Srinivas

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.2

Question 1.
Use deductive reasoning to answer the following.

i) Human beings are mortal. Jeevan is a human being. Based on these two statements, what can’ you conclude about Jeevan?
Solution:
From the above statements we can deduce that Jeevan is mortal as it is given that all humans are mortal and Jeevan is a human.

ii) All Telugu people are Indians. X is an Indian. Can you conclude that X belongs to Telugu people?
Solution:
No. X may belong to any other language like Tamil, Kannada, Malayali…. etc.

iii) Martians have red tongues. Gulag is a Martian. Based on these two statements, what can you conclude about Gulag?
Solution:
Gulag had red, tongue.

iv) What is the fallacy in the Raju’s reasoning in the cartoon below

Solution:
All smarts need not be a President. There could be some other persons who are smart too.

Question 2.
Once again you are given four cards. Each card has a number printed on one side and a letter on the other side. Which are the only two cards you need to turn .over to check whether the following rule holds ?
“If a card has a consonant on one side, then it has an odd number on the other side”.

Solution:
You need to turn over B and 8 only. If B has an even number then the rule has broken. Similarly if 8 has a consonant on the other side then also the rule has been broken.

Question 3.
Think of this puzzle. What do you need to find a chosen number from this square ? Four of the clues below are true but do nothing to help in finding the number. Four of the clues are necessary for finding it.
Here are eight clues to use:
a) The number is greater than 9.
b) The number is not a multiple of 10.
c) The number is a multiple of 7.
d) The number is odd.
e) The number is not a multiple of 11.
f) The number is less than 200.
g) Its ones digit is larger than its tens digit.
h) Its tens digit is odd.
What is the number ?

Solution:

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.1

Question 1.
State whether the following sentences are always true, always false or ambiguous. Justify your answer.
i) There are 27 days in a month.
Solution:
Always false. Generally 30 days or 31 days make a month except February.

ii) Makarasankranthi fells on a Friday.
Solution:
Ambiguous. Makarasankranthi may fall on any day of the week.

iii) The temperature in Hyderabad is 2°C.
Solution:
Ambiguous. Sometimes the temperature may go down to 2°C in winter.

iv) The earth is the only planet where life exist.
Solution:
We can’t say always true. To the known fact, so far we can say this.

v) Dogs can fly.
Solution:
Always false,’as dogs can never fly.

vi) February has only 28 days.
Solution:
Ambiguous. A leap year has 29 days for February.

Question 2.
State whether the following statements are true or false. Give reasons for your answers.

i) The sum of the interior angles of a quadrilateral is 350°.
Solution:
False. Sum of the interior angles of a quadrilateral is 360°.

ii) For any real number x, x² > 0
Solution:
True. This is true for all real numbers,

iii) A rhombus is a parallelogram.
Solution:
True. In a rhombus, both pairs of opposite sides are parallel and hence every rhombus is a parallelogram.

iv) The spm of two even numbers is even.
Solution:
True. This is true for any two even numbers.

v) Square numbers can be written as the sum of two odd numbers.
Solution:
Ambiguous. Since square of an odd number can’t be written as sum of two odd numbers.

Question 3.
Restate the following statements with appropriate conditions, so that they become true statements.

i) All numbers can be represented in prime factorization.
Solution:
Any natural number greater than 1 can be represented in prime factorization.

ii) Two times a real number is always even.
Solution:
Two times a natural number is always even.

iii) For any x, 3x + 1 > 4.
Solution:
For any x > 1; 3x + 1 > 4.

iv) For any x, x³ ≥ 0.
Solution:
For any x > 0; x³ ≥ 0.

v) In every triangle, a median is also an angle bisector.
Solution:
In an equilateral triangle, a median is also an angle bisector.

Question 4.
Disprove, by finding a suitable counter example, the statement
x² > y² for all x > y.
Solution:
If x = – 8 and y = – 10
Here x > y
x² = (- 8)² = 64 and y² = (- 10)² = 100
But x² > y² is false here. [ ∵ 64 < 100]
(This can be proved for any set of nega-tive numbers or a negative number and a positive number)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles InText Questions

Activity

Question
Let us now do the following activity. Mark a point on a sheet of paper. Taking this point as centre draw a circle with any radius. Now increase or decrease the radius and.again draw some more circles with the same centre. What do you call the circles obtained in this activity ?[Page No. 261]

Circles having a common centre are called concentric circles.

Do This

Question 1.
In the figure which circles are congruent to the circle A ?
[Page No. 262]

Solution:
Circle ‘E’ is congruent to circle A.

