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SCERT AP 7th Class Science Study Material Pdf 2nd Lesson Nature of Substances Textbook Questions and Answers.

AP State Syllabus 7th Class Science 2nd Lesson Questions and Answers Nature of Substances

7th Class Science 2nd Lesson Nature of Substances Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Taste of an acid is ______.
2. The pH of a substance is 0. It indicates that the substance is ______ in nature.
3. Blue litmus paper turns ______ colour in tamarind solution.
4. The nature of an antacid is ______
5. Acid + base → ______ + ______
Answer:
1. sour
2. acidic
3. red
4. basic
5. salt + water

II. Choose the correct answer.

1. Colour of Turmeric solution in acids
a) Blue
b) Red
c) Purple
d) No change
Answer:
d) No change

2. Example to an acid.
a) Vinegar
b) Baking soda
c) Caustic soda
d) None
Answer:
a) Vinegar

3. Main component of soap is
a) An Acid
b) A base
c) a & b
d) None
Answer:
b) A base

4. If you add baking soda to lemon juice ______ gas will release.
a) Hydrogen
b) Oxygen
c) Carbon Dioxide
d) Sulphur Dioxide

5. To treat acidic nature of the soil, farmers add ______ to his agricultural field.
a) lemon juice
b) calcium oxide
c) sodium chloride
d) sulphur dioxide
Answer:
b) calcium oxide

III. Matching

Answer:

IV. Answer the following questions.

Question 1.
Write differences between acids and bases.
Answer:

Question 2.
Give examples to different types of acid and base indicators.
Answer:
Substances which are used to test acids or bases are called acid base indicators. There are many types of indicators, some of them are

1. Natural indicators
2. Synthetic indicators
3. Olfactory indicators
4. Universal indicators.

1) Natural indicators:
Indicators obtained from natural sources are called as Natural indicators. Ex: litmus, turmeric, hibiscus, red cabbage etc.

2) Synthetic indicators:
An indicator prepared from artificial sources is known as Syn-thetic indicator. We use many synthetic indicators like methyl orange, phenolphtha-lein, in our laboratory.

3) Olfactory indicators:
Substances which change their smell when mixed with acid or base are known as olfactory indicators. Ex: onion, vanilla and clove oil.

4) Universal indicators:
It is a mixture of different indicators. It shows different colours in different solutions, “universal indicator” contains thymol blue, methyl red, bromothymol blue and phenolphthalein.

Question 3.
One substance is slippery to touch and bitter in taste, another substance is sour in taste. If you add both, which substances will form?
Answer:

1. Substance that is slippery to touch and bitter in taste is Base.
2. Substance that is sour in taste is Acid.
3. When acids react with base neutralisation reaction takes place.
4. Salt and water will form in neutralisation reaction.
   Acid + Base → Salt + Water

Question 4.
Guess, How do you test acetic acid, if there is no availability of indicators?
Answer:

1. Acetic acid is a colour less liquid with pungent smell and sour taste.
2. It is used in preservation of pickles and other food substances as it is sour.
3. So, It can be identify by testing its smell and taste.

Question 5.
Anitha’s mother was filled mango pickle in a ceramic jar and stored, so many doubts were arisen her mind. What would be those doubts?
Answer:

1. Why do pickles store only in ceramic jars?
2. What materials are used in making ceramic jars?
3. What will happen if pickles are stored in metal containers?
4. Sometimes pickles spoil quickly. Why?
5. Do all pickles contain oil?
6. Why is it not suggestable to use aluminium spoons for pickles?

Question 6.
To conduct a neutralization reaction, what materials are required?
Answer:
Materials required to conduct a neutralization reaction are dilute hydrochloric acid (any acid), Sodium hydroxide solution (any base), Phenolphthalein indicator, Conical flask, Dropper.

Question 7.
Draw a neat diagram of arrangement of apparatus to conduct an activity that acid reacts with metal and liberates hydrogen gas.
Answer:

Question 8.
How do you appreciate the role of bases in helping of the persons suffering from acidity problems?
Answer:

1. Our stomach produces gastric juice with hydrochloric acid.
2. It helps us in digestion of food.
3. But sometimes secretion of excess acid causes acidity or indigestion.
4. It leaves a burning sensation and pain in the stomach.
5. Antacids help us to get relief from acidity.
6. Antacids contain bases, eg: aluminum hydroxide, milk of magnesia.
7. The bases in the antacids neutralize gastric juice and give us relief.
8. Thus bases do a great help to the persons suffering from acidity problems.

Question 9.
What measures do you follow to prevent acid rains?
Answer:
Air pollution is a major cause of acid rains. So acid rains can be prevented by controlling the air pollution. To control air pollution

1. Reduce the usage of personal vehicles.
2. Reduce the usage of fossil fuels.
3. Encourage the usage of alternative sources of energy.
4. Use electric vehicles.
5. Use solar energy.
6. Minimize use of fire and fire products.
7. Treat and purify the industrial emissions before letting in to air.
8. Plantation and reforestation should be done as much as possible.

7th Class Science 2nd Lesson Nature of Substances InText Questions and Answers

7th Class Science Textbook Page No. 20

Question 1.
Name any other substance which tastes like tamarind?
Answer:
Lemon juice is another substance which has sour taste like tamarind.

Question 2.
Why are tamarind and lemon juice sour in taste?
Answer:
Substances like tamarind and lemon contain some substances which give sour taste. These type of substances with sour taste are known as acids.

7th Class Science Textbook Page No. 21

Question 3.
What do you observe when an acid is poured on the bathroom floor?
Answer:
Generally, hydrochloric acid is used for bathroom cleaning. It liberates thick fumes and pungent smell when poured on the floor.

Question 4.
Have you seen batteries being used in vehicles? Which chemical is used in batteries?
Answer:
Sulphuric acid is used in batteries.

Question 5.
Have you ever drunk a soda or a cool drink?
Answer:
Yes. It contain carbonic acids.

Question 6.
Which substances do you use to wash your hands and clean your teeth?
Answer:
We use soap to wash our hands and for bathing. We use tooth paste to clean our teeth.

Question 7.
What are the chemicals present in soap and tooth paste?
Answer:
Soap and tooth paste contain bases.

7th Class Science Textbook Page No. 22

Question 8.
See the main components on the tooth paste tube. What chemicals names do you find?
Answer:
Calcium carbonate, aluminium hydroxide and sodium bicarbonate etc.

Question 9.
Is there any substance other than an acid and a base? Discuss with your Mends.
Answer:
The substance which is neither an acid nor a base is known as neutral substance. Pure water is a neutral substance.

Question 10.
List out some acids, bases and neutral substances available in your house.
Answer:
Acids:
Vinegar, lemon juice, tomato juice, bathroom cleaner (hydrochloric acid), cool drinks, curd, etc.

Bases :
Bath soap, cloth soap, baking soda, tooth paste, dish wash liquid etc.

Neutral: Water

7th Class Science Textbook Page No. 23

Question 11.
Is it possible to test all the acids and bases simply by tasting or touching them?
Answer:

1. Not possible.
2. Some acids like hydrochloric acid, sulphuric acid etc. are very harmful and strong acids. We cannot test them by touch or taste.

Question 12.
In such cases, how can we test them whether they are acids or bases?
Answer:
By using acid base indicators.

Question 13.
What are the different types of indicators?
Answer:
There are many types of indicators, some of them are

1. Natural indicators
2. Synthetic indicators
3. Olfactory indicators
4. Universal indicators.

7th Class Science Textbook Page No. 24

Question 14.
Can we use any flower as an indicator?
Answer:
We can use hibiscus flowers as an indicator. We can also use Indigofera tinctoria (Neeli chettu) flowers as indicator.

7th Class Science Textbook Page No. 26

Question 15.
Why it is so? What makes the hydrochloric acid dangerous than vinegar?
Answer:

1. Hydrochloric is an strong acid whereas vinegar is a weak acid.
2. pH of hydrochloric is very less than that of vinegar.

Question 16.
How can we measure the strength of an acid or a base?
Answer:
Strength of acid or base solutions is measured in pH scale.

7th Class Science Textbook Page No. 27

Question 17.
What happens when a metal piece is dropped in an acid?
Answer:
If you drop a metal piece in an acid it reacts with acid and releases hydrogen gas.

7th Class Science Textbook Page No. 28

Question 18.
Pickles are not stored in aluminium or steel or copper vessels. Why?
Answer:

1. Pickles contain acids.
2. These acids react with metal containers and releases toxic substances.
3. So, we should not store them in metal containers.
4. Generally, they are stored in ceramic or glass containers which are not react.

Question 19.
What will happen when metals are placed in bases?
Answer:
Bases like sodium hydroxide react with metals and release hydrogen gas.

7th Class Science Textbook Page No. 29

Question 20.
What will happen if we mix an acid and a base?
Answer:
If we add an acid to a base, they neutralize each other.

7th Class Science Textbook Page No. 30

Question 21.
How is acidity caused? What is the remedy for it?
Answer:

1. Excess secretion of gastric juice (hydrochloric acrd) in the stomach causes acidity problems.
2. Antacids help the patients to get relief from acidity.

Question 22.
What are antacids?
Answer:
Antacids are bases. They neutralize the gastric juice in the stomach.

7th Class Science Textbook Page No. 31

Question 23.
How can you help the farmers in finding out whether their agriculture fields are acidic or basic?
Answer:
We can help the farmers by testing the soil in their fields.

Activities and Projects

1. Prepare different greeting cards by using indicators.
Answer:

I had prepared a greeting card using turmeric indicator.

I took a white card board and cut it in the shape of a greeting card.

Applied turmeric paste on it and allowed to dry.

I draw designs and wrote wishes on it with soap water using a point brush.

It turns in to reddish brown.

Thus, I prepared greeting card using indicator.

Question 2.
Prepare beetroot indicator and test some acids and bases. Write a report.
Answer:
I have crushed beetroot and collected its juice.

This juice is filtered and used as an acid-base indicator.

When acids are added to this indicator there is no change in colour.

But when bases are added to this, it changed its colour from dark red to yellow. With this I concluded that the beetroot indicator is used to test the bases.

