
7x² + 10x + 3x² + 12x + 12 = 0
10x² + 22x + 12 = 0
5x² + 11x + 6 = 0
(x + 1) (5x + 6) = 0
x + 1 = 0 or 5x + 6 = 0
x = -1 or 5x = -6
x = –\(\frac{6}{5}\)
y = -(x + 2)
x = -1 ⇒ y = – (-1 + 2) = – 1
⇒ co-ordinates of A are (-1, -1)
x = –\(\frac{6}{5}\) ⇒ y = -(\(\frac{6}{5}\)+ 2) = –\(\frac{4}{5}\)
⇒ co-ordinates of B are (-\(\frac{6}{5}\), –\(\frac{4}{5}\))
Suppose the diagonals AB, OC intersect in O’ O’ bisects AB and OC.
Suppose co-ordinates of C are (x, y)
Midpoint of OC = Midpoint of AB
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 13
∴ The vertices are O(0, 0), A(-1, -1)
C(-\(\frac{11}{5}\), –\(\frac{9}{5}\)), B(-\(\frac{6}{5}\), –\(\frac{4}{5}\))

Question 9.
Find the centroid and the area of the triangle formed by the following lines.
(i) 2y² – xy – 6x² = 0, x + y + 4 = 0
(ii) 3x² – 4xy + y² = 0, 2x – y = 6
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 14
Solution:
(i) Combined equation of OA, OB is
2y² – xy – 6x² = 0 ………… (1)
Equation of AB is x + y + 4 = 0
y = – (x + 4) ………… (2)
Substituting in (1)
2(x + 4)² + x (x + 4) – 6x² = 0
2(x² + 8x + 16) + x² + 4x – 6x² = 0
2x² + 16x + 32 + x² + 4x – 6×2 = 0
-3x² + 20x + 32 = 0
3x² – 20x – 32 = 0
(3x + 4) (x-8) = 0
3x + 4 = 0 or x – 8 = 0
x = –\(\frac{4}{3}\) or 8

Case (i) : x = –\(\frac{4}{3}\)
y = – (x + 4)
= -(\(\frac{-4}{3}\) + 4) = –\(\frac{8}{3}\)
Co – ordinates of A are (-\(\frac{4}{3}\), –\(\frac{8}{3}\))

Case (ii) : x = 8
y = -(x + 4) = – (8 + 4) = – 12
Co-ordinates of B are (8, – 12)
Suppose G is the centroid of ∆ AOB
Co-ordinates of G are
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 15
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 16

(ii) Combined equation of OA, OB is
3x² – 4xy + y² = 0 ……………. (1)
Equation of AB is 2x – y = 6
y = 2x – 6 ……………. (2)
Substituting in (1)
3x² – 4x (2x – 6) + (2x – 6)² = 0
3x² + 8x² + 24x + 4x² + 36 – 24x = 0
– x + 36 = 0
x² – 36 = 0
(x + 6) (x – 6) = 0
x + 6 = 0 or x – 6 = 0
x = – 6 or 6
y = 2x – 6
x = 6 ⇒ y = 12 – 6 = 6
Go -ordinates of A are (6, 6)
x = -6 ⇒ y = – 12 – 6 = -18
Co-ordinates of B are (-6, -18)
Co-ordinates of G ate
(\(\frac{0+6-6}{3}\), \(\frac{0+6-18}{3}\)) = (0, -4)
∆OAB = \(\frac{1}{2}\)|x1y2 – x2y1|
= \(\frac{1}{2}\)|16 (-18) – (-6). 6|
= \(\frac{1}{2}\)|- 108 + 36|
= \(\frac{1}{2}\) . 72 =36 sq.units

Question 10.
Find the equation of the pair of lines intersecting at (2, -1) and
(i) perpendicular to the pair
6x² – 13xy – 5y² = 0 and
(ii) parallel to the pair
6x² – 13xy – 5y² = 0.
Solution:
Equation of OA, OB is 6x² – 13xy – 5y² = 0
(i) Equation of the pair of lines through (x1 y1) and perpendicular to
ax² + 2hxy + by² = 0 is
b(x – x1)² – 2h(x – x1) (y – y1) + a (y – y1)² = 0
Equation of the perpendicular pair of lines is
-5(x – 2)² + 13(x – 2) (y + 1) + 6(y+ 1)² =0
-5(x² – 4x + 4) + 13(xy + x – 2y – 2) + 6(y² + 2y + 1) = 0
-5x² + 20x – 20 + 13xy + 13x – 26y – 26 + 6y² + 12y + 6 = 0
-5x² + 13xy + 6y² + 33x – 14y – 40 = 0
or 5x² – 13xy – 6y² – 33x + 14y + 40 = 0

(ii) Equation of the pair of lines through (x1, y1) and parallel to ax² + 2hxy + by² = 0 is
a(x – x1)² + 2h (x – x1) (y – y1) + b (y – y1)² = 0
Equation of the pair of parallel lines is
6(x- 2)² – 13(x – 2) (y + 1) – 5(y + 1)² = 0
6(x² – 4x + 4) – 13(xy + x – 2y – 2) – 5(y² + 2y + 1) = 0
6x² – 24x + 24 – 13xy – 13x + 26y + 26 – 5y² – 10y – 5 = 0
6x² – 13xy – 5y² – 37x + 16y + 45 = 0.

Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a)

Question 11.
Find the equation of the bisector of the acute angle between the lines
3x – 4y + 7 = 0 and 12x + 5y – 2 = 0
Solution:
Given lines
3x – 4y + 7 = 0 ………….. (1)
12x + 5y – 2 = 0 ………… (2)
The equations of bisector’s angles between (1) & (2) is
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 17
13 (3x – 4y + 7) ± 5 (12x + 5y- 2) = 0
(39x – 52y + 51) ± (60x + 25y – 10) = 0

(i) 39x- 52y + 51 + 60x + 25y- 10 = 0
99x-27y + 41 = 0 ……… (3)

(ii) (39x – 52y + 51) – (60x + 25y- 10) = 0
39x – 52y + 51 – 60x – 25y +10 = 0
– 21x – 77y + 61 =0
21x + 77y- 61 = 0 ……… (4)
Let ‘θ’ be the angle between (1), (4)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 18
∴ (4) is obtuse angle bisector, then other one (3) is the accute angle bisector.
∴ 99x – 27y + 41 = 0 is the accute angle bisector.

Question 12.
Find the equation of the bisector of the obtuse angle between the lines x + y – 5 = 0 and x – 7y + 7 = 0
Solution:
Given lines
x + y- 5 = 0 ………. (1)
x – 7y + 7 = 0 ……..(2)
The equations of bisectors of angles between (1), (2) is
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 19
⇒ (5x + 5y-25)±(x-7y + 7) = 0

(i) 5x + 5y – 25 + x – 7y + 7 = 0
6x – 2y – 18 = 0
3x-y-9 = 0 ………. (3)

(ii) (5x + 5y – 25) – (x – 7y + 7) = 0
4x+ 12y – 32 = 0
x + 3y – 8 = 0 ……… (4)
Let ‘θ’ be the angle between (1), (4)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 20
∴ (4) is the ocute angle bisector, then other one 3x – y – 9 = 0 is the obtuse angle bisector.

III.