Question 2.
What measure of the circles make them congruent ? [Page No. 262]
Solution:
Radius of circles determines their congruency.

Activity

Take a thin circular-sheet and fold it to half and open. Again fold it along any other half and open. Repeat this activity for several times. Finally when you open it, what do you observe?
[Page No. 262]
Solution:
Student Activity.

Activity

Take a circular paper. Fold it along any diameter such that the two edges coincide with each other. Now open it and again fold it into half along another diameter. On open¬ing, we find two diameters meet at the centre ‘O’. There forms two pairs of vertically opposite angles which are equal. Name the end points of the diameter as A, B, C and D.

Draw the chords \(\overline{\mathrm{AC}}, \overline{\mathrm{BC}}, \overline{\mathrm{BD}}\) and \(\overline{\mathrm{AD}}\).
Now take cut-out of the four segments namely 1, 2, 3 and 4.
If you place these segments pair wise one above the other the edges of the pairs (1,3) and (2.4) coincide with each other.
Is \(\overline{\mathrm{AD}}=\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}=\overline{\mathrm{BD}}\) ?
Though you have seen it in this par¬ticular case, try it out for other equal angles too. The chords will all turn out to be equal because of the following theorem. [Page No. 265]

Activity

Take a circle shaped paper and mark centre ‘O’: Fold it into two unequal parts and open it. Let the crease represent a chord AB, and then make a fold such that ‘A’ coincides with B. Mark the point of intersection of the two folds as D. Is AD = DB?

∠ODA = ?; ∠ODB = ?. Measure ‘the angles between the creases. They are right- angles. So. we can make a hypothesis “the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord”.
[Page No. 267]

Try This

In a circle with centre ‘O’, \(\overline{\mathbf{A B}}\) is a chord and M is its midpoint. Now prove that \(\overline{\mathbf{O M}}\) is perpendicular to AB.
(Hint : Join OA and OB consider tri-angles OAM and OBM)
[Page No. 267]
Solution:

‘O’ is the centre of the circle.
AB is a chord, M is its midpoint.
Join A, B to’O’.
Now in ΔOMA and ΔOMB
OA = OB (radii)
OM = OM (common)
MA = MB (given)
∴ ΔOMA s ΔOMB (SSS congruence)
∴ ∠OMA = ∠OMB (C.P.C.T)
But ∠OMA and ∠OMB are linear pair
∴∠OMA = ∠OMB = 90°
i.e., OM ⊥ AB.

Question
If three points are eollinear, how many circles can be drawn through these points ? Now try to draw a circle passing through these, three points.
[Page No. 268]
Solution:
If three points are eollinear, we can’t draw a circle passing’ through these points.

Activity

Draw a big circle on a paper and take a cut-out of it. Mark its centre as ‘O’. Fold it in half. Now make another fold near semi-circular edge. Now unfold it. You will get two congruent folds of chords. Name them as AB and CD. Now make perpendicular folds pass¬ing through centre ‘O’ for them. Using di¬vider compare the perpendicular distances of these chords from the centre.

Repeat the above activity by folding congruent chords. State your observations as a hypothesis.
“The congruent chords in a circle are at equal distance from the centre of the circle”. [Page No. 269]

Try This

In the figure, O is the centre of the circle and AB = CD. OM is perpen-dicular on \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{O N}}\) is perpen-dicular on \(\overline{\mathbf{C D}}\). Then prove that OM = ON. [Page No. 269]

Solution:
O’ is the centre of the circle.
Chords AB = CD
OM ⊥ AB; ON ⊥ CD
In ΔOMB and ΔONC
OB = OC [∵ radii]
BM = CN \(\left[\because \frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}\right]\)
∠OMB = ∠ONC [ ∵90° each]
∴ ΔOMB ≅ ΔONC [R.H.S congruence]
∴ OM = ON (CPCT)

Activity

Take a circle paper. Mark four points A, B, C and D on the circle paper. Draw cyclic quadrilateral ABCD and measure its angles and record it in the table. Repeat this activity for three more times

What do you infer from the table ?
Solution:
Student Activity

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 14th Lesson Probability Exercise 14.1

Question 1.
A die has six faces numbered from 1 to 6. It is rolled and number on the top face is noted. When this is treated as a random trial.
(Random trial: All possible outcomes are known before hand and the exact outcome can’t be predicted, then only the experiment is treated as a random experiment or a trial.)

a) What are the possible outcomes ?
Solution:
The possible outcomes are 1, 2, 3, 4, 5 and 6.

b) Are they equally likely ? Why ?
Solution:
Yes. All the outcomes are equally likely since every event has equal chance of occurrence or no event has priority to occur.