Test with beetroot indicator

Question 3.
Visit various agricultural fields and collect soils samples. Do the soil test. Make a report.
Answer:
I have visit nearest agricultural fields and collected soil from the fields. I took 10 g of soil from each sample in a beaker and added 500 ml of water to it. I stirred it well and filtered the solution. Then tested the filtered solution with a pH paper and tabulated the results. I gave suggestions to the farmers to neutralize their soils basing on the results.

Activities

Activity – 1

Question 1.
Taste some substances in our house and identify their nature through taste from your previous experiences. Fill the findings in the table.

Answer:

→ Which substances have sour taste in the above table? Why?
Answer:
Tomato juice, curd, raw mango, amla and orange juice have sour taste, because they contain acids. . .

Activity – 2

Question 2.
Take a soap into your palm and add some water. Now rub with another palm.
→ How do you feel while rubbing?
Answer:
You can feel the slippery nature of the soap.

Repeat this activity with tooth paste now. Observe your feeling.

Tooth paste is also slippery in nature.

These substances contain slippery chemicals they are called Bases. Bases are slip¬pery in nature and also bitter to taste.

Activity – 3

Question 3.
Take a table spoon of turmeric powder in a plate. Add little water and make it into a paste. Take a white paper and apply the paste over it on both sides and let it dry. After drying, cut the paper into strips. Now turmeric paper strips are ready for use as indicator.
→ What is the colour of these strips?
Answer:
They are yellow in colour.
Testing with turmeric indicator

1. Take soap solution in a plate. Dip one turmeric paper strip in the soap solution.
2. Take it out and observe the colour of the strip.
3. Repeat the activity with lime water and lemon juice.

→ Do you observe any change in colour of tumeric strips?
Note your observations in the table.
Answer:

The colour of turmeric strip will change into reddish brown in soap water and lime water, but its colour will not change in lemon juice.

Test with hibiscus indicator :
Take lemon juice, soda water, lime water, glucose solution, sugar solution, soap water etc.into test tubes. Add sufficient amount of hibiscus indicator in each test tube.Observe the change in the colour of the substance and record in the table.

7th Class Science Textbook Page No. 24

Question 4.
How do you prove that hibiscus could be used as an acid base indicator?
(OR)
How do you prepare a hibiscus indicator?
Answer:

1. Take some hibiscus (china rose or mandara) petals.
2. Take warm water into a beaker and place the petals in the beaker.
3. Keep it until the water has completely turned into violet colour.
4. Filter the solution with a strainer.
5. Now hibiscus indicator is ready to test.

Test with indicator:

1. Take lemon juice, soda water, lime water, glucose solution, sugar solution, soap water etc. into test tubes.
2. Add sufficient amount of hibiscus indicator in each test tube.
3. Observe the change in the colour of the substance and record in the table.

Conclusion :
In acids and bases the hibiscus indicator changes its colour. Hence, it can used as an acid base indicator.

Lab Activity – 1

Question 5.
Test the given solution with different indicators and prepare a report.
(OR)
How do you test different indicators change their colours in different solutions?
Answer:
1) Take the solutions a) Dilute hydrochloric acid b) Sodium hydroxide c) Acetic acid d) Salt solution e) Sugar solution f) Soap solution in test tubes.

2) Test them with a) Red litmus b) Blue litmus c) Methyl orange d) Phenolphthalein indicators.
Table

3) Blue litmus turns red in acids.
4) Red litmus turns blue in bases.
5) Methyl orange turns red in acids and yellow in bases.
6) Phenolphthalein turns pink in bases, but does not change its colour in acids.
7) In neutral substances all the above indicators do not change their colour.

Activity – 4

Question 6.
How do you find the strength and nature of the following substances?
1) Dilute hydrochloric acid
2) Vinegar
3) Water
4) Sodium hydroxide
5) Ammonium hydroxide.
Answer:
Take dilute hydrochloric acid, vinegar, water, sodium hydroxide and ammonium hydroxide in test tubes.;Now take universal indicator and add 2 drops in each test tube. You can observe that the solutions in the test tubes change into different colours. Now compare these colours with pH colour chart which was given on the indicator bottle.
Try to find pH values or the strengths Of acids and bases based on the colours they show and write them in the table.
Table

Lab Activity – 2

Question 7.
How do you prove that the acid reacts with metals and produces hydrogen gas?
Answer:
Aim: Acid Reacts with metals and produces hydrogen gas.

Materials required:
Conical flask, hydrochloric acid, zinc pieces, incense stick, match box.

Procedure:
Take 5g of zinc metal pieces in a conical flask. Pour 20 ml of dilute hydrochloric acid into it. Now, observe what happens. The zinc pieces reacted with the hydrochloric acid and releases a gas.

How can we identify this gas?
Test for hydrogen gas :
Now, introduce a incense stick into mouth of the conical flask. The flame of incense stick will put off with a pop sound. This is a test for hydrogen gas. Hydrochloric acid reacts with zinc metal and forms zinc chloride, releases hydrogen gas. This reaction can be written as a word equation,
Zinc+hydrochloric acid → zinc chloride+hydrogen

Conclusion:
Hence, we can conclude that acids react with metals and release hydrogen gas.

Activity – 5

Question 8.
How do you prove that bases react with metal and releases hydrogen gas ?
Answer:
Aim: Base reacts with metal and produces hydrogen gas.

Material required :
Conical flask, sodium hydroxide solution, zinc pieces, incense stick.

Procedure:

1. Take 5gr of zinc metal pieces in a conical flask,
2. Pour 20 ml of sodium hydroxide solution in it.
3. Heat the conical flask.

Observation :
The zinc pieces reacts with base and releases gas.

Test for hydrogen gas :

1. Now, introduce a incense stick into the mouth of the conical flask.
2. The flame of incense stick will put off with a pop sound.
3. This is a test for hydrogen gas.

Conclusion :
Bases react with metals and produce hydrogen gas.
Note : All bases do not react with metals.

Activity – 6

Question 9.
How do you prove that the carbon dioxide gas releases when an egg shell reacts with an acid?
(OR)
What happens when egg shells are dipped in an acid? How do you prove it?
Answer:
Take some crushed egg shells in a test tube and pour dilute Hydrochloric acid until the egg shells completely sink. Egg shell is made of calcium carbonate. Bring a burning match stick near the mouth of test tube.

The gas that put off the burning match stick is carbon dioxide. It was released by the action of acid with calcium carbonate.

Activity – 7

Question 10.
How do you perform a neutralization reaction in your lab?
Answer:

1. Take sodium hydroxide solution in a conical flask and observe its colour. Now add 2-3 drops of phenolphthalein indicator to it.
2. Now observe the colour. It is pink in colour.
3. Using a dropper add dilute hydrochloric acid drop by drop to this solution and stir gently.
4. Continue adding of the acid, till the pink colour disappears.
5. The solution now is no more basic as it was neutralized by acid.
6. We know that the colour of phenolphthalein indicator is pink in bases and colourless in acids and neutral substance.
   
7. The substances are salt and water because acids react with bases to form salts and water,
8. Neutralization is a chemical reaction where acid and base reacts with each other to forms salt and water.
9. Word equation of the above reaction is
   Hydrochloric acid + sodium hydroxide → Sodium chloride + water
10. Now add one or more drops of hydrochloric acid to the solution. The pink colour disappears, as it has again become acidic.
11. We can conclude that acids react with bases to produce salt and water.

Activity – 8

Question 11.
How do you test soil of agricultural field?
Answer:
Visit nearest agricultural fields and collect soil from the fields.

Take 10 g of soil in a beaker. Add 500 ml of water to it and stir well. Filter the solution. Now test the filtered solution with universal indicator or a pH paper and check your results.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.4

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.4 Textbook Questions and Answers

Question 1.
Find the slope of the line joining the two given points.
i) (4,-8) and (5,-2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+8}{5-4}\)
= 6

ii) (0, 0) and (√3, 3)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{\sqrt{3}-0}\)
= \(\frac{3}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\)
= √3

iii) (2a, 3b) and (a, -b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-b-3b}{a-2a}\)
= \(\frac{-4b}{-a}\)
= \(\frac{4b}{a}\)

iv) (a, 0) and (0, b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{b-0}{0-a}\)
= \(\frac{-b}{a}\)

v) A (-1.4, -3.7), B (-2.4, 1.3).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\)
= -5

vi) A (3, -2), B (-6, -2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+2}{-6-3}\)
= 0

vii) A (-3\(\frac{1}{2}\), 3), B (-7, 2\(\frac{1}{2}\)).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{2 \frac{1}{2}-3}{-7+3 \frac{1}{2}}\)
= \(\frac{-\frac{1}{2}}{-3 \frac{1}{2}}\)
= \(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1}{7}\)

viii) A(0, 4), B(4, 0)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-4}{4-0}\)
= \(\frac{-4}{4}\)
= -1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.5

10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers

Question 1.
For each geometric progression find the common ratio ‘r’, and then find a_n.
i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
Answer:
Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….

ii) 2, -6, 18, -54, …….
Answer:
Given G.P. = 2, -6, 18, -54, …….
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-6}{2}\) = -3
a_n = a . r^n-1 = 2 × (-3)^n-1
∴ r = -3; a_n = 2(-3)^n-1

iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-3}{-1}\) = 3
a_n = a . r^n-1 = (-1) × 3^n-1
∴ r = 3; a_n = (-1) × 3^n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
Given G.P. = 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)
a_n = a . r^n-1 = 5 × \(\left(\frac{2}{5}\right)^{n-1}\)
∴ r = \(\frac{2}{5}\); a_n = 5\(\left(\frac{2}{5}\right)^{n-1}\)

Question 2.
Find the 10^th and nth term of G.P.: 5, 25, 125,…..
Answer:
Given G.P.: 5, 25, 125,…..
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{25}{5}\) = 5
a_n = a . r^n-1 = 5 × (5)^n-1 = 5^1+n-1 = 5ⁿ
a₁₀ = a . r⁹ = 5 × 5⁹ = 5¹⁰
∴ a₁₀ = 5¹⁰; a_n = 5ⁿ

Question 3.
Find the indicated term of each geometric progression.
i) a₁ = 9; r = \(\frac{1}{3}\); find a₇.
Answer:
a_n = a . r^n-1

ii) a₁ = -12; r = \(\frac{1}{3}\); find a₆.
Answer:
a_n = a . r^n-1

Question 4.
Which term of the G.P.
i) 2, 8, 32,….. is 512?
Answer:
Given G.P.: 2, 8, 32,….. is 512
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{8}{2}\) = 4
Let the n^th term of G.P. be 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5^th term of the given G.P.

ii) √3, 3, 3√3, …….. is 729?
Answer:
Given G.P.: √3, 3, 3√3, …….. is 729
a = √3, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{3}{\sqrt{3}}\) = √3
now a_n = a . r^n-1 = 729
⇒ (√3)(√3)^n-1 = 729
⇒ (√3)ⁿ = 3⁶ = (√3)¹²
⇒ n = 12
So 12th term of GP √3, 3, 3√3, …….. is 729.