Question 1.
Show that the lines represented by (lx + my)² – 3(mx – ly)² = 0 and lx + my + n = 0 form an equilateral triangle with area \(\frac{n^{2}}{\sqrt{3}\left(l^{2}+m^{2}\right)}\).
Solution:
Combined equation of A and is
(lx + my)² – 3(mx – ly)² = 0
l²x² + m²y² + 2lmxy – 3m²x² – 3l²y² + 6 lmxy = 0
(l² – 3m²) x² + 8lmxy + (m² – 3l²) y² = 0
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 21

Combined equation of the bisectors of OA
and OB is h (x² – y²) = (a – b) xy
4 lm (x² – y²) = (l² – 3m² – m² + 3l²) xy
4 lm (x² – y²) = 4(l² – m²) xy
lmx² – (l² – m²)xy – lmy² = 0
(lx – my) (mx – ly) = 0
lx + my = 0 and mx – ly = 0
∴ The bisectors mx – ly = 0 is perpendicular to AB whose equation lx + my + n = 0
OAB is an scales triangle and ∠AOB = 60°
OAB is an equalated tringle
P = Length of the X lan prove P and AB
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 22

Question 2.
Show that the straight tines represented by 3x² + 48xy + 23y² = 0 and 3x – 2y + 13 = 0 form an equilateral triangle of area \(\frac{13}{\sqrt{3}}\) sq.umts.
Solution:
Combined equation of OA, OB is
3x² + 48xy + 23y² = 0 …………. (1)
Equation of AB is 3x – 2y + 13 = 0 …….. (2)
(1) can be written as
(9x² – 12xy + 4y²) – 3(4x² + 12xy + 9y²) = 0
i.e., (3x – 2y)² – 3(2x + 3y)² = 0
⇒ [(3x – 2y) + √3(2x +3y)] [(3x – 2y) – √3(2x+3y)] = 0
⇒ [(3 + 2√3)x+ (3√3 – 2)y] [(3 – 2√3)x – (3√3 + 2)y]=0
Equation of OA is
(3 + 2√3)x – (3√3 – 2)y = 0 ………….. (1)
Equation of OB is
(3 – 2√3)x – (3√3 +2)y =0 ………… (2)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 23
∴ OAB is an equilateral triangle.
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 24

Question 3.
Show that the equation of the pair of lines bisecting the angles between the pair of bisectors of the angles between the pair of lines ax² + 2hxy + by² = 0 is (a – b) (x² – y²) + 4hxy = 0
Solution:
Equation of the given lines is
ax² + 2hxy + by² = 0
Equation of the pair of bisectors is
h(x² – y²) = (a – b)xy …………. (1)
hx² – hy² – (a – b) xy = 0
∴ A = h, B = -h, 2H = -(a – b)
Equation of the pair of bisectors of (1) is
H(x² – y²) = (A – B) xy
–\(\frac{(a-b)}{2}\)(x² – y²) = 2hxy
-(a – b) (x² – y²) = 4hxy
or (a – b) (x² – y²) + 4hxy = 0
∴ Equation of the pair of bisectors of the pair of bisectors of ax² + 2hyx + by² = 0 is
(a – b) (x² – y²) + 4hxy = 0.

Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a)

Question 4.
If one line of the pair of lines ax² + 2hxy + by² = 0 bisects the angle between the co-ordinate axes, prove that (a + b)²= 4h².
Solution:
The angular bisectors of the co-ordinate axes are:
y = ±x
Case (i) :
y = x is one of the lines of
ax² + 2hxy + by² = 0
x²(a + 2h + b) = 0
a + 2h + b = 0 ………….. (1)

Case (ii) :
y = – x is one of the lines of
ax² + 2hxy + by² = 0
x² (a – 2h + b) = 0
a – 2h + b = 0 ………. (2)
Multiplying (1) and (2), we get
(a + b + 2h).(a + b – 2h) = 0
(a + b)² – 4h² = 0
(a + b)² = 4h².

Question 5.
If (α, β) is the centroid of the triangle formed by the lines ax² + 2hxy + by² =0 and lx + my = 1, prove that
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 25
Solution:
Combined equation of OA, OB is
ax² + 2hxy + by² = 0 …………. (1)
Equation of AB is lx + my = 1
my = 1 – lx
\(y=\frac{1-1 x}{m}\) …………. (2)
Substituting in (1)
ax² + 2hx\(\frac{(1-b)}{m}\) + b\(\frac{(1-1 x)^{2}}{m^{2}}\) = 0
am²x² + 2hmx(1 – lx) + b(1 + l²x² – 2lx) = 0
am²x² + 2hmx – 2hlmx² + b + bl²x² – 2blx = 0
(am² – 2hlm + bl²)x² – 2(bl – hm)x + b = 0
Suppose coordinates of A are (x1, y1) and B are (x2, y2)
x1 + x2 = \(\frac{2(b /-h m)}{a m^{2}-2 h / m+b l^{2}}\) …………. (3)
A and B are points on
lx + my = 1
lx1 + my1 = 1
lx2 + my2 = 1
l(x1 + x2) + m(y1 + y2) = 2
m(y1 + y2) = 2 – l(x1 + x2)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 26

Co-ordinates of the vertices are
O(0, 0), A(x1, y1), B(x2, y2)
Co-ordinates of G are
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 27
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 28

Question 6.
Prove that the distance from the origin to the orthocentre of the triangle formed by the lines \(\frac{x}{\alpha}+\frac{y}{\beta}\) = 1 and ax² + 2hxy + by² = 0 is (α² + β²)1/2 \(\left|\frac{(a+b) \alpha \beta}{a \alpha^{2}-2 h \alpha \beta+b \beta^{2}}\right|\).
Solution:
Let ax² + 2hxy + by² = 0 represent the lines
l1x + m1y = 0 ………… (1)
l2x + m2y = 0 ………… (2)
∴ (l1x + m1y) (l2x + m2y) = ax² + 2hxy + by²
Comparing both sides
l1l2 = a, m1m2 = b, l1m2 + l2 m1 = 2h
Given line is lx + my =1 ……….. (3)
Clearly the origin O is the point of intersection of (1) & (2)
Let A be the point of intersection of (1) & (3)
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 29
By the method of cross multiplication,
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 30
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 31
Let B be the point of intersection of (2) & (3)
Let P be the orthocentre of ∆ OAB.
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 32
⇒ yl2(l1α – m1β) – αβl1l2 = m2(x(l1α – m1β) + m1αβ)
⇒ (l1α – m1P) (m2x – l2y) = m1m2αβ + l1l2αβ
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 33
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 34

Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a)

Question 7.
The straight line lx + my + n = 0 bisects an angle between the pair of lines of which one is px + qy + r = 0. Show that the other line is (px + qy + r) (l² + m²) – 2(lp + mq) (lx + my + n) = 0.
Solution:
lx + my + n – 0 is a bisector and let (a, P) be any point on it so that
lα + mβ + n = 0 …………….. (1)
The other line will pass through the intersection of given lines and given bisector and hence by p +λq = 0
Its equation is
(px + qy + r) + λ(lx + my + n) = 0 …………….. (2)
Also px + qy + r = 0
If (α, β) be a point on the bisector then its perpendicular distance from the lines (2) and (3) is same.
Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(a) 35
Putting lα + mβ + n = 0 by (1) in the above and cancelling pα + qβ + r and then squaring both sides, we get
(p + lλ)² + (q + mλ)² = p² + q² or
2λ(pl + qm)+ λ²(l² + m²) = 0
∴ λ = -2\(\frac{p l+Q m}{l^{2}+m^{2}}\)
Substitute X value in (2),
(px + qy + r) + \(\left(\frac{-2(p /+Q m)}{l^{2}+m^{2}}\right)\) lx + my + n = 0
⇒ (px + qy + r)(l² + m²) -2(pl + qm) (lx + my + n ) = 0

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions and Answers.