c) Find the probability of a composite number turning up on the top face.
Solution:
Even : Turning up of a composite number
Possible outcomes = 4, 6
No. of possible outcomes = 2
Total outcomes = 1, 2, 3, 4, 5 and 6
Number of total outcomes = 6
Probability = \( \frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }} \\\)
= \(\frac{2}{6}=\frac{1}{3}\)

Question 2.
A coin is tossed 100 times and the following outcomes are recorded. Head : 45 times; Tails : 55 times from the experiment.
a) Compute the probability of each outcome.
Solution:
Head = 45 times; Tails = 55 times
Total = 100
P(H),
Probability of getting Head = \(\frac{45}{100}\)
P(T),
Probability of getting Tail = \(\frac{55}{100}\)
\(\left[P=\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\right]\)

b) Find the sum of the probabilities of all outcomes.
Solution:
P(H) + P(T) = \(\frac{45}{100}+\frac{55}{100}=\frac{100}{100}\) = 1

Question 3.
A spinner has four colours as shown in the figure. When we spin it once, find

a) At which colour, is the pointer more likely to stop ?
Solution:
Red = 5 sectors; Blue = 3 sectors;
Green = 3 sectors; Yellow = 1 sector
Total = 5 + 3 + 3+1 = 12 sectors
∴ Pointer is more likely to stop at Red.

b) At which colour, is the pointer less likely to stop ?
Solution:
Yellow; as only one sector is shaded in Yellow.

c) At which colours, is the pointer equally likely to stop ?
Solution:
Blue and Green have equal chances; as they are shaded in equal number of sectors.

d) What is the chance the pointer will stop on white ?
Solution:
No chance. Since no sector is shaded in white.

e) Is there any colour at which the pointer certainly stops ?
Solution:
No; as the experiment is a random ex-periment.

Question 4.
A bag contains five green marbles, three blue marbles, two red marbles and two yellow marbles. One marble is drawn out randomly. M)
a) Are the four different colour outcomes equally likely ? Explain.
Solution:
No. As they are not in equal number, they have different chances of occurrence.

b) Find the probability of drawing each colour marble, i.e., P (green), P (blue), P(red) and P (yellow).
Solution:

c) Find the sum of their probabilities.
Solution:
P(Green) + P(Blue) + P(Red) +P(Yellow)
= \(\frac{5}{12}+\frac{3}{12}+\frac{2}{12}+\frac{2}{12}\)
= \(\frac{5+3+2+2}{12}=\frac{12}{12}\) = 1

Question 5.
A letter is chosen from English alphabet. Find the probability of the letters being
a) A vowel
b) A letter comes after P
c) A vowel or a consonant
d) Not a vowel
Solution:
Total letters ;= 26 [A, B, C ….. Z]
Probability = \(\frac{\text { No. of favourables }}{\text { Total no. of outcomes }}\)
a) Vowels = a, e, i, o, u [5]
∴ (vowels) = \(\frac{5}{26}\)

b) Letter after P = 10
[Q, R, S, T, U, V, W, X, Y, Z]
Probability of a letter that comes after
’P’ = \(\frac{10}{26}\) = \(\frac{5}{13}\)

c) A vowel or a consonant
Vowel or consonants = 26
[all letters from A to Z]
P(vowel or consonant) = \(\frac{26}{26}\) = 1

d) Not a vowel = 21
[other than A, E, I, O, U]
∴ P(not a vowel) = \(\frac{21}{26}\)

Question 6.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98,5.04,5.07,5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Total number of bags = 11
No. of bags with a weight more than 5 kg = 7
[5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07]
∴ Probability = \(\frac{\text { favourable outcomes }}{\text { total outcomes }}\)
P(E) = \(\frac{7}{11}\)

Question 7.
An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained is given in the following table.

Find the probabilities of the following events for a driver chosen at random from the city:

i) The driver being in the age group 18-29 years and having exactly 3 accidents in one year.
Solution:
Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 +
360 + 45 + 35 + 15 + 9) = 2000
Event: The driver being in the age group (18 – 29) years and having exactly 3 accidents
= 61
P(E) = \(\frac{\text { No. of favourables }}{\text { Total outcomes }}=\frac{61}{2000}\)

ii) The driver being in the age group of 30 – 50 years and having one or more accidents in a year.
Solution:
Favourable outcomes = 125 + 60 + 22 + 18 = 225
Total outcomes = 2000
P(E) = \(\frac{225}{2000}=\frac{9}{80}\)

iii) Having no accidents in the year.
Solution:
Favourable outcomes = 440 + 505 + 360 = 1305
Total outcomes = 2000
P(E) = \(\frac{1305}{2000}=\frac{261}{400}\)