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)?
Answer:
Given G.P.: \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)

Let \(\frac{1}{2187}\) be the n^th term of the G.P., then
a_n = a . r^n-1 = \(\frac{1}{2187}\)

[∵ bases are equal, exponents are also equal]
7^th term of G.P is \(\frac{1}{2187}\).

Question 5.
Find the 12^th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a₈ = 192 and r = 2
a_n = a . r^n-1
a₈ = a . (2)^8-1 = 192

= 3 × 2¹⁰ = 3 × 1024 = 3072.

Question 6.
The 4^th term of a geometric progression is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.
Answer:
Given: In a G.P.

Now substituting r = \(\frac{2}{3}\) in equation (1)
we get,

Question 7.
If the geometric progressions 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),….. have their n^th term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……

Given that n^th terms are equal
a_n = a . r^n-1

⇒ 3^n-1+n-1 = 81 × 81
⇒ 3^2n-2 = 3⁴ × 3⁴
⇒ 3^2n-2 = 3⁸ [∵ a^m . aⁿ = a^m+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = \(\frac{10}{2}\) = 5
The 5^th terms of the two G.P.s are equal.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2

10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).

Answer:
i) There are no zeroes as the graph does not intersect the X – axis.
ii) The number of zeroes is one as the graph intersects the X – axis at one point only.
iii) The number of zeroes is three as the graph intersects the X – axis at three points.
iv) The number of zeroes is two as the graph intersects the X – axis at two points.
v) The number of zeroes is four as the graph intersects the X – axis at four points.
vi) The number of zeroes is three as the graph intersects the X – axis at three points.

Question 2.
Find the zeroes of the given polynomials,
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Answer:
i) Given p(x) = 3x
Let p(x) = 0
So, 3x = 0
x = \(\frac{0}{3}\) = 0,
Zeroes of p(x) = 3x is zero.
∴ No. of zeroes is one.

ii) Given p(x) = x² + 5x + 6 is a quadratic polynomial.
It has atmost two zeroes.
To find zeroes, let p(x) = 0
⇒ x² + 5x + 6 = 0
⇒ x² + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2 (x + 3) = 0
⇒ (x + 3) (x + 2) = 0
⇒ x + 3 = 0 or x + 2 = 0
⇒ x = -3 or x = -2
Therefore the zeroes of the polynomial are -3 and -2.

iii) Given p(x) = (x + 2) (x + 3)
It is a quadratic polynomial.
It has atmost two zeroes.
Let p(x) = 0
⇒ (x + 2) (x + 3) = 0
⇒ (x + 2) = 0 or (x + 3) = 0
⇒ x = -2 or x = -3
Therefore the zeroes of the polynomial are -2 and – 3.

iv) Given p(x) = x⁴ – 16 is a biquadratic polynomial. It has atmost two zeroes.
Let p(x) = 0
⇒ x⁴ – 16 = 0
⇒ (x²)² – 4² = 0
⇒ (x² – 4) (x² + 4) = 0
⇒ (x + 2) (x – 2) (x² + 4) = 0
⇒ (x + 2) = 0 or (x – 2) = 0 or (x² + 4) = 0
⇒ x = -2 (or) x = 2 (or) x² = -4
Therefore the zeroes of the polynomial are 2, – 2, we do not consider √-4 since it is not real.

Question 3.
Draw the graphs of the given polynomial and find the zeroes. Justify the answers,
i) p(x) = x² – x – 12
ii) p(x) = x² – 6x + 9
iii) p(x) = x² – 4x + 5
iv) p(x) = x² + 3x – 4
v) p(x) = x² – 1
Answer:
i) Given polynomial p(x) = x² – x – 12.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-3, 0) and (4, 0).
So, the zeroes of the polynomial are -3 and 4.
Justification:
Given p(x) = x² – x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4) (x + 3) = 0
⇒ x – 4 = 0 and x + 3 = 0
x = 4 and x = – 3

ii) Given polynomial p(x) = x² – 6x + 9
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (3, 0).
So, the zeroes of the given polynomial are same i.e., 3.
Justification:
Given p(x) = x² – 6x + 9
⇒ x² – 3x – 3x + 9 = 0
⇒ x(x – 3) – 3(x – 3) = 0
⇒ (x – 3) (x – 3) = 0
⇒ x – 3 = 0 and x – 3 = 0
x = 3 and x = 3

iii) Given polynomial p(x) = x² – 4x + 5
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph does not cut the X – axis at any point.
So, the quadratic polynomial p(x) has no zeroes.
Justification: For the given p(x) = x² – 4x + 5 not possible to split in factors.

iv) Given polynomial p(x) = x² + 3x – 4.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-4, 0) and (1, 0).
So, the zeroes of the polynomial are -4 and 1.
Justification:
Given p(x) = x² + 3x – 4 = 0
⇒ x² + 4x – x – 4 = 0
⇒ x(x + 4)- 1(x + 4) = 0
⇒ (x + 4) (x – 1) = 0
⇒ x + 4 = 0 and x – 1 = 0
x = – 4 and x = 1

v) Given polynomial p(x) = x² – 1
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-1, 0) and (1,0).
So, the zeroes of the polynomial are – 1 and 1.
Justification:
Given p(x) = x² – 1 = 0
⇒ p(x) = (x + 1) (x – 1) = 0 [∵ a² – b² = (a + b) (a – b)]
⇒ x + 1 = 0 and x – 1 = 0
x = -1 and x = 1

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1 ?
Answer:
Given polynomial p(x) = 4x² + 3x – 1
Given zeroes are \(\frac{1}{4}\) and -1
Let x = \(\frac{1}{4}\)

Let x = -1
⇒ p(-1) = 4(-1)² + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0
∴ P(\(\frac{1}{4}\)) = 0 and p(-1) = 0
So these values are zeroes of the polynomial p(x).

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1

10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers

Question 1.
A joker’s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:

Radius of the cap (r) = 7 cm
Height of the cap (h) = 24 cm
Slant height of the cap (l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{7^{2}+24^{2}}\)
= \(\sqrt{49+576}\)
= √625
= 25
∴ l = 25 cm.
Lateral surface area of the cap = Cone = πrl
L.S.A. = \(\frac{22}{7}\) × 7 × 25 = 550 cm².
∴ Area of the sheet required for 10 caps = 10 x 550 = 5500 cm².

Question 2.
A sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.
Answer:

Radius of the cylinder, r = 7 cm
Height of the cylinder, h = 35 cm
T.S.A. of the cylinder with lids at both ends = 2πr(r+h)
= 2 × \(\frac{22}{7}\) × 7 × (7 + 35)
= 2 × \(\frac{22}{7}\) × 7 × 42 = 1848 cm².
Area of thin paper required for 100 cylinders = 100 × 1848
= 184800 cm²
= \(\frac{184800}{100 \times 100}\) m²
= 18.48 m².

Question 3.
Find the volume of right circular cone with radius 6 cm. and height 7 cm.
Answer:

Base radius of the cone (r) = 6 cm.
Height of the cone (h) = 7 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7
= 264 c.c. (Cubic centimeters)
∴ Volume of the right circular cone = 264 c.c.

Question 4.
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.
Answer:
Base of cylinder and cone be the same.

CSA / LSA of cylinder = 2πrh
CSA of cone = πrl
The lateral surface area of a cylinder is equal to the curved surface area of cone.
∴ 2πrh = πrl
⇒ \(\frac{h}{l}=\frac{\pi r}{2 \pi r}\)
⇒ \(\frac{h}{l}\) = \(\frac{1}{2}\)
∴ h : l = 1 : 2

Question 5.
A self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm², then how many caps can be manufactured from that paper sheet?
Answer:

Radius of the cap (conical cap) (r) = 3 cm
Height of the cap (h) = 4 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
(by Pythagoras theorem)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
C.S.A. of the cap = πrl
= \(\frac{22}{7}\) × 3 × 5
≃ 47.14 cm²
Number of caps that can be made out of 1000 cm² = \(\frac{1000}{47.14}\) ≃ 21.27
∴ Number of caps = 21.

Question 6.
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.
Answer:
Given dimensions are:
Cone:
Radius = r
Height = h
Volume (V) = \(\frac{1}{3}\)πr²h

Cylinder:
Radius = r
Height = h
Volume (V) = πr²h

Ratio of volumes of cylinder and cone = πr²h : \(\frac{1}{3}\)πr²h
= 1 : \(\frac{1}{3}\)
= 3 : 1
Hence, their volumes are in the ratio = 3 : 1.

Question 7.
A solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?
Answer:

Diameter of the cylinder (d) = 7 cm
Radius of the base (r) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 11 cm
Volume of the cylinder V = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 11 = 423.5 cm³
∴ Total volume of 50 rods = 50 × 423.5 cm³ = 21175 cm³.

Question 8.
A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use π = 3.14)
Answer:

Diameter of the heap (conical) (d) = 12 cm
∴ Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Height of the cone (h) = 8 m
Volume of the cone, V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 8
= 301.71 m³.

Question 9.
The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height?
Answer:
C.S.A. of a cone = πrl = 4070 cm²
Diameter of the cone (d) = 70 cm
Radius of the cone = r = \(\frac{d}{2}\) = \(\frac{70}{2}\) = 35 cm
Let its slant height be ‘l’.
By problem,
πrl = 4070 cm²
\(\frac{22}{7}\) × 35 × l = 4070
110 l = 4070
l = \(\frac{4070}{110}\) = 37 cm
∴ Its slant height = 37 cm.