8th Class Maths 3rd Lesson Construction of Quadrilaterals InText Questions and Answers

Do this

Question 1.
Take a pair of sticks of equal length, say 8 cm. Take another pair of sticks of equal length, say 6 cm. Arrange them suitably to get a rectangle of length 8 cm and breadth 6 cm. This rectangle is created with the 4 available measurements. Now just push along the breadth of the rectangle. Does it still look alike? You will get a new shape of a rectangle Fig (ii), observe that the rectangle has now become a parallelogram. Have you altered the lengths of the sticks? No! The measurements of sides remain the same.
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 7AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 8
Give another push to the newly obtained shape in the opposite direction; what do you get? You again get a parallelogram again, which is altogether different Fig (iii). Yet the four measurements remain the same. This shows that 4 measurements of a quadrilateral cannot determine its uniqueness. So, how many measurements determine a unique quadrilateral? Let us go back to the activity!
You have constructed a rectangle with two sticks each of length 8 cm and other two sticks each of length 6 cm. Now introduce another stick of length equal to BD and put it along BD (Fig iv). If you push the breadth now, does the shape change? No!
It cannot, without making the figure open. The introduction of the fifth stick has fixed the rectangle uniquely, i.e., there is no other quadrilateral (with the given lengths of sides) possible now. Thus, we observe that five measurements can determine a quadrilateral uniquely. But will any five measurements (of sides and angles) be sufficient to draw a unique quadrilateral? (Page No. 60)
Answer:
Yes, any 5 individual measurements are needed to construct a quadrilateral.

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

Question 2.
Equipment (Page No. 61)
You need: a ruler, a set square, a protractor.
Remember: To check if the lines are parallel.
Slide set square from the first line to the second line as shown in adjacent figures.
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 9
Now let us investigate the following using proper instruments. For each quadrilateral,
a) Check to see if opposite sides are parallel.
b) Measure each angle.
c) Measure the length of each side.
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 10
Record your observations and complete the table below.
Answer:
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 1
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 2

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

Question 3.
Can you draw the angle of 60°?    (Page No. 63)
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 3
Answer:
Using a scale and compass,
we can construct 60°.
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 4

Question 4.
Construct the parallelogram above (Refer text book page no: 75) BELT by using other properties of parallelogram. (Page No. 75)
Answer:
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 5
We can construct a parallelogram using the measurements of a side, a diagonal and an angle.
BE = 5 cm ⇒ LT = 5 cm
∠B = 110° ⇒ ∠E = 180° – 110° = 70°
TE= 7.2 cm

Try these

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

Question 1.
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm?    (Page No. 70)
Answer:
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 11
In a parallelogram BATS, opposite sides are equal.
BA = ST = 5 cms
AT = BS 6 cms
AS = 6.5 cms
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 12
∴ So, we can construct BATS parallelogram. It needs only three measurements.

Question 2.
A student attempted to draw a quadrilateral PLAY given that PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm. But he was not able to draw it why ?
Try to draw the quadrilateral yourself and give reason. (Page No. 70)
Answer:
In a quadrilateral PLAY
PL = 3 cm LA = 4 cm AY = 4.5 cm
PY = 2 cm LY = 6 cm
Here YP + PL < YL [∵ 2 + 3 < 6 ⇒ 5 < 6]
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 13
But in a △YPL, the sum of two sides is less than the third side.
∴ We are unable to construct a quadrilateral PLAY [∵ YL > YP]
[∵ The arcs do not intersect which are drawn from L and P, also Y, P, L are collinear points]

Think, discuss and write

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

Question 1.
Is every rectangle a parallelogram? Is every parallelogram a rectangle?    (Page No. 63)
Answer:
Yes, every rectangle is a parallelogram. But every parallelogram is not a rectangle.

Question 2.
Uma has made a sweet chikki. She wanted it to be rectangular. In how many different ways can she verify that it is rectangular?    (Page No. 63)
Answer:
If the sweet chikki is to made into a rectangular shape, she has to verify the following shapes:

  1. Quadrilateral
  2. Trapezium
  3. Parallelogram

Question 3.
Can you draw the quadrilateral ABCD with AB = 4.5 cm, BC = 5.2 cm, CD = 4.8 cm and diagonals AC = 5 cm, BD = 5.4 cm by constructing △ABD first and then fourth vertex ‘C’ ? Give reason.       (Page No. 72)
Answer:
We cannot construct △ABD. So, if we start first from △ABD, it is impossible to construct □ ABCD.
[∵ The length of \(\overline{\mathrm{AD}}\) is not given]

Question 4.
Construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm. Justify your result.      (Page No. 72)
Answer:
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 14
PQ = 3 cm
RS = 3 cm
PS = 7.5 cm
PR = 8 cm
SQ = 4 cm
With the given measurements △PQS is not possible to construct.
∵ PQ + QS < PS
The arcs which drawn from P and Q are not intersecting.
∴ We can’t obtain vertex ‘S’.
∴ Without vertex ‘S’ we can’t get a quadrilateral PQRS.

Question 5.
Can you construct the quadrilateral PQRS, if we have an angle of 100° at P instead of 75°? Give reason. (Page No. 74)
Answer:
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 15
PQ = 4 cm,
QR = 4.8 cm,
ZP = 100°,
ZQ = 100°,
ZR = 120°
∴ We can construct a quadrilateral with the given measurements.
Since the sum of 4 angles is equal to 360°.
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 16

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

Question 6.
Can you construct the quadrilateral PLAN if PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15° and ∠A = 140°?
(Draw a rough sketch in each case and analyse the figure). State the reasons for your conclusion. (Page No. 74)
Answer:
PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15°, ∠A = 140°
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 17
∴ With the given measurements it is not possible to construct a quadrilateral.

Question 7.
Do you construct the given quadrilateral ABCD with AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75° by taking BC as base instead of AB? If so, draw a rough sketch and explain the various steps involved in the construction. (Page No. 77)
Answer:
AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75°
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 18
Construction Steps:

  1. Construct a line segment with radius 4.5 cms as \(\overline{\mathrm{BC}}\)
  2. With the centres B and C draw two rays with 100°, 75° respectively.
  3. With the centres B and C, two arcs are drawn with radius 5 cm and 6 cm respectively. The arcs and the rays are intersected.
  4. Let the intersecting points be keep as A, D.
  5. Join A, D.
  6. ∴ ABCD quadrilateral is formed.
    AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 19

AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions

Question 8.
Can you construct the given AC = 4.5 cm and BD = 6 cm quadrilateral (rhombus) taking BD as a base instead of AC? If not give reason. (Page No. 79)
Answer:
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 20
We can construct a rhombus taking BD as base instead of base AC.

Question 9.
Suppose the two diagonals of this rhombus are equal in length, what figure do you obtain? Draw a rough sketch for it. State reasons. (Page No. 79)
Answer:
In a rhombus if the two diagonals are equal then it becomes a square.
∴ ABCD is a square.
[∵ AB = BC = CD = DA Also AC = BD]
AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals Questions 21

AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 5th Lesson Attaining the Age of Adolescence

8th Class Biology 5th Lesson Attaining the Age of Adolescence Textbook Questions and Answers

Improve Your Learning

Question 1.
How is adolescence different from childhood?
Answer:

Adolescence Childhood
1. It is Independent nature and very self conscious. 1. It depends upon parental assistance for basic needs.
2. Adolescents seek company of friends to share their feelings. 2. Children are learning through experimenting and communicating with other.
3. Taking decisions by critical thinking. Don’t like the supervision of elders. 3. Adults supervise and support the development process of child.
4. Lot of Stress and strain. 4. No stress, make new friends and gain new skills.
5. Rate of growth is more 5. Comparatively less.

Question 2.
Write short notes on the following.
a) Secondary sexual characters
b) Adam’s Apple.
Answer:
a) Secondary Sexual characters:

  1. In adolescence age some external changes have seen in boys and girls.
  2. These are called secondary sexual characters.
  3. Example: in boys facial hair, moustaches and beards begin to grow. Hair starts growing on the chest of boys.
  4. In girls breast begin to develop.
  5. In both boys and girls hair grows in the arm pits.

b) Adam’s Apple:

  1. The Adam’s apple is actually a partial growth of our voice box or larynx.
  2. The larynx is made up of nine cartilages, one of which is the largest, called thyroid cartilage.
    AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 1
  3. Due to the elongation of the thyroid cartilage the Adam’s apple is formed. It protrudes out in front of the neck.
  4. This is caused mainly by male hormone. Testosterone during adolescence.
  5. As a result muscles or chords attached to the cartilage get loosened and thickened.
  6. When air passes through these chords a hoarse sound is produced.
  7. This is the reason for disturbance in voice in the stage of adolescence. At the end of this stage voice get perfect.