Andhra Pradesh AP Board 5th Class Telugu Solutions 2nd Lesson సాయం Textbook Exercise Questions and Answers.

AP State Syllabus 5th Class Telugu Solutions Chapter 2 సాయం

చిత్రం చూడండి. ఆలోచించి మాట్లాడండి.

ప్రశ్నలకు జవాబులు చెప్పండి.

ప్రశ్న 1.
చిత్రంలో ఏం జరుగుతున్నది?
జవాబు:
చిత్రంలో ఒక విద్యార్థిని తాతగారిని రోడ్డు దాటిస్తున్నది.

ప్రశ్న 2.
చిత్రంలో ఎవరెవరు ఉన్నారు? ఏమేం చేస్తున్నారు ?
జవాబు:
చిత్రంలో ఒక ఆటో డ్రైవరు ద్విచక్ర వాహనదారుడు, ఒక విద్యార్థిని ఒక పెద్దాయన ఉన్నారు. పెద్దాయన మరియు విద్యార్థిని రోడ్డు దాటుతున్నారు. వాహన దారులు అందుకు ఆగారు.

ప్రశ్న 3.
మీరెప్పుడైనా ఎవరికైనా సహాయం చేసారా ? అప్పుడు మీకేమనిపించింది.
జవాబు:
చేసాను. ఆ సమయంలో నాకు చాలా ఆనందం అనిపించింది. నా ఇంట్లో వాళ్ళు – స్నేహితులు నన్ను మెచ్చుకుంటారనిపించింది. అది నాకు చాలా గొప్ప విషయం అనిపించింది. తోటివారికి సహాయంచేసే అవకాశం నాకు కలిగింది కదా! అని సంతోషం కలిగింది. అందరూ నాగురించి గొప్పగా చెప్తుంటే – నా తల్లిదండ్రులు ఆనందించారు కదా! అనిపించింది.

ఇవి చేయండి

వినడం – ఆలోచించి మాట్లాడటం

ప్రశ్న 1.
రవికి పక్షులంటే ఇష్టంకదా ! మీకు ఏ ఏ పక్షులంటే ఇష్టం ?
జవాబు:
నాకు రామచిలుక, పావురం, నెమలి.

ప్రశ్న 2.
‘సాయం’ కథను సొంత మాటల్లో చెప్పండి.
జవాబు:
రవి మంచి బాలుడు. ఇతరులకు సాయంచేసే గుణం కలవాడు. రవికి పక్షులంటే చాలా ఇష్టం.

ఒకరోజు తన ఇంటిముందు ఒక చిత్రం జరిగింది. ఎక్కడినుండో ఒక పిచ్చుకల 7గుంపు ఎగురుకుంటూ వచ్చి రవి – ఇంటిముందు ఉన్న కరంటు తీగమీద వాలాయి. కొద్ది సేపటికి అన్నీ ఎగిరి పోయాయి. కాని అందులో ఒక పిచ్చుక కాలు తీగలో ఇరుక్కుపోవటంవల్ల ఎగరలేక అక్కడే ఉండిపోయింది. దాని బాధచూసి రవికి బాధ కలిగింది. అమ్మకు చెప్పాడు. మనం ఏం చేయగలం? అని వెళ్ళిపోయింది. నాన్నకు చెప్పాడు – ఎవరైనా కరంటాఫీసులో వాళ్ళకు చెప్తే వాళ్ళు కరెంటాపి, స్తంభం ఎక్కి పిచ్చుకను కాపాడతారు. అంతేకానీ మనమేంచేయగలం ? నువ్వు స్కూల్ కి వెళ్ళమని చెప్పాడు

రవి నేరుగా నాన్న చెప్పిన ప్రకారం చేసి – కరంటు వాళ్ళని పిలిచి ఆ పిచ్చుకను కాపాడాడు. ఆ పిచ్చుక ఆనందంగా గాలిలో ఎగిరిపోతుంటే ఎంతో ఆనందించాడు. (ఇతరులకు చేసిన సాయం ఎంతో ఆనందాన్నిస్తుంది.)

ప్రశ్న 3.
మీ స్నేహితులకు మీరు చేసిన ఏదైనా సాయం గురించి చెప్పండి.
జవాబు:
ఈ రోజు నేను పాఠశాలకొస్తూ దారి పక్కనే ఒక ముసలి అవ్వను చూసాను. కాళ్ళకు దెబ్బ తగిలిందేమో లేవలేకపోతోంది, చూడలేకపోతోంది. కళ్ళు కనపడవనుకుంటా. అవ్వ ఆకలి అరపు విని జాలి కలిగింది. అందరూ అవ్వను చూస్తూ వెళ్తున్నారే కాని సాయం చేయడం లేదు. నా జేబులో డబ్బులు ఉన్నాయి. నాన్న నన్ను కొనుక్కోమని ఇచ్చినవి.

అవి తీసి అవ్వచేతిలో పెట్టాను. తడుముకుంటూ అవి చూసుకుని అవ్వ చాలా ఆనందించింది. అవ్వ మొహంలో ఆనందంచూసి నాకు చాలా ఆనందం కలిగింది. ఇంతవరకూ ఎప్పుడూ కలగని ఆనందం కలిగింది. అప్పుడు అనుకున్నాను. ఇతరులకు చేసిన సాయం ఎంతో ఆనందాన్నిస్తుంది కదా ! అని.

పదజాలం

ఆ) ఈ కింది ఒత్తు ఉన్న పదాలను రాయండి.
క్క _____________
గ్గ _____________
చ్చ _____________
జ్జ _____________

ట్ట _____________
త్త _____________
ద్ద _____________
ప్ప _____________

మ్మ _____________
య్య _____________
ర్ర _____________
ల్ల _____________
జవాబు:
క్క అక్క
గ్గ మొగ్గ
చ్చ పచ్చిక, వెచ్చని
జ్జ ముజ్జగము

ట్ట బుట్ట
త్త అత్త
ద్ద గ్రద్ద
ప్ప చిప్ప

మ్మ అమ్మ
య్య అయ్య
ర్ర  కర్ర
ల్ల చిల్లర

ప్రాజెక్టు పని

మీ పాఠశాల గ్రంథాలయంలో ఇతరులకు సాయం చేసే అంశం ఉన్న కథలు సేకరించండి. మీ తరగతిగదిలో ప్రదర్శించండి.
జవాబు:
విద్యార్థి కృత్యం

సాధారణ చేద్దాం

గువ్వ కొరకు మేనుకోసి యా శిబిరాజు
వార్త విడువరాక కీర్తి కెక్కె
ఓగు నెంచబోవ రుషకారి నెంతురు.
ఈ విశ్వదాభిరామ వినుర వేమ !

భావం :
తన దేహమునే కోసి ఇచ్చి పావురాన్ని కాపాడి శిబి చక్రవర్తి కీర్తి పొందాడు. – చెడ్డవారిని ఎవరూ తలుచుకోరు. ఉపకారం చేసే వారిని అందరూ గుర్తుంచుకుంటారు.
– వేమన
జవాబు:
విద్యార్థులు పద్యాన్ని – భావాన్ని పద విభాగంతో చదవటం నేర్చుకుని భావయుక్తంగా ధారణ చేయాలి. అందుకు ఉపాధ్యాయులు సహకరించాలి. అందులోని నీతిని విషయాన్ని వంటపట్టించుకోవాలి.

కవి పరిచయం

కవి : జాక్ కోప్
కాలము : 1913 నుండి 1991 వరకు
విశేషాంశాలు : జాక్ కోప్ దక్షిణాఫ్రికా నవలా రచయిత. కథారచయిత. కవి సంపాదకుడు. ఈ పాఠం ఒక అనువాద కథ.

పదాలు – అర్థాలు

దృశ్యం = చూడదగినది
కష్టం = ఇబ్బంది
ఆత్రం = తొందర
అవధులు = హద్దులు
గుంపు = సమూహం
ఆసక్తి = అపేక్ష

చదువు – అర్థం చేసుకో – ఆనందించు. పరీక్షల కోసం కాదు.

అనకు కనకు వినకు

అది సబర్మతీ ఆశ్రమం. గాంధీజీ ఒక్కరే చిరుచాప మీద కూర్చొని ఉన్నారు. ఆయన అప్పుడే భగవద్గీత పారాయణం చేసి పుస్తకాన్ని ముందున్న చెక్క పెట్టెమీద పెట్టారు. నిన్న ! ఎవరో తనకు బహుమానంగా తెచ్చిన బొమ్మ వైపు దీక్షగా చూస్తున్నారు బాపూజీ. మహాదేవ దేశాయ్ లోపలికి వచ్చాడు.

తదేక ధ్యానంగా ప్రతిమవైపు చూస్తున్న బాపూజీని చూచాడు. ఎకాగ్రతతో ఉన్న గాంధీజీకి వెనుకనుంచి ఎవరో నవ్వుతున్నట్లు అనిపించింది. బాపూ వెనక్కు తిరిగి చూచారు. మహదేవ దేశాయ్ నిలబడి నవ్వుతూ కనిపించాడు.

“మహదేవ్ భాయీ! రా ! కూర్చో! ఎందుకు నవ్వుతున్నావు?” అన్నాడు గాంధీజీ. ! మిమ్మల్ని మీరు మరచి ఆ పెట్టెమీద బోమ్మనే చూస్తుంటే నాకునవ్వు వచ్చింది బాపూజీ!” అని ! దేశాయ్ మళ్లీ పకపకా నవ్వటం మొదలెట్టాడు. గాంధీజీ దేశాయ్ భుజం మీద చెయ్యి వేసి తట్టుతూ “అసలీ బొమ్మను చూశావా ? ఎంత గొప్పగా ఉన్నదో!” అన్నాడు.

“అవును మూడు కోతులూ చాలా అందంగా ఉన్నాయి” అన్నాడు దేశాయ్. “కోతులు కాదు, అవి నా గురువులు” అని, గాంధీజీ ఆ కోతుల బొమ్మను మహాదేవ దేశాయ్ కి చూపిస్తూ ! “ఇదుగో చూడు ఈ బొమ్మలో మూడు కోతులు కన్నిస్తున్నాయి కదా! అవి మూడు పనులు | చేస్తూ మనకు మూడు విధాలైన ఉపదేశాలు ఇస్తున్నాయి. “మొదటి బొమ్మ చూడు రెండు చేతులతో ! నోరుమూసుకొని కూర్చుంది. అంటే చెడు మాట్లాడవద్దు అని, రెండో బొమ్మ రెండు చేతులతో రెండు కళ్ళూ ! – మూసుకొని కూర్చుంది. అంటే చెడు చూడకు అని అర్థం” అన్నాడు గాంధీజీ.

గాంధీజీ మాటలు విని మహాదేవ దేశాయ్ ఆశ్చర్య చకితుడయ్యాడు.

“మూడో బొమ్మ వైపు చూచి ఏం చేస్తున్నదో చెప్పు” అన్నాడు మళ్లీ గాంధీజీ. “అది తన రెండు చేతులతో గట్టిగా రెండు చెవులు మూసుకుంది” అన్నాడు దేశాయ్.

“దానర్థం ఏమిటంటే చెడు మాటల్ని వినవద్దు అని అది మనల్ని హెచ్చరిస్తున్నది. మొత్తం మీద ఈ మూడు కోతులూ ‘చెడు అనకు, చెడు కనకు, చెడు వినకు” అని మానవులకు హితోపదేశం చేస్తున్నాయని నా ఉద్దేశం. ఏమంటావ్?” అన్నాడు.

“బాపూజీ! మీరు మట్టిలో నుంచి మాణిక్యాలు వెలికి తీయగలరు. అందుకనే మీరు మహాత్ములు. మీ భావాన్ని అందుకోలేక నవ్వినందుకు నన్ను మన్నించండి” అని బాపూజీ పాదాలు స్పృశించాడు మహాదేవ్జీ.

కవి పరిచయం

కవి : జంధ్యాల పాపయశాస్త్రి
కాలము : (4-08-1912 – 21-06-1992)
రచించినవి : తెలుగుబాల శతకం
ఖండకావ్యాలు: ఉదయశ్రీ, విజయశ్రీ, కరుణశ్రీ, అరుణ కిరణాలు
బిరుదు : కరుణశ్రీ

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.4

10th Class Maths 6th Lesson Progressions Ex 6.4 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved in the form a G.P.?
i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10% .
Answer:
Given: Sharmila’s yearly salary = Rs. 5,00,000.
Rate of annual increment = 10 %.

Here, a = a₁ = 5,00,000
a₂ = 5,00,000 × \(\frac{11}{10}\) = 5,50,000
a₃ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,05,000
a₄ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,65,000

Every term starting from the second can be obtained by multiplying its pre¬ceding term by a fixed number \(\frac{11}{10}\).
∴ r = common ratio = \(\frac{11}{10}\)
Hence the situation forms a G.P.

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Answer:
Given: Bricks needed for the bottom step = 100.
Each successive step needs 2 bricks less than the previous step.
∴ Second step from the bottom needs = 100 – 2 = 98 bricks.
Third step from the bottom needs = 98 – 2 = 96 bricks.
Fourth step from the bottom needs = 96 – 2 = 94 bricks.
Here the numbers are 100, 98, 96, 94, ….
Clearly this is an A.P. but not G.P.

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.

Answer:
Given: An equilateral triangle whose perimeter = 24 cm.
Side of the equilateral triangle = \(\frac{24}{3}\) = 8 cm.
[∵ All sides of equilateral are equal] ……. (1)
Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = \(\frac{8}{2}\) = 4 cm
[∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.]
Similarly, the side of third triangle = \(\frac{4}{2}\) = 2 cm
∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm,
a = 8

Thus each term starting from the second; can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).
∴ The situation forms a G.P.

Question 2.
Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.
i) a = 4 ; r = 3.
Answer:
The terms are a, ar, ar², ar³, ……..
∴ 4, 4 × 3, 4 × 3² , 4 × 3² , ……
⇒ 4, 12, 36, 108, ……

ii) a = √5 ; r = \(\frac{1}{5}\)
Answer:
The terms are a, ar, ar², ar³, ……..

iii) a = 81 ; r = –\(\frac{1}{3}\)
Answer:
The terms of a G.P are:
a, ar, ar², ar³, ……..

⇒ 81, -27, 9,

iv) a = \(\frac{1}{64}\); r = 2.
Answer:
Given: a = \(\frac{1}{64}\); r = 2.

∴ The G.P is \(\frac{1}{64}\), \(\frac{1}{32}\), \(\frac{1}{16}\), …….

Question 3.
Which of the following are G.P. ? If they are G.P, write three more terms,
i) 4, 8, 16, ……
Answer:
Given: 4, 8, 16, ……
where, a₁ = 4; a₂ = 8; a₃ = 16, ……
\(\frac{a_{2}}{a_{1}}=\frac{8}{4}=2\)
\(\frac{a_{3}}{a_{2}}=\frac{16}{8}=2\)
∴ r = \(\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}\) = 2
Hence 4, 8, 16, … is a G.P.
where a = 4 and r = 2
a₄ = a . r³ = 4 × 2³ = 4 × 8 = 32
a₅ = a . r⁴ = 4 × 2⁴ = 4 × 16 = 64
a₆ = a . r⁵ = 4 × 2⁵ = 4 × 32 = 128

ii) \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….
Answer:
Given: t₁ = \(\frac{1}{3}\), t₂ = –\(\frac{1}{6}\), t₃ = \(\frac{1}{12}\), ….
\(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….

Hence the ratio is common between any two successive terms.
∴ \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), ……. is G.P.
where a = \(\frac{1}{3}\) and r = –\(\frac{1}{2}\)

iii) 5, 55, 555, ……..
Answer:
Given: t₁ = 5, t₂ = 55, t₃ = 555, ….

∴ 5, 55, 555, …….. is not a G.P.

iv) -2, -6, -18, ……
Given: t₁ = -2, t₂ = -6, t₃ = -18

∴ -2, -6, -18, is a G.P.
where a = -2 and r = 3
a_n = a . r^n-1 =
a₄ = a . r³ = (-2) × 3³ = -2 × 27 = -54
a₅ = a . r⁴ = (-2) × 3⁴ = -2 × 81 = -162
a₆ = a . r⁵ = (-2) × 3⁵ = -2 × 243 = -486

v) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), …….
Answer:

i.e., \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), ….. is not a G.P.

vi) 3, -3², 3³, ……
Given: t₁ = 3, t₂ = -3², t₃ = 3³, ……


i.e., every term is obtained by multiplying its preceding term by a fixed number -3.
3, -3², 3³, …… forms a G.P,
where a = 3; r = -3
a_n = a . r^n-1
a₄ = a . r³ = 3 × (-3)³ = 3 × (-27) = -81
a₅ = a . r⁴ = 3 × (-3)⁴ = 3 × 81 = 243
a₆ = a . r⁵ = 3 × (-3)⁵ = 3 × (-243) = -729

vii) x, 1, \(\frac{1}{x}\), …….
Answer:
Given: t₁ = x, t₂ = 1, t₃ = \(\frac{1}{x}\), ……

Hence x, 1, \(\frac{1}{x}\), …. forms a G.P.
where a = x; r = \(\frac{1}{x}\)

viii) \(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\), …….
Answer:
Given: t₁ = \(\frac{1}{\sqrt{2}}\), t₂ = -2, t₃ = \(\frac{8}{\sqrt{2}}\), ……

ix) 0.4, 0.04, 0.004, ……..
Answer:
Given: t₁ = 0.4, t₂ = 0.04, t₃ = 0.004, ……

∴ 0.4, 0.04, 0.004, …….. forms a G.P.
where a = 0.4; r = \(\frac{1}{10}\)

Question 4.
Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
Answer:
Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6.
∴ r = \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) = \(\frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}\)
⇒ \(\frac{x+2}{x}\) = \(\frac{x+6}{x+2}\)
⇒(x + 2)² = x(x + 6)
⇒ x² + 4x + 4 = x² + 6x
⇒ 4x – 6x = – 4 = -2x = -4
∴ x = 2

Andhra Pradesh AP Board 5th Class English Solutions 2nd Lesson My Sweet Memories Textbook Exercise Questions and Answers.

AP State Syllabus 5th Class English Solutions Chapter 2 My Sweet Memories

Activity 1

Read the following text and answer the questions.

a) Visiting new places, meeting new people and knowing about different cultures is a part of holidays fine. Look at the pictures of some famous places that are popular among tourists. Match the pictures with their names. One has been done for you.

Answer:

b) Holidays help us to relax and refresh. They are always packed with fun, People love holidays for different reasons. Rank them according to your choice and order of importance.
I have holidays because I can

• have lots of fun   (   )
• spend a lot of time with my family.   (   )
• visit new places and meet new people.   (    )
• know more about different cultures. (    )
• relax and forget about studies. (   )

Answer:

• have lots of fun  ( 2 )
• spend a lot of time with my family.   ( 4 )
• visit new places and meet new people.  ( 1 )
• know more about different cultures.    ( 5 )
• relax and forget about studies.   ( 3 )

Comprehension :

Activity 2

I. Answer the following questions.

Question 1.
Name some of the activities of the village children during their vacation.
Answer:
The village children go to relatives’ places. They play in the fields. They swim in the ponds and canals. They meet new friends. They watch films on TV or in cinema halls.

Question 2.
Did Santhosh enjoy his morning walk? How do you know?
Answer:
Santhosh enjoyed his morning walk. We know this from sentences like ‘The cool breeze was pleasant’. The air was filled with sweet songs of birds.

Question 3.
How did Santhosh spend his time when he went to the canal?
Answer:
Santhosh played for some time in cool water. But he could not swim. He sat on a big smooth rock and watched his friends swimming like fish.

Question 4.
How did Santhosh know that the canal was so clean?
Answer:
Santhosh was able to see the pebbles and colourful fish in the canal. So, Santhosh knew that the canal was clean.

Question 5.
Why did Santhosh forget to watch television or play video / mobile games?
Answer:
The trip to the village was very enjoyable. So, Santhosh did not watch television or video / mobile games.

Question 6.
Is T.V. or video/mobile game the only way to pass time? Can we do something else?
Answer:
T.V. or Video / mobile game is not the only way to pass time. We can read books and play with friends. There are many other ways to pass the time.