Question 3.
List out the changes in the body that take place at the age of adolescence.
(or)
What are the changes that could be observed in adolescence phase?
Answer:

  1. During this adolescence changes occur in external, internal parts of the body.
  2. They show interest to spend time with peers.
  3. Girls voice becomes soft, pimples also may appear on the face by the activation of oil and sweat glands.
  4. There is growth and maturity in reproductive system.
  5. In boys voice becomes coarse. Pimples, acne may appear on the face. Facial hair like mustache begin to grow.

Question 4.
Match the following:
1. Testes                                  ( )           a. Estrogen
2. Endocrine gland                  ( )           b. Pituitary
3. Menarche                            ( )           c. Sperm
4.Female hormone                  ( )           d. First menstruation
Answer:
1. c
2. b
3. d
4. a

Question 5.
Why acne and pimples are common in adolescents?
Answer:

  1. Naturally in adolescence boys and girls feel worried of their pimples and acne.
  2. The reason is the secretions of sweat and sebaceous glands are very active in adolescence.
  3. Because of increased activity of these glands in the skin, boys and girls get acne and pimples.

Question 6.
What can you suggest to your classmates to keep himself/herself clean and healthy?
Answer:

  1. It would better to have bath atleast twice in a day.
  2. All parts of the body and inner wears should be washed and cleaned every day.
  3. If cleanliness is not maintained, there are chances of having fungal, bacterial and other unwanted infections.
  4. Girls should take special care of cleanliness during the time of menstrual cycle.
  5. Making use of disposable napkins.

Question 7.
If you have chance to talk with a doctor, what questions you would ask about adolescent emotions and changes in the body ?
Answer:
If I have chance to talk with a doctor, I would ask about

  1. How to develop positive emotions like bravery, self confidence, happiness, satisfaction, appreciation gratitude, concern and forgiveness.
  2. And also how to over come the negative emotions like anger, bitterness, dissatisfaction, sadness, anxiety, fear, shame and guilt; which are needed in adolescence.

Question 8.
Some mobile phones have auditory meter to measure frequency of produced sound. By using this phone measure your friend’s voice frequency one from each class VI to X. Report your findings.
Answer:

Name Class His/Her Voice Frequency
Madhavi VI 50 decibels
Kalyan VII 52 decibels
Ravi VIII 54 decibels
Hemanth IX 55.5 decibels
Jalaja VI 48 decibels
Madhu VII 48.5 decibels
Padmaja VIII 49 decibels
Sailaja IX 49.5 decibels

Question 9.
Write five suggestions to improve the performance of Red Ribbon club of your school.
Answer:

  1. To instill life skills.
  2. To ensure that every college going youth is equipped with conceptual knowledge about various basic health aspects.
  3. To increase the capacity of educational system in teaching various basic health aspects.
  4. To motivate youth and build their capacity as peer educators.
  5. To promote voluntary blood donations.

Question 10.
Prepare a three minute speech on behavioural changes in adolescents.
Answer:

  1. Adolescence is the growing age where we may observe some changes in behaviour.
  2. They are very fast in taking decisions.
  3. They do not want to be forced to do any work, behave peculiarly sometimes fast and sometimes frigid.
  4. Adolescents prefers to spend more times before the mirror and like to use perfumes.
  5. They do not want to listen to parents suggestions and feels friends are correct but not parents.
  6. They search for identify from teachers and peer groups.
  7. They want more independence in taking decisions.
  8. Sometimes they feel shy and sometimes feel happy.
  9. They try to get romantic relationships.
  10. They are more inclined towards unhealthy habits.
  11. The adolescents have attraction towards opposite sex.
  12. The mind of an adolescent is full of zealous acts and urge to find reasons for several things around.
  13. Emotionally they are in a turbulent state all the time they get new thoughts for their life activities.
  14. An adolescent feel insecure while trying to adjust to the changes in the body and the mind.
  15. They seek company of friends to share their feelings even if they are of opposite sex.

Question 11.
Nature prepares human body to reproduce her generations. What do you think of it?
Answer:

  1. In females, the reproductive phase of life begins usually around 10 to 12 years of age.
  2. And generally it lasts till the age of 40 – 50 years.
  3. The ova begin to mature with the onset of adolescence.
  4. One ovum matures and is released by one of the ovaries once in about 28 to 30 days.
  5. During this period the wall of the uterus becomes thick so as to receive a fertilized egg.
  6. This results in pregnancy and childs’ birth.
  7. If fertilization does not occur, the released egg and thickened lining of the uterus will be released with some amount of blood in woman.
  8. This is called menstrual cycle.
  9. Thus nature prepares the female human body to reproduce generation after generation to continue human life on the earth.
  10. This is the secret of nature and is nature’s wonderful phenomena.

Question 12.
You know that early marriage is a social taboo. Prepare some slogans to prevent this. (OR)
You know about that child-marriages are social evil. Your school students are conducting a rally to educate the society. Prepare some slogans on this.
Answer:

  1. Avoid child marriage – Prevent childhood.
  2. Let a child be a child – stop child marriage.
  3. Child marriage – a loosing game.
  4. Stop child marriage – stop child abuse
  5. Childhood is not for motherhood.
  6. Let girls be girls but not brides.

Question 13.
13 years old Swaroop always think of his height. Can he improve his height? What do you suggest him?
Answer:
The suggestion is to take nutritious food and to do body exercise regularly to improve his height.

Question 14.
Are you angry with your parents. How do you wish your parents to be?
Answer:
When insulted or threatened unfairly by parents we get angry on them. In our opinion a parent is

  1. to be a good advisor to give advice how to control stress and strain, which is needed by the adolescent.
  2. to be like a guide to give guidance how to behave with opposite sex.
  3. to be like a friend to give good suggestions.
  4. to be like a wellwisher and always stand behind us to lead us for bright future.

Question 15.
What are your expectations about your parents and teachers?
Answer:
Parents and teachers play an important role to develop the adolescents in to healthy, productive young ones to the nation. Parents feel to develop their children in to better ones than themselves. They must have good courage, confidence, boldness, and free to solve the problems. They should not be tense and worse. Parent is to be a friend and guide towards adolescents.
Teacher is not only a master is to be a captain or a leader. Every adolescent needs mental support. Teacher is the only people who give suitable suggestions to make them free from all mental stress.

8th Class Biology 5th Lesson Attaining the Age of Adolescence InText Questions and Answers

Question 1.
Some of you also may behave like this, Why?
Answer:
At the age of 13 – 19 years, some changes like voice becomes hoarse, not caring to follow the suggestions and advises of parents, shows restlessness and growing tall.
Because at this age children will be entering into a period which is called as Adolescence, where some changes in the behaviour is seen.

Question 2.
Have you noticed that you are growing?
Answer:
At the age of adolescence growth in height takes place about 18 years of age both boys and girls reach their maximum height.

Question 3.
Have you reached the age of“Adolescence”?
a) Is mustache growing on your upper lip?
b) Did your voice change?
c) Are hairs growing under arm pit?
d) Are there pimples or acne on your face?
e) Are you taking care of your face by applying powder and combing your hairs frequently?
f) Are you feeling shy when talking with opposite sex?
g) Are you not interested to play with opposite sex which you have done earlier?
h) Are you showing restlessness while your parents suggest to do something?
Answer:
For above all the questions the answer is ‘yes’ during adolescence changes occur in external, internal parts of the body.