II. Choose the correct answer and write the corresponding letters in the brackets:

Question 1.
Santhosh went to __________.
(a) Ramapuram
(b) Krishnapuram
(c) Rangapuram
Answer:
(a) Ramapuram

Question 2.
The paddy fields were __________ in colour.
(a) white
(b) green
(c) blue
Answer:
(b) green

Question 3.
____________ helped his father in milking the cows.
(a) Peter
(b) Sruthi
(c) Aman
Answer:
(c) Aman

Question 4.
The girls were playing near a ___________ tree.
(a) banyan
(b) coconut
(c) neem
Answer:
(a) banyan

Question 5.
Santhosh returned home on _________.
(a) April 24th
(b) April 2 8th
(c) June 11^th
Answer:
(b) April 2 8th

Vocabulary

Activity 3

Let’s revisit the following sentences from the reading text.

1. I got a seat near the window.
   I have tasted the first mango of the season.

The opposite words for the underlined words are far and last respectively. Such words which give the opposite meanings are called antonyms. Here are some antonyms for you to understand.

less × more
fresh × stale
beautiful × ugly
hard × soft
thick × thin
narrow × broad

Rewrite the following sentences by changing the underlined word with its opposite. One has been done for you.
Example: Our journey started today.
Our journey ended today.

Question 1.
The weather was pleasant in the morning.
Answer:
The weather was unpleasant in the morning.

Question 2.
The river was so clean.
Answer:
The river was so unclean.

Question 3.
I sat on a smooth rock.
Answer:
I sat on a rough rock.

Question 4.
The breakfast was delicious.
Answer:
The breakfast was unpalatable.

Question 5.
The mango was tasty.
Answer:
The mango was tasteless.

Activity 4

Santhosh saw many things in Ramapuram. Write the words associated with the village. One has been done for you.

Answer:

Grammar

Activity 5

Read the following sentences carefully.

“There is no one at home. And nothing is there to do. It is very boring.
Let’s go out and do something.” Pallavi said to Santhosh.

In the above sentences, the under lined words i.e., ‘no one’, ‘nothing’ and ‘something’ do not refer to any specific person or thing. Such words are called Indefinite Pronouns.

These pronouns are used to talk about pcople, places or things without saying exactly who, where, or what they are.

Here is a list of indefinite pronouns referring to people, places and things.

Now complete the sentences, choosing one of the words.

Question 1.
I lost my watch. I’ve looked for it ____________ (everywhere / anywhere / nowhere)
Answer:
everywhere

Question 2.
____________ broke the window, (nothing/everywhere/someone)
Answer:
Someone

Question 3.
It’s a secret. Don’t tell ______________ (anyone /something/no one)
Answer:
anyone

Question 4.
I’m looking for my glasses. I can’t find them _____________ (everything/anywhere / something)
Answer:
anywhere

Question 5.
The problem is very difficult ___________ knows the answer. (something / anywhere / Nobody)
Answer:
Nobody

Activity 6

Let’s revisit the following sentences from the lesson.

• Our journey started today.
• Pallavi showed me how to play with a tyre and a stick.

In the above sentences, the underlined words are the past forms of the verbs ‘start’ and ‘show’ respectively. The above actions really took place and the speaker expressed the same using the above past form of the verbs. But sometimes, we need to express certain actions that did not take place.

e.g. It rained yesterday.
It did not rain yesterday.
The negative of the verb form rained is written as did not rain
i.e. did not rain = did + not + rain or didn’t + rain
(in contracted form)
Here are some more examples.

I. Now change the following into their negatives.

Answer:

II. Pallavi has a list of things to complete by last Monday. But today is Thursday. She completed some (✓) and did not complete some (✗). Read the list and write sentences in the space given below. One has been done for you.

Answer:

Writing

Activity 7

Read a sample diary entry of Santhosh on his birthday.

Tuesday, 10 December, 2019.

Dear Diary,
Today is my birthday! I woke up at around six and had breakfast with my family My parents hugged and gave me the largest bar of chocolate I have ever seen! They also gave me a new dress, a football and a board game. My parents had invited my friends to my birthday party in the evening. I received a potted plant, a storybook, and a set of colour pencils as gifts. My mother had baked a strawberry cake that tasted delicious. We also had biryani and ice cream. I enjoyed every moment of the day. I think this is the best birthday I have ever had!

Santhosh

Now write your diary entry describing how you celebrated Deepawali. Use the hints given below.
(Deepawali – house decoration – new clothes – Lakshmi pooja- colourful lights – rangoli patterns – burning crackers – enjoyed)
Answer:

16 June, 2020.

Dear Diary,

Today is Deepawali. We celebrated this festival very well. It is to celebrate the death of Narakasura. Deepavali follows Naraka Chaturdasi. It is believed that Sri Krishna killed Narakasura. The day on which Narakasura died is called Naraka Chaturdasi.

In our home, we performed Lakshmi Pooja. And later, we burst fire works flower pots, pencils and sparklers. The fireworks made the festival unforgettable.

Activity 8

Paragraph writing on a given topic:

Look at the following paragraph.

In Delhi, a balloon race is held every year. On the day of the race, one can see a large number of huge and colourful balloons. They are floating in the air above the houses and the tall buildings with people standing in the baskets or gondolas, as they are called, looking down upon us. That’s really a wonderful – experience one could see.

A paragraph is a group of sentences that tells us about a single idea.

A paragraph consists of :

• topic sentence that tells the reader the main idea of a paragraph.
  (In Delhi, ………….. every year.)
• Supporting details that give details about the main idea.
  (On the day …………….. upon us.)
• A concluding sentence that brings the paragraph to a clear end.
  (That’s really …………… see.)

Now write a paragraph about ‘How you spent an interesting weekend’.

I spent an interesting weekend in Delhi. I visited all important places in Delhi. The places include Red Fort, Jantar Mantar, Parliament House etc. I went to Agra to see the Taj Mahal. That was an unforgettable experience. From there, we went to Fateh Pur Sikri. I enjoyed the trip very much.

Write your sentences under the correct headings.
Answer:
Topic : I spent an interesting weekend in Delhi. I visited all important places in Delhi.

Supporting details: The places include Red Fort, Jantar Mantar, Parliament House etc. I went to Agra to see the Taj Mahal. That was an unforgettable experience.

Concluding sentence: From there, we went to Fateh Pur Sikri, I enjoyed the trip very much.

Listening and Responding

Activity 9

We use ‘may‘ to say that something is possible.

1. It may rain today. (It is possible that it will rain today.)
2. She may come tomorrow. (There is a possibility of her coming tomorrow)
3. The train may be late, (There is a possibility of train coming late. )
   
4. I may go to Delhi next week. (There is a possibility of my going to Deihi next week.)
5. He may get good marks. (There is a possibility of his getting good marks.)

Now work in pairs and role play the following conversation. Give possible answer for each question. You can ask some more questions.

Answer:

Sing and Enjoy

Poem

Written in March

The cock is crowing.
The stream is flowing.
The small birds twitter,
The lake doth glitter
The green- field sleeps in the sun;

The oldest and youngest Are at work with the strongest;
The cattle are grazing.
There’s joy in the mountains;
There’s life in the fountains;
Small clouds are sailing,
Blue sky prevailing;
The rain is over and gone!
– William Wordsworth.

Reading & Responding

Written in March

Summary :

The cock is crowing. The stream is flowing. The small birds twitter. The lake glitters. The green field sleeps in the sun. All are working. The cattle are grazing. Mountains are joyful. Fountains arelively. Small clouds are sailing. The sky is blue. The rain is over and gone.

సారాంశం

కోడిపుంజు కూస్తోంది. సెలయేరు ప్రవహిస్తోంది. చిన్న చిన్న పక్షులు కిచ కిచ లాడుతున్నాయి. సెలయేరు తళ తళ మెరుస్తోంది. ఆకుపచ్చని పొలం సూర్యుడి కాంతిలో నిద్రిస్తోంది. చిన్నా, పెద్దా అందరూ పనిచేస్తున్నారు. పశువులు మేత మేస్తున్నాయి. పర్వతాలు ఆనందిస్తున్నాయి. ఫౌంటెన్లలో జీవం తొణికిసలాడుతోంది. చిన్న చిన్న మేఘాలు ఆకాశంలో పయనిస్తున్నాయి. ఆకాశం అంతా నీలంగా ఉంది. వర్షం (వర్షకాలం) అయిపోయింది.

Glossary

crowing = the characterestic loud cry of a cock కోడిపుంజు అరుపు
stream = a small, narrow river: సెలయేరు
twitter = a short sound made by birds repeatedly: పిట్టల కూత
cattle = A group of animals with horing and cloven hoofs: పశువులు
prevailing = existing at a particular time; ఒకానొక సమయంలో ఉండు
doth (old English) = does/do

Activity 10.

I. Find the rhyming words rom the lesson and add a few more.

Question 1.
crowing, flowing, __________, __________
Answer:
sailing, prevailing,

Question 2.
twitter, ____________, ____________, ____________, ____________
Answer:
glitter, litter, batter, matter, splitter

Question 3.
youngest, ____________, ____________, ____________, ____________
Answer:
strongest, congest, largest, longest

Question 4.
mountain, ____________, ____________, ____________, ____________
fountain, maintain, certain, plantain

Question 5.
sailing, ____________, ____________, ____________, ____________
Answer:
boiling, failing, mailing, scaling

II. Answer the following questions.

Question 1.
What are the cattle doing?
Answer:
The cattle are grazing.

Question 2.
What glitters in the poem?
Answer:
The lake glitters in the poem.

Question 3.
Who work along with the strongest?
Answer:
The oldest and the youngest work with the strongest.

Question 4.
How does the poet describe the nature in this poem?
Answer:
The poet describes the nature and its action through different creatures and things.

Question 5.
Describe mountains and fountains in your own words.
Answer:
Mountains are very big. They are a natural land form. They rise above the surrounding land in a limited area. Mountains are steeper than hills. Most mountains occur in huge Mountain ranges. Fountains are a piece of architecture that pours water into a basin. It is mainly decorative in nature. Some times they are used to supply drinking water also. They are used to decorate city parks and squares. Now-a-days musical fountains are an attraction.