8th Class Biology 5th Lesson Attaining the Age of Adolescence Activities

Activity – 1

Question 1.
Observing growth rate.
a) Observe the below table and given graph, answer the following questions.
AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 2
i) What have you observed from the above table?
Answer:
We have observed the height attained by boys and girls in different ages.

ii) When does growth in height nearly stop?
Answer:
In boy growth in height nearly stops at the age of 18, in girls it stops at 17 years.

iii) Which period of age according to you is the fastest growing period for girls?
Answer:
The fastest growing period for girls is between 14-17 years.

iv) Which period of age is the fastest growing period for boys?
Answer:
The fastest growing period for boys is between 16 to 18 years.

v) Who do grow faster? How can you say?
Answer:
The girls grow faster than the boys. By seeing the above graph, about 17 years of age, girls reach their maximum height.

b) Sneha is 13 years old with 125 cm tall. At the end of the growth period likely to be use the following formulae and calculate the maximum height that Sneha will reach.
AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 3
Answer:
AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 4
Sneha’s present height = 125 cm
At the end of the growth period Sneha is likely to be 131.5 cm.

c) Table – 1 shows that girls grow faster than boys in their adolescent period. From a group with six students in your class. Measure the height and calculate the future heights in the following table:
Answer:
AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 5

Activity – 2

Question 2.
Form five groups in your class. Select at least 15 students. Collect body measurement data of the selected 15 students.
Find an average body measurements for boys and girls separately.
Answer:
AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 6

Activity – 3

Question 3.
Read the following check list. Put tick (✓) mark which points reflect your behaviour.
Answer:
Check List:
AP Board 8th Class Biology Solutions Chapter 5 Attaining the Age of Adolescence 7

Think & Discuss

I. Read the following information and answer the following questions.
Some sections of people in our society believe that during the period of menstruation women are untouchable. So, they are asked to keep a distance from others. During this time girls may be ristricted from taking bath, cooking food or going to school. In that case they may lag behind in their studies. In some sections of the society even women are also forced to stay in the huts built at the outskirts of the village.

Question 1.
In what way this kind of discrimination is harmful for girls and women?
Answer:

  1. During the period of menstruation women are treated as untouchable by some sections of people in our society.
  2. Generally during this period they feel weak and uncomfortable physically.
  3. By this kind of discrimination, mentally also they get hurt and feel that why they have born as woman.
  4. By separation, it would be known to all by this the girl may feels shy, unable to move freely with others and she becomes dull in studies.

Question 2.
Several researches have been done to prove that all these are myths and there is no scientific reason behind these. The blood and egg that is discarded would give rise to a baby if fertilization took place. This is a biological phenomena.
So how can it be impure or unclean?
Answer:

  1. If fertilization took place, the uterus receives a fertilized egg and this results in pregnancy and would give rise a baby which develops in the uterus.
  2. If fertilization does not occur this cause bleeding in woman, which is the unwanted and waste to the uterus. So it can be treated as impure or unclean and may create some problems in uterus.
  3. During menstrual period proper care regarding health and hygiene is needed rather than following myths.

Question 3.
If young generation is trapped into such unhealthy habits, what will be the
future of our country ? What are its effects?
Answer:

  1. If young generation is trapped into unhealthy habits like consuming tobacco (gutkha, cigarettes, cigar, beedi) they addicted to such social evil.
  2. Todays children are tomorrows citizens so it should be avoided.
  3. A famous psychiatrist Stanly Hall stated that adolescence is the age of stress and strain. By getting proper guidance from teachers, parents and elders, the adolescents be able to lead a happy meaning full life and they will save the future of the country.

AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions and Answers.

8th Class Maths 12th Lesson Factorisation InText Questions and Answers

Do this

Question 1.
Express the given numbers in the form of product of primes. [Page No. 267]
(i) 48      (ii) 72      (iii) 96
(i) 48
Answer:
48 = 2 × 2 × 2 × 2 × 3
AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions 1

AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions

ii) 72
Answer:
72 = 2 × 2 × 2 × 3 × 3
AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions 2

iii) 96
Answer:
96 = 2 × 2 × 2 × 2 × 2 × 3
AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions 3

Question 2.
Find the factors of following:      [Page No. 268]
(i) 8x2yz     (ii) 2xy (x + y)       (iii) 3x + y3z
Answer:
i) 8x2yz = 2 × 2 × 2 × x × x × y × z
ii) 2xy (x + y) = 2 × x × y × (x + y)
iii) 3x + y3z = (3 × x) + (y × y × y × z)

Question 3.
Factorise:      [Page No. 270]
(i) 9a2 – 6a
(ii) 15a3b – 35ab3
(iii) 7lm – 21lmn
Answer:
(i) 9a2 – 6a = 3 × 3 × a × a – 2 × 3 × a
= 3 × a (3a – 2)
∴ 9a2 – 6a = 3a (3a – 2)

ii) 15a3b – 35ab3
= 3 × 5 × a × a × a × b – 7 × 5 × a × b × b × b
= 5 × a × b [3 × a × a – 7 × b × b]
= 5ab [3a2 – 7b2]

iii) 7lm – 21lmn
= 7 × l × m7 × 3 × m × n × l
= 7 × l × m [1 – 3n]
= 7lm [1 – 3n]

AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions

Question 4.
Factorise:
i) 5xy + 5x + 4y + 4
ii) 3ab + 3a + 2b + 2 [Pg. No. 271]
Answer:
i) 5xy + 5x + 4y + 4
= (5xy + 5x) + (4y + 4)
= 5x(y + 1) + 4(y + 1)
= (y + 1) (5x + 4)

ii) 3ab + 3a + 2b + 2
= [3 × a × b + 3 × a] + [2 × b + 2]
= 3 × a [b + 1] + 2 [b + 1]
= (b + 1) (3a + 2)

Think, Discuss and Write

While solving some problems containing algebraic expressions in different operations, some students solved as given below. Gan you identify the errors made by them? Write correct answers.     [Page No. 279]
Question 1.
Srilekha solved the given equation as shown below.
3x + 4x + x + 2x = 90
9x = 90 Therefore x = 10 What could say about the correctness of the solution?
Can you identify where Srilekha has gone wrong?
Answer:
Srilekha’s solution is wrong,
∵ 3x + 4x + x + 2x = 90
10x = 90
x = \(\frac{90}{10}\)
∴ x = 9

Question 2.
Abraham did the following.      [Page No. 280]
For x = -4, 7x = 7 – 4 = -3.
Answer:
Abraham’s solution is wrong.
∴ If x = -4
⇒ 7x = 7(-4) = -28

AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions

Question 3.
John and Reshma have done the multiplication of an algebraic expression by the following methods: verify whose multiplication is correct.      [Page No. 280]
AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions 4
Answer:
AP Board 8th Class Maths Solutions Chapter 12 Factorisation InText Questions 5
∴ John’s solutions are wrong and Reshma’s solutions are correct.

Question 4.
Harmeet does the division as (a + 5) ÷ 5 = a + 1 His friend Srikar done the same (a + 5) ÷ 5 = a/5 + 1 and his friend Rosy did it this way (a + 5) ÷ 5 = a Can you guess who has done it correctly? Justify!     [Page No. 280]
Answer:
The solutions of Harmeet, Rosy are wrong.
(a + 5) ÷ 5 = \(\frac{a+5}{5}\)
= \(\frac{a}{5}\) + \(\frac{5}{5}\)
= \(\left(\frac{a}{5}+1\right)\)
∴ Srikar had done it correctly.

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

Andhra Pradesh AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం Textbook Exercise Questions and Answers.

AP State Syllabus 4th Class Maths Solutions Chapter 1 గుర్తుకు తెచ్చుకుందాం

Textbook Page No. 1

1. హర్షిత తన నాయనమ్మ నాగమ్మతో కలిసి బొమ్మల దుకాణానికి వెళ్ళింది. బొమ్మలపై రాసి ఉన్న ధరలను పరిశీలిస్తున్నది. మీరు కూడా వాటి పై ఉన్న ధరలను పరిశీలించండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 1
ఈ క్రింది ప్రశ్నలకు సమాధానాలు రాయండి.