Project Work

Activity 11

Which country would you like to visit when you growup ? Collect the details and write about the country.
1. Map of that country, its national flag.
2. Stamps of the country, its currency.
3. Some famous monuments.
Answer:
I would like to visit the United States of America. When I grow up. United States of America is a federal republic. It consists of 50 states. It is the third largest nation by area. The capital of USA is Washington DC. English and Spanish are main languages spoken there. It is considered one of the richest countries in the world.

The National Flag of the USA : The national flag of the USA consists of white stars on a blue canton with a field of 13 alternating stripes, the fifty stars stand for 50 states. The 13 stripes stand for the original 13 states. The width length ratio of the flag is 10 to 19.

Some famous monuments : We can see many historical monuments through out the United States. Such monuments represent important events and significant people in American history.

Arlington National cemetery is a military cemetary of US. It is located in Virginia. Lincoln memorial is located in Washington D.C. It commemorates the great president of the United States, Abraham Lincoln. Washington monument is an Iconic structure built to honour the first President of the United States. Mount Rush more national memorial represents four huge sculptures of the heads of four inferential US presidents. The sculptures are carved on the granite face of mount rushmore.

Riddles

Question 1.
They come out at night without being called, and are lost in the day without being stolen, what are they ?

Answer:
Stars.

Question 2.
When I am alive I do not speak. Anyone who wants to take me captive cuts off my head. Then bites my bare body. I do not harm anyone unless they cut me first. Then I soon, make them cry. Who am I?

Answer:

Summary :

My Sweet Memories

Santhosh is a school going boy. His vacation started. He went to his grand parents’ village in the vacation. He wrote his experiences in his diary.

Santhosh packed his clothes and toys. He went to his grand parents’ village by bus. Santhosh sat near the window. There were different trees with flowers on bothsides of the road. There were many trees around the grand parents’ house. The mango tree was full of mangoes. He tasted the first mango.

Next day, he went out for a walk in the morning. The farmers were working in the fields. The air was filled with sweet songs of birds. Then, he played with a tyre and a stick, He met pallavi and her friends.

Santhosh went to the canal for a bath. Some children were climbing trees. Some were playing. He watched some boys swimming. He liked the” scenery.

He came back to his home in city. He enjoyed the trip very much. He never watched television. He never played video/mobile games. He liked the trip and waited for the next night to the village.

సారాంశం

సంతోష్ స్కూల్లో చదువుకునే ఒక విద్యార్థి. అతని సెలవు ప్రారంభం అయ్యాయి. సెలవుల్లో వాళ్ల తాతగారి ఊరు వెళ్ళాడు. సెలవుల్లో తన అనుభవాలని డైరీలో రాశాడు.

సంతోష్ తన బట్టలు, బొమ్మలు సర్దుకున్నాడు. బస్సెక్కి తన తాతగారి ఊరికి బయలుదేరాడు. బస్సులో కిటికీకి దగ్గరగా కూర్చొన్నాడు. రోడ్డుకి ఇరువైపులా రకరకాల చెట్లు పూలతో నిండి ఉన్నాయి. తాతయ్యగారి ఇంటిచుట్టూతా బోలెడు మామిడిపళ్లు ఉన్నాయి. అతడు ఆ సంవత్సరం మొట్టమొదటి మామిడి కాయ రుచి చూశాడు. అది ఎంత రుచిగా ఉందో!

తరువాత రోజు ఉదయాన్నే లేచి కొంతదూరం నడిచాడు. రైతులు పొలాల్లో పని చేయడం చూశాడు. పక్షులు పాడే తియ్యనైన పాటలతో అక్కడ గాలి నిండి ఉంది. ఆ రోజు సైకిల్ టైరు, కర్ర ముక్కతో ఎలా ఆడుకోవచ్చో నేర్చుకున్నాడు. పల్లవి అనే ఒక బాలికని, ఆమె స్నేహితులని కలిశాడు.

సంతోష్ స్నానం చెయ్యడానికి కొలవదగ్గరకి వెళ్లాడు. అక్కడ, కొంత మంది పిల్లలు చెట్లు ఎక్కుతున్నారు. కొంతమంది ఆడుతున్నారు. అతడు కొంతమంది బాలురు ఈత కొడుతూ ఉండడం చూశాడు. అతనికి ఆ దృశ్యం ఎంతో నచ్చింది.
అతడు నగరానికి తిరిగి వచ్చాడు. ఆ పర్యటన అతడికెంతో నచ్చింది. అతడు టెలివిజన్ చూడలేదు. వీడియో, మొబైల్ గేమ్స్ ఆడలేదు. అతడు ఆ పర్యటనని ఎంతో ఇష్టపడ్డాడు. తరువాత పర్యటన ఎప్పుడెప్పుడా అని ఎదురు చూడసాగాడు.

Glossary

vacation = holidays ; సెలవలు
experiences = observation ; అనుభవం
excited = thrilled, happy ; ఆనందించు
numerous = many ; చాలా
lush = attractive : ఆకర్షణీయమైన
scarecrows = human like objects ; దిష్టిబొమ్మలు
delighted = happy ; ఆనందంతో
Fruits : jackfruit ; పనస
pomegranate = దానిమ్మ

laden = weighed down with load ; బరువెక్కిన
juicy = full of juice ; రసముతో నిండిన
breeze = a gentle wind ; చిరుగాలి
swaying = move to cause to move slowly ; మెల్లగా కదులు, కదుల్చు
threshing = separating grain ; గింజలను వేరుచేయు
enchanting = attractive ; ఆకర్షణీయమైన
delicious = tasty ; రుచికరమైన
splashing = sound made by striking something into water ; నీటిలోకి ఏదైనా విసరడంద్వారా వచ్చే చప్పుడు
pebbles = small smooth round stones ; గులకరాళ్లు

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.3

10th Class Maths 8th Lesson Similar Triangles Ex 8.3 Textbook Questions and Answers

Question 1.
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Answer:

Let △PQR is a right angled triangle, ∠Q = 90°
Let PQ = a, QR b and
PR = hypotenuse = c
Then from Pythagoras theorem we can
say a² + b² = c² ……… (1)
△PSR is an equilateral triangle drawn on hypotenuse
∴ PR = PS = RS = c,
Then area of triangle on hypotenuse
= \(\frac{\sqrt{3}}{4}\)c² ……… (2)
△QRU is an equilateral triangle drawn on the side ‘QR’ = b
∴ QR = RU = QU = b
Then area of equilateral triangle drawn on the side = \(\frac{\sqrt{3}}{4}\)b² …….. (3)
△PQT is an equilateral triangle drawn on another side ‘PQ’ = a
∴ PQ = PT = QT = a
Area of an equilateral triangle drawn an another side ‘PQ’ = \(\frac{\sqrt{3}}{4}\)a² …….. (4)
Now sum of areas of equilateral triangles on the other two sides

= Area of equilateral triangle on the hypotenuse.
Hence Proved.

Question 2.
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
Answer:

Let PQRS is square whose side is ‘a’ units then PQ = QR = RS = SP = ‘a’ units.
Then the diagonal
\(\overline{\mathrm{PR}}\) = \(\sqrt{a^{2}+a^{2}}\) = a√2 units.
Let △PRT is an equilateral triangle, then PR = RT = PT = a√2
∴ Area of equilateral triangle constructed on diagonal

Let △QRZ is another equilateral triangle whose sides are
\(\overline{\mathrm{QR}}\) = \(\overline{\mathrm{RZ}}\) = \(\overline{\mathrm{QZ}}\) = ‘a’ units
Then the area of equilateral triangle constructed on one side of square = \(\frac{\sqrt{3}}{4}\)a² ……. (2)
∴ \(\frac{1}{2}\) of area of equilateral triangle on diagonal = \(\frac{1}{2}\left(\frac{\sqrt{3}}{2} a^{2}\right)\) = \(\frac{\sqrt{3}}{4}\)a² = area of equilateral triangle on the side of square.
Hence Proved.

Question 3.
D, E, F are midpoints of sides BC, CA, AB of △ABC. Find the ratio of areas of △DEF and △ABC.
Answer:

Given in △ABC D, E, F are the midpoints of the sides BC, CA and AB.
In △ABC, EF is the line join of mid-points of two sides AB and AC of △ABC.
Thus FE || BC [∵ \(\frac{AF}{FB}\) = \(\frac{AE}{EC}\) Converse of B.P.T.]
Similarly DE divides AC and BC in the same ratio, i.e., DE || AB.
Now in □ BDEF, both pairs of opposite sides (BD || EF and DE || BF) are parallel.
Hence □ BDEF is a parallelogram where DF is a diagonal.
∴ △BDF ≃ △DEF ……… (1)
Similarly we can prove that
△DEF ≃ △CDE ……… (2)
[∵ CDFE is a parallelogram]
Also, △DEF ≃ △AEF …….. (3)
[∵ □ AEDF is a parallelogram]
From (1), (2) and (3)
△AEF ≃ △DEF ≃ BDF ≃ △CDE
Also, △ABC = △AEF + △DEF + △BDF + △CDE = 4 . △DEF
Hence, △ABC : △DEF = 4 : 1.

Question 4.
In △ABC, XY || AC and XY divides the triangle into two parts of equal area. Find the ratio of \(\frac{AX}{XB}\).
Answer:

Given: In △ABC, XY || AC.
XY divides △ABC into two points of equal area.
In △ABC, △XBY
∠B = ∠B
∠A = ∠X
∠C = ∠Y
[∵ XY || AC; (∠A, ∠X) and ∠C, ∠Y are the pairs of corresponding angles]
Thus △ABC ~ △XBY by A.A.A similarity condition.
Hence \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XBY}}=\frac{\mathrm{AB}^{2}}{\mathrm{XB}^{2}}\)
[∵ The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

⇒ \(\frac{AX}{XB}\) + 1 = √2
⇒ \(\frac{AX}{XB}\) = √2 – 1
Hence the ratio \(\frac{AX}{XB}\) = \(\frac{√2 – 1}{1}\).

Question 5.
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Answer:

Given: △ABC ~ △XYZ
R.T.P: \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XYZ}}=\frac{\mathrm{AD}^{2}}{\mathrm{XW}^{2}}\)
Proof : We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Hence the ratio of areas of two similar triangles is equal to the squares of ratio of their corresponding medians.

Question 6.
△ABC ~ △DEF. BC = 3 cm, EF = 4 cm and area of △ABC = 54 cm². Determine the area of △DEF.
Answer:

Given: △ABC ~ △DEF
BC = 3 cm
EF = 4 cm
△ABC = 54 cm²
∴ △ABC ~ △DEF, we have
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
[∵ The ratio of two similar triangles is equal to the ratio of the squares of the corresponding sides].
\(\frac{54}{\Delta \mathrm{DEF}}\) = \(\frac{3^{2}}{4^{2}}\)
∴ △DEF = \(\frac{54 \times 16}{9}\) = 96 cm²

Question 7.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm and BP = 3 cm, AQ =1.5 cm, CQ = 4.5 cm. Prove that area of △APQ = \(\frac{1}{16}\) (area of △ABC).
Answer:

Given: △ABC and \(\overline{\mathrm{PQ}}\) – a line segment meeting AB in P and AC in Q.
AP = 1 cm; AQ =1.5 cm;
BP = 3 cm; CQ = 4.5 cm.
\(\frac{AP}{PQ}\) = \(\frac{1}{3}\) ……… (1)
\(\frac{AQ}{QC}\) = \(\frac{1.5}{4.5}\) = \(\frac{1}{3}\) ……..(2)
From (1) and (2),
\(\frac{AP}{BP}\) = \(\frac{AQ}{CQ}\)
[i.e., PQ divides AB and AC in the same ratio – By converse of Basic pro-portionality theorem]
Hence, PQ || BC.
Now in △APQ and △ABC
∠A = ∠A (Common)
∠P = ∠B [∵ Corresponding angles for the parallel lines PQ and BC]
∠Q = ∠C
∴ △APQ ~ △ABC [∵ A.A.A similarity condition]
Now, \(\frac{\Delta \mathrm{APQ}}{\Delta \mathrm{ABC}}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}\)
[∵ Ratio of two similar triangles is equal to the ratio of the squares of their corresponding sides].
= \(\frac{1^{2}}{(3+1)^{2}}\) = \(\frac{1}{16}\) [∵ AB = AP + BP = 1 + 3 = 4 cm]
∴ △APQ = \(\frac{1}{16}\) (area of △ABC) [Q.E.D]

Question 8.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

Answer:
Given: △ABC ~ △DEF
△ABC = 81 cm²
△DEF = 49 cm²
AX = 4.5 cm
To find: DY
We know that,
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{AX}^{2}}{\mathrm{DY}^{2}}\)
[∵ Ratio of areas of two similar triangles is equal to ratio of the squares of their corresponding altitudes]

∴ DY = 3.5 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.3

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers

Question 1.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding: (use π = 3.14)
i) Minor segment ii) Major segment
Answer:

Angle subtended by the chord = 90° Radius of the circle = 10 cm
Area of the minor segment = Area of the sector POQ – Area of △POQ
Area of the sector = \(\frac{x}{360}\) × πr²
\(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segment = 78.5 – 50 = 28.5 cm²
Area of the major segment = Area of the circle – Area of the minor segment
= 3.14 × 10 × 10 – 28.5
= 314 – 28.5 cm²
= 285.5 cm²

Question 2.
A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle.
(use π = 3.14 and √3 = 1.732)
Answer:
Radius of the circle r = 12 cm.
Area of the sector = \(\frac{x}{360}\) × πr²
Here, x = 120°

\(\frac{120}{360}\) × 3.14 × 12 × 12 = 150.72
Drop a perpendicular from ‘O’ to the chord PQ.
△OPM = △OQM [∵ OP = OQ ∠P = ∠Q; angles opp. to equal sides OP & OQ; ∠OMP = ∠OMQ by A.A.S]
∴ △OPQ = △OPM + △OQM = 2 . △OPM
Area of △OPM = \(\frac{1}{2}\) × PM × OM

= 18 × 1.732 = 31.176 cm
∴ △OPQ = 2 × 31.176 = 62.352 cm²
∴ Area of the minor segment
= (Area of the sector) – (Area of the △OPQ)
= 150.72 – 62.352 = 88.368 cm²

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (use π = \(\frac{22}{7}\))
Answer:
Angle made by the each blade = 115°
Total area swept by two blades
= Area of the sector with radius 25 cm and angle 115°+ 115° = 230°
= Area of the sector = \(\frac{x}{360}\) × πr²
= \(\frac{230}{360}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96
≃  1255 cm²

Question 4.
Find the area of the shaded region in figure, where ABCD is a square of side 10 cm. and semicircles are drawn with each side of the square as diameter (use π = 3.14).

Answer:
Let us mark the four unshaded regions as I, II, III and IV.

Area of I + Area of II
= Area of ABCD – Areas of two semicircles with radius 5 cm
= 10 × 10 – 2 × \(\frac{1}{2}\) × π × 5²
= 100 – 3.14 × 25
= 100 – 78.5 = 21.5 cm²
Similarly, Area of II + Area of IV = 21.5 cm²
So, area of the shaded region = Area of ABCD – Area of unshaded region
= 100 – 2 × 21.5 = 100 – 43 = 57 cm²

Question 5.
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles. (use π = \(\frac{22}{7}\))

Answer:
Given,
ABCD is a square of side 7 cm.
Area of the shaded region = Area of ABCD – Area of two semicircles with radius \(\frac{7}{2}\) = 3.5 cm
APD and BPC are semicircles.
= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5
= 49 – 38.5
= 10.5 cm²
∴ Area of shaded region = 10.5 cm

Question 6.
In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm., find the area of the shaded region, (use π = \(\frac{22}{7}\)).
Answer:
Given, OACB is a quadrant of a Circle.
Radius = 3.5 cm; OD = 2 cm.
Area of the shaded region = Area of the sector – Area of △BOD

= 9.625 – 3.5 = 6.125 cm²
∴ Area of shaded region = 6.125 cm².

Question 7.
AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 30°, find the area of the shaded region. (use π = \(\frac{22}{7}\)).

Answer:
Given, AB and CD are the arcs of two concentric circles.
Radii of circles = 21 cm and 7 cm and ∠AOB = 30°
We know that,
Area of the sector = \(\frac{x}{360}\) × πr²
Area of the shaded region = Area of the OAB – Area of OCD

∴ Area of shaded region = 102.66 cm²

Question 8.
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each, {use π = 3.14)
Answer:
Mark two points P, Q on the either arcs.
Let BD be a diagonal of ABCD
Now the area of the segment

= 28.5 + 28.5 = 57 cm²

Side of the square = 10 cm
Area of the square = side × side
= 10 × 10 = 100 cm²
Area of two sectors with centres A and C and radius 10 cm.
= 2 × \(\frac{\pi r^{2}}{360}\) × x = 2 × \(\frac{x}{360}\) × \(\frac{22}{7}\) × 10 × 10
= \(\frac{1100}{7}\)
= 157.14 cm²
∴ Designed area is common to both the sectors,
∴ Area of design = Area of both sectors – Area of square
= 157 – 100 = 57 cm²
(or)
\(\frac{1100}{7}\) – 100 = \(\frac{1100-700}{7}\)
= \(\frac{400}{7}\)
= 57.1 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.3

10th Class Maths 11th Lesson Trigonometry Ex 11.3 Textbook Questions and Answers

Question 1.
Evaluate:
i) \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
Answer:
Given that \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
= \(\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}\) [∵ cot (90 – θ) = tan θ]
= \(\frac{\tan 36^{\circ}}{\tan 36^{\circ}}\)
= 1

ii) cos 12° – sin 78°
Answer:
Given that cos 12° – sin 78°
= cos 12° – sin(90° – 12°) [∵ sin (90 – θ) = cos θ]
= cos 12° – cos 12° = 0

iii) cosec 31° – sec 59°
Answer:
Given that cosec 31° – sec 59°
= cosec 31° – sec (90° – 31°) [∵ sec (90 – θ) = cosec θ]
= cosec 31° – cosec 31° = 0

iv) sin 15° sec 75°
Answer:
Given that sin 15° sec 75°
= sin 15° . sec (90° – 15°)
= sin 15° . cosec 15° [∵ sec (90 – θ) = cosec θ]
= sin 15° . \(\frac{1}{\sin 15^{\circ}}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

v) tan 26° tan 64°
Answer:
Given that tan 26° tan 64°
= tan 26° . tan (90° – 26°)
= tan 26° . cot 26° [∵ tan (90 – θ) = cot θ]
= tan 26° . \(\frac{1}{\tan 26^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1

Question 2.
Show that
i) tan 48° tan 16° tan 42° tan 74° = 1
Answer:
L.H.S. = tan 48° tan 16° tan 42° tan 74°
= tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°)
= tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ]
= tan 48° . tan 16° . \(\frac{1}{\tan 48^{\circ}}\) . \(\frac{1}{\tan 16^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1 = R.H.S.
∴ L.H.S. = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0
Answer:
L.H.S. = cos 36° cos 54° – sin 36° sin54°
= cos (90° – 54°) . cos (90° – 36°) – sin 36° . sin 54° [∵ cos (90 – θ) = sin θ]
= sin 54° . sin 36° – sin 36° . sin 54°
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A.
Answer:
Given that tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
⇒ A = \(\frac{108^{\circ}}{3}\) = 36°
Hence the value of A is 36°.

Question 4.
If tan A = cot B where A and B. are acute angles, prove that A + B = 90°.
Answer:
Given that tan A = cot B
⇒ cot (90° – A) = cot B [∵ tan θ = cot (90 – θ)]
⇒ 90° – A = B
⇒ A + B = 90°

Question 5.
If A, B and C are interior angles of a triangle ABC, then show that \(\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\cot \frac{\mathbf{C}}{2}\)
Answer:
Given A, B and C are interior angles of right angle triangle ABC then A + B + C = 180°.
On dividing the above equation by ‘2’ on both sides, we get 180°

On taking tan ratio on both sides

Hence proved.

Question 6.
Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
We have sin 75° + cos 65°
= sin (90° – 15°) + cos (90° – 25°)
= cos 15° + sin 25° [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]