ప్రశ్న 1.
మీరెప్పుడైనా బొమ్మల దుకాణానికి వెళ్ళారా?
జవాబు:
అవును, నేను బొమ్మల దుకాణానికి వెళ్ళాను.

ప్రశ్న 2.
బొమ్మల దుకాణంలో మీరేమేమి చూశారు ?
జవాబు:
నేను బొమ్మల దుకాణంలో రకరకాల బొమ్మలను చూశాను.

ప్రశ్న 3.
పటంలో ఎన్ని కారు బొమ్మలు ఉన్నాయి ?
జవాబు:
పటంలో 6 కారు బొమ్మలు ఉన్నాయి.

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 4.
తెల్ల టెడ్డీ బొమ్మ ధర ఎంత ?
జవాబు:
తెల్ల టెడ్డీ బొమ్మ ధర ₹200.

ప్రశ్న 5.
ఆకుపచ్చ కారు ఎంత ?
జవాబు:
ఆకుపచ్చ కారు ధర ₹150.

అభ్యాసం – 1.0

1. కింది వానికి విస్తరణ రూపం రాయండి.

అ) 8
జవాబు:
ఎనిమిది

ఆ) 20
జవాబు:
ఇరవై

ఇ) 35
జవాబు:
ముప్పై ఐదు

ఈ) 46
జవాబు:
వలభై ఆరు

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఉ) 100
జవాబు:
వంద

ఊ) 101
జవాబు:
నూట ఒకటి

ఋ) 150
జవాబు:
నూట యాభై

బ) 200
జవాబు:
రెండు వందలు

ఎ) 375
జవాబు:
మూడు వందల డెబ్బై ఐదు

ఏ) 425
జవాబు:
నాలుగు వందల ఇరవై ఐదు

ఐ) 802
జవాబు:
ఎనిమిది వందల రెండు

ఒ) 892
జవాబు:
ఎనిమిది వందల తొంభై రెండు

ఓ) 956
జవాబు:
తొమ్మిది వందల యాభై ఆరు

2. క్రింది వానిని సంఖ్యా రూపంలో రాయండి.

అ) ఆరు
జవాబు:
6

ఆ) పద్దెనిమిది
జవాబు:
18

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ) యాభై రెండు
జవాబు:
52

ఈ) డెబ్బై ఐదు
జవాబు:
75

ఉ) నాలుగు వందల డెబ్బై.
జవాబు:
470

ఊ) ఆరువందల నాలుగు
జవాబు:
604

ఋ) ఎనిమిది వందల ఒకటి
జవాబు:
801

ఋ) రెండు వందల ఇరవై రెండు
జవాబు:
222

3. కింది సంఖ్యలలోని గీత గీయబడిన అంకెల స్థానం, స్థాన విలువలు రాయండి.

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 2
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 3

4. కింది వానికి విస్తరణ రూపం రాయండి.

అ) 56
జవాబు:
50 + 6

ఆ) 62
జవాబు:
60 + 2

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ) 83
జవాబు:
80 + 3

ఈ) 87
జవాబు:
80 + 7

ఉ) 95
జవాబు:
90 + 5

ఊ) 110
జవాబు:
100 + 10 + 0

ఋ) 175
జవాబు:
100 + 70 + 5.

బూ) 325
జవాబు:
300 + 20 + 5

ఎ) 1,450
జవాబు:
1000 + 400 + 50+ 0

ఏ) 3752
జవాబు:
3,000 + 700+ 50 + 2

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఐ) 5,927
జవాబు:
5,000 + 900 + 20 + 7

5. కింది వానికి సంక్షిప్త రూపం రాయండి.

అ) 20+5
జవాబు:
25

ఆ) 40 + 7
జవాబు:
47

ఇ) 80 + 2
జవాబు:
82

ఈ) 300 + 20
జవాబు:
320

ఉ) 600 + 40 + 8
జవాబు:
648

ఊ) 900 + 90 +9
జవాబు:
999

ఋ) 3000 + 400 + 20 + 5
జవాబు:
3,425

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

బూ) 5000 + 20 + 7
జవాబు:
5,027

అభ్యాసం – 1.1

1. కింది వానిని కూడండి.

అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 4
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 5

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 6
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 7

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 8
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 9

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఈ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 10
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 11

ఉ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 12
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 13

ఊ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 14
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 15

2. ఈ క్రింది కూడికలను చేయండి

అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 16
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 17

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 18
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 19

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 20
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 21

ఈ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 22
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 23

ఉ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 24
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 25

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఊ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 26
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 27

3. ఈ కింది ఇవ్వబడిన ఖాళీలలో సరియైన సంఖ్యను రాయండి.

అ) 526 + 326 + 94 = ………………
జవాబు:
526 + 326 + 94 =        946       

ఆ) 829 + 408 = …………….. + 829
జవాబు:
829 + 408 =       408         + 829

ఇ) ……………….. + 396 = 396
జవాబు:
     0         + 396 = 396

4. ఈ కింది సంఖ్యలను సమీప పదులకు సవరించి రాయండి.

అ) 56
జవాబు:
56కి సమీప పదులలో 60

ఆ) 79
జవాబు:
79కి సమీప పదులలో 80

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ) 42
జవాబు:
42కి సమీప పదులలో 40

ఈ) 91
జవాబు:
91కి సమీప పదులలో 90

ఉ) 28
జవాబు:
28కి సమీప పదులలో 30

5. ఈ కింది సంఖ్యలను సమీప వందలకు సవరించి రాయండి.

అ) 235
జవాబు:
235 కి సమీప వందలలో 200

ఆ) 374
జవాబు:
374 కి సమీప వందలలో 400

ఇ) 929
జవాబు:
929 కి సమీప వందలలో 900

ఈ)562
జవాబు:
562 కి సమీప వందలలో 600

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఉ) 810
జవాబు:
810 కి సమీప వందలలో 800

ప్రశ్న 6.
ఒక తోటలో 235 మామిడి, 652 జమ మరియు 120 కొబ్బరి చెట్లు కలవు. తోటలోని మొత్తం చెట్లెన్ని?
జవాబు:
తోటలో మామిడి చెట్లు సంఖ్య = 235
తోటలో జామ చెట్లు సంఖ్య = 652
తోటలో కొబ్బరి చెట్లు సంఖ్య = 120
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 28
తోటలో మొత్తం చెట్లు సంఖ్య = 1007

ప్రశ్న 7.
ఒక పాఠశాలలో బాలికల సంఖ్య బాలుర సంఖ్య కన్నా 92 ఎక్కువ. బాలికలు 358 మంది అయిన పాఠశాలలోని మొత్తం విద్యార్థుల సంఖ్య ఎంత ?
జవాబు:
పాఠశాలలో బాలికల సంఖ్య = 358
బాలికల సంఖ్య బాలుర సంఖ్య కన్నా 92 ఎక్కువ
∴ బాలుర సంఖ్య = 358 – 92 = 266
∴ పాఠశాలలోని మొత్తం మొత్తం విద్యార్థుల సంఖ్య = 358 + 266
= 624 మంది

Textbook Page No. 5

ప్రయత్నించండి

కింద ఇవ్వబడిన ఖాళీలలో సరియైన సంఖ్యలను రాయండి.

అ) 5+3 = 3 + _____________
జవాబు:
5+3 = 3 +       5       

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఆ) 82 + 40 = __________ + 82
జవాబు:
82 + 40 =       40       + 82

ఇ) _______________ + 596 = 596
జవాబు:
      0          + 596 = 596

అభ్యాసం – 1.2

1. ఈ కింది తీసివేతలను చేయండి.

అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 29
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 30

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 31
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 32

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 33
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 34

ఈ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 35
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 36

ఉ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 37
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 38

ఊ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 39
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 40

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఋ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 41
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 42

బూ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 43
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 44

2. తీసివేయండి.

అ) 62 నుంచి 59
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 45

ఆ) 92 నుంచి 86.
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 46

ఇ) 536 నుంచి 192
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 47

ఈ) 928 నుంచి 485
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 48

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 3.
205 మరియు 62 ల బేధం ఎంత ?
జవాబు:
205 మరియు 62 ల బేధం =
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 49

ప్రశ్న 4.
653 నుండి ఎంత తీసివేసిన 268 వస్తుంది?
జవాబు:
653 కు268 కి గల బేధము 385
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 50
∴ 653 నుండి 268 తీసిన 385 వచ్చును.

ప్రశ్న 5.
246కు ఎంత కలిపిన 859 వస్తుంది ?
జవాబు:
859 మరియు 246 మధ్య బేధము = 613
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 51
∴ 246 కు 613 ను కలిపిన 859 వచ్చును.

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 6.
రెండు సంఖ్యల మొత్తం 453. వాటిలో ఒక సంఖ్య 285 అయిన రెండవ సంఖ్య ఎంత ?
జవాబు:
ఒక సంఖ్య = 285
రెండు సంఖ్యల మొత్తం = 453
453, 285 ల బేధం
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 52
∴ రెండవ సంఖ్య 168.

ప్రశ్న 7.
రెండు సంఖ్యల భేదం 568. వాటిలో ఒక సంఖ్య 796 అయిన రెండవ సంఖ్య ఎంత?
జవాబు:
ఒక సంఖ్య 796
రెండు సంఖ్యల భేదం = 568.
∴ రెండవ సంఖ్య = 796 – 568
= 228

అభ్యాసం – 1.3

1. ఈ క్రింది గుణకారాలను చేయండి.

అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 53
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 54

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 55
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 56

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 57
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 58

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఈ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 59
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 60

ఉ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 61
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 62

ఊ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 63
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 64

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఋ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 65
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 66

2. ఈ కింది లబ్దాలను కనుగొనండి:

అ) 395 × 7 = __________
జవాబు:
2765

ఆ) 402 × 9 = ___________
జవాబు:
3618

ఇ) 534 × 4 = ____________
జవాబు:
2136

ఈ) 826 × 5 = ___________
జవాబు:
4130

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఉ) 498 × 0 = ___________
జవాబు:
0

ఊ) 0 × 35 = ___________
జవాబు:
0

ప్రశ్న 3.
ఒక పెన్ను ₹ 25. అటువంటి 9 పెన్నుల ధర ఎంత?
జవాబు:
ఒక పెన్ను ధర = ₹ 25
9 పెన్నుల ధర = 9 × ₹ 25 = ₹ 225

ప్రశ్న 4.
ఒక మామిడి పండ్ల బుట్ట బరువు 36 కి.గ్రా. అటువంటి 10 మామిడి పండ్ల బుట్టల బరువు ఎంత?
జవాబు:
మామిడి పండ్ల బుట్ట బరువు = 36 కి.గ్రా.
10 మామిడి పండ్ల బుట్టల బరువు
= 10 × 36 కి.గ్రా.
= 360 కి.గ్రా.

ప్రశ్న 5.
ఒక బియ్యం బస్తా బరువు 24 కి.గ్రా. 478 బస్తాల బియ్యం బరువెంత ?
జవాబు:
ఒక బియ్యం బస్తా బరువు = 24 కి.గ్రా.
478 బియ్యం బస్తాల బరువు = 478 × 24
= 11,742 కి.గ్రా.

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 6.
ఒక సంఖ్య మరియు 5ల లబ్దం ‘0’. ఆ సంఖ్యను కనుక్కోండి.
జవాబు:
5 మరియు ‘0’ల లబ్దం = 5 × 0 = 0

అభ్యాసం – 1.4

1. ఇవి చేయండి :

అ) 6 ÷ 2
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 67

ఆ) 8 ÷ 4
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 68

ఇ) 9 ÷ 3
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 69

ఈ) 24 ÷ 6
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 70

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఉ) 45 ÷ 3
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 71

ఊ) 96 ÷ 4
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 72

ఋ) 224 ÷ 7
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 73

బూ) 845 ÷ 8
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 74

2.

అ) 40 ÷ 4 = ?
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 75

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఆ) 60 ÷ 10 = ?
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 76

ప్రశ్న 3.
90లో ఎన్ని 9లు ఉన్నాయి ?
జవాబు:
90 ÷ 9 = 10 .
∴ 90 లో 9 లు 10 ఉన్నాయి.

4. ఈ కింది భాగహారాల నుంచి భాగ ఫలాలను కనుక్కోండి.

అ) 69 ÷ 3
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 77
∴ భాగఫలం = 23

ఆ) 76 ÷ 4
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 78
∴ భాగఫలం = 19

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ) 96 ÷ 2
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 79
∴ భాగఫలం = 48

ఈ) 846 ÷ 3
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 80
∴ భాగఫలం = 282

ఉ) 925 ÷ 5
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 81
∴ భాగఫలం = 185

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 5.
ఈ క్రింది పట్టికను పూరించండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 82
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 83

ప్రశ్న 6.
57 సెం.మీ పొడవు గల రిబ్బనను ఎన్ని 3 సెం.మీ. ముక్కలుగా కత్తిరించవచ్చు ?
జవాబు:
మొత్తం పొడవు గల రిబ్బన్ = 57 సెం.మీ.
ఒక్కొక్క రిబ్బన్ ముక్క పొడవు = 3 సెం.మీ.
కత్తిరించిన ముక్కల సంఖ్య = 57 ÷ 3
= 19 సెం.మీ.

ప్రశ్న 7.
ఒక వ్యక్తి 12 చాక్లెట్లను 4 గురు పిల్లలకు సమానంగా పంచిన, ఒకొక్క పిల్లవాడికి ఎన్నెన్ని చాక్లెట్లు వస్తాయి?
జవాబు:
మొత్తం చాక్లెట్లు = 12
మొత్తం పిల్లల సంఖ్య = 4
ఒక్కొక్క పిల్లవానికి పంచిన చాక్లెట్లు
= 12 ÷ 4
= 3 చాక్లెట్లు

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 8.
91 రోజులలో వారాలు ఎన్ని ?
జవాబు:
మొత్తం రోజులు = 91
వారంలో రోజుల సంఖ్య = 7
91 రోజులలో ఉన్న వారాల సంఖ్య
= 91 ÷ 7
= 13 వారాలు

అభ్యాసం – 1.5

ప్రశ్న 1.
ఎక్కువ బరువు కలిగిన వస్తువుకు సున్న చుట్టండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 84
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 85

ప్రశ్న 2.
ఈ కింది వాహనాలను వాటి బరువులను బట్టి ఆరోహణ క్రమంలో అమర్చండి.
అ) సైకిలు
ఆ) బస్సు
ఇ) మోటారు సైకిల్
ఈ) కారు
జవాబు:
సైకిల్ < మోటార్ సైకిలు < కారు < బస్సు

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 3.
సరైన కొలత ప్రమాణం (కిలోగ్రాం లేక, గ్రాంలలో) రాయండి.
జవాబు:
బియ్యం బస్తా – కిలోగ్రాం లలో
రబ్బరు – గ్రాంలలో
పుస్తకాల సంచి  – కిలోగ్రాంలో

అభ్యాసం – 1.6

ప్రశ్న 1.
ఈ కింది వానిలో ఏవి మీటర్లలో మరియు ఏవి ‘సెంటీమీటర్లలో కొలుస్తారో గుర్తించండి.
అ) మీ తరగతి గది నల్లబల్ల పొడవు
ఆ) పెన్సిలు పొడవు
ఇ) జెండా స్తంభం పొడవు
ఈ)నీ చేతివేలు పొడవు
జవాబు:
అ) మీ తరగతి గది నల్లబల్ల పొడవు – మీటర్లు
ఆ) పెన్సిలు పొడవు – సెంటీమీటర్లు
ఇ) జెండా స్తంభం పొడవు – మీటర్లు
ఈ)నీ చేతివేలు పొడవు – సెంటీమీటర్లు

ప్రశ్న 2.
ఈ కింది పొడవులను ఆరోహణ క్రమంలో అమర్చండి.
a) 8మీ.
b) 10 సెం.మీ.
c) 5మీ.
d) 20 సెం.మీ.
జవాబు:
8 మీ. > 5 మీ. > 20 సెం.మీ. > 10 సెం.మీ.

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 3.
ఏవైనా మూడు వస్తువులు మీటర్లలో మరియు మూడు వస్తువులు సెంటీ మీటర్లలో కొలిచేవి రాయండి.
జవాబు:
మీటర్లలో

  1. మంచం
  2. కిటికీలు
  3. గ్యాస్ పొయ్యి

సెంటీమీటర్లలో

  1. దువ్వేన
  2. పెన్సిల్
  3. రబ్బరు

అభ్యాసం – 1.7

ప్రశ్న 1.
లీటర్లలో కొలిచే కొన్ని పదార్థాలమ రాయండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 86
జవాబు:
పాలు, నీరు మరియు నూనెలను లీటర్లలలో కొలుస్తారు.

ప్రశ్న 2.
ఎక్కువ పరిమాణం గల వస్తువులకు టిక్ పెట్టండి.
మగ్గు ( )
బక్కెట్టు ( )
నీళ్ళసీసా ( )
నీళ్ళ ట్యాంకు ( )
జవాబు:
మగ్గు ( )
బక్కెట్టు ( )
నీళ్ళసీసా ( )
నీళ్ళ ట్యాంకు (✓)

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 3.
ఈ కింది వాటిని లీటర్లలో సుమారుగా అంచనా వేసి, చెప్పండి.
అ) ఒక రోజుకు ఒక వ్యక్తి త్రాగే నీరు
ఆ) ఒకసారి ఒక వ్యక్తి స్నానానికి కావలసిన నీరు
ఇ) దంతధావనానికి కావలసిన నీరు
ఈ) ఒక మొక్కకు పోయడానికి కావలసిన నీరు
జవాబు:
అ) 5 లీటర్లు
ఆ) 24 లీటర్లు
ఇ) 2 లీటర్లు
ఈ) 7 లీటర్లు

అభ్యాసం – 1.8

1. గడియారాలలో చూపబడిన సమయాన్ని చదివి రాయండి.

అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 87
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 88

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 89
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 90

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 91
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 92

ఈ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 93
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 94

2. ఈ కింది సమయాలను గడియారాలలో సూచించండి.

అ) 9 : 45
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 95

ఆ) 1 : 15
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 96

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ) 6 : 30
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 97

ఈ) 11 : 20
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 98

Textbook Page No. 11

ఇవి చేయండి.

కింది పట్టికను పూర్తిచేయండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 99
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 100

అభ్యాసం – 1.9

1. ఈ కింది ఆకారాల పేర్లను కింద ఇవ్వబడిన పెట్టెలలో రాయండి. 

అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 101
జవాబు:
త్రిభుజం

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 102
జవాబు:
చతురస్రం

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 103
జవాబు:
దీర్ఘచతురస్రం

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఈ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 104
జవాబు:
వృతం

ప్రశ్న 2.
ఈ కింది పట్టికను పూరించండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 105
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 106

ప్రశ్న 3.
ఈ కింది పట్టికను పూరించండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 107
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 108

అభ్యాసం – 1.10

1. ఈ కింది సంఖ్యలకు గణన చిహ్నాలు రాయండి.

అ) 4 = __________
జవాబు:
||||

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఆ) 3 = ___________
జవాబు:
|||

ఇ) 1 = ____________
జవాబు:
|

ఈ) 2 = ____________
జవాబు:
||

2. ఈ కింది గణన చిహ్నాలకు సంఖ్యలు రాయండి.

a) || = ________
జవాబు:
3

b) | = ________
జవాబు:
1

c) || = ________
జవాబు:
2

d) |||| = ________
జవాబు:4

ప్రశ్న 3.
ఈ కింది పట్టికను పూరించండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 109

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 110

Textbook Page No. 13

ఇవి చేయండి

సమాన భాగాలుగా విభజించబడిన పటాలను గుర్తించండి.
అ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 111
జవాబు:

ఆ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 112
జవాబు:

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ఇ)
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 113
జవాబు:

అభ్యాసం – 1.11

ప్రశ్న 1.
సమభాగాలుగా విభజించబడిన పటాలను టిక్ (✓) చేయండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 114
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 115

ప్రశ్న 2.
ఈ కింది బొమ్మలలో సగ (1/2) భాగాన్ని షేడ్ చేయండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 116
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 117

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 3.
ఈ కింది బొమ్మలలో పావు (1/4) భాగాన్ని షేడ్ చేయండి.
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 118
జవాబు:
AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం 119

బహుళైచ్ఛిక ప్రశ్నలు

ప్రశ్న 1.
524 యొక్క సంఖ్యా నామం
A) ఐదువందల ఇరవై నాలుగు
B) ఐదు వందల ఇరవై
C) ఐదు వందల నాలుగు
D) ఏదీకాదు
జవాబు:
A) ఐదువందల ఇరవై నాలుగు

ప్రశ్న 2.
ఏడువందల ఇరవై నాలుగు యొక్క సంఖ్యాగుర్తును కనుగొనుము.
A) 720
B) 724
C) 742
D) 702
జవాబు:
B) 724

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 3.
352 సంఖ్యలో 3 యొక్క స్థాన విలువ
A) 30
B) 3000
C) 300
D) 3
జవాబు:
C) 300

ప్రశ్న 4.
5000 + 30 + 8 యొక్క సంక్షిప్త రూపం
A) 5038
B) 538
C) 5308
D) ఏదీకాదు
జవాబు:
A) 5038

ప్రశ్న 5.
845ను దగ్గరి వందలకు రాయగా
A) 900
B) 800
C) 850
D) ఏదీకాదు
జవాబు:
B) 800

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 6.
960 మరియు 456 ల భేదం ?
A) 504
B) 540
C) 489
D) 450
జవాబు:
A) 504

ప్రశ్న 7.
ఒక పెన్సిల్ ఖరీదు – 6 అయితే 27 పెన్సిళ్ళు ధర ఎంత ?
A) 162
B) 621
C) 261
D) ఏదీకాదు
జవాబు:
A) 162

ప్రశ్న 8.
64 లో ఎన్ని 8లు ఉన్నాయి ?
A) 8
B) 16
C) A
D) 6
జవాబు:
A) 8

ప్రశ్న 9.
జెండా స్తంభం యొక్క పొడవును దేనితో కొలుస్తారు?
A) గ్రాముల్లో
B) కిలోగ్రాముల్లో
C) మీటర్లు
D) సెంటీమీటర్లు
జవాబు:
C) మీటర్లు

AP Board 4th Class Maths Solutions 1st Lesson గుర్తుకు తెచ్చుకుందాం

ప్రశ్న 10.
1 లీటరు = ………….. ml.
A) 10
B) 1000
C) 100
D) ఏదీకాదు
జవాబు:
B) 1000

ప్రశ్న 11.
1 గంట = __________ నిమిషాలు
A) 06
B) 100
C) 10
D) 60
జవాబు:
D) 60

ప్రశ్న 12.
త్రిభుజానికి ఎన్ని శీర్షాలు ఉంటాయి
A) 4
B) 1
C) 2
D) 3
జవాబు:
D) 3