Question 8.
What is the probability that a randomly thrown dart hits the square board in shaded region ( Take π = \(\frac{22}{7}\) and express in percentage )
Solution:

Radius of the circle r = 2 cm
Area of the circle A = πr² = \(\frac{22}{7}\) × 2 × 2 = \(\frac{88}{7}\) cm²
Side of the square = 2 × radius
= 2 × 2 = 4cm
Area of the square = S² = 4 × 4 = 16 cm²
Area of the shaded region = Area of the square – Area of the circle

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 11 Areas InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas InText Questions

Activity

Question

Observe Figure I and II. Find the area of both. Are the areas equal?
Trace these figures on a sheet of paper, cut them. Cover fig. 1 with fig. II. Do they cover each other completely? Are they congruent?
Observe fig. Ill and IV. Find the areas of both. What do you notice?
Are they congruent?
Now, trace these figures on sheet of paper. Cut them let us cover fig. Ill by fig. IV by coinciding their bases (length of same side).
As shown in figure V are they covered completely?
We conclude that Figures I and II are congruent and equal in area. But figures III and IV are equal in area but they are not congruent

Think, Discuss and Write

Question 1.
If 1 cm represents 5 in, what would be an area of 6 cm² represents ?
[Page No. 247]
Solution:
1 cm² = 5 m
1 cm² = 1 cm × 1 cm = 5m × 5m = 25m²
∴ 6 cm² = 6 × 25 m² = 150 m²

Question 2.
Rajni says 1 sq. m = 100² sq. cm. Do you agree ? Explain.
Solution:
No
1 sq. m = 100 cm²

Think, Discuss and Write

Question
Which of the following figures lie on the same base and between the same parallels? In such cases, write the common base and two parallels. [Page No. 249]

Solution:
a) In figure (a) ΔPCD and □ ABCD lie on same base CD and between the same parallels AB//CD.
b) No,
c) ΔTRQ and □ PQRS lie on the same base QR and between the same par-allels PS//QR.
d) ΔAPD and □ ABCD lie on the same base AD and between the same par-allels AD//BC.
e) No.

Activity

Question
Take a graph sheet and draw two par-allelograms ABCD and PQCD on it as show in the Figure, [Page No. 250]

The parallelograms are on the same base DC and between the same parallels PB and DC. Clearly the part DCQA is common between the two parallelo-grams. So if we can show that ΔDAP and ΔCBQ have the same area then we can say ar(PQCD) = ar(ABCD)

Activity

Draw pairs of triangles one the same base or ( equal bases) and between the same parallels on the graph sheet as shown in the Figure.
Let AABC and ADBC be the two triangles lying on the same base BC and between parallels BC and FE.
Draw CE II AB and BF II CD. Parallelograms AECB and FDCB are on the same base BC and are between the same parallels BC and EF.

Thus ar (AECB) = ar (FDCB).
We can see ar (ΔABC) = \(\frac { 1 }{ 2 }\) ar (parallelogram AECB) …………….(i)
and ar (ΔDBC) = \(\frac { 1 }{ 2 }\) ar (parallelogram FDCB) ……………..(ii)
From (i) and (ii), we get ar (ΔABC) = ar (ΔDBC)
You can also find the areas of ΔABC and ΔDBC by the method of counting the squares in graph sheet as we have done in the earlier activity and check the areas are whether same.[Page No. 254]

Think, Discus and Write

Draw two triangles ABC and DBC on the same base and between the same parallels as shown in the figure with P as the point of intersection of AC and BD. Draw CE//BA and BF//CD such that E and F lie on line AD.
Can you show ar(ΔPAB) = ar(ΔPDQ) ?[Page No. 254]

[Hint: These triangles are not congruent but have equal areas.
Solution:
□ ABCE = 2 × ΔABC
[∵ ΔABC; □ABCE lie on the same base BC and between the same parallels BC // AE]
ΔABC = \(\frac { 1 }{ 2 }\) × □ ABCE ……………(1)
Also □ BCDF = 2 × ΔBCD…………..
[∵ΔBCD and □ BCDE lie on the same base BC and between , the same parallels BC//DE]
ΔBCD = \(\frac { 1 }{ 2 }\) × □ BCDF ……………… (2)
But □ABCE = □BCDF
[ ∵ □ABCE and □BCDF lie on the same base BC and between the same parallels BC//FE]
From (1) & (2); ΔABC = ΔBCD
ΔPAB + ΔPBC = ΔPBC + ΔPDC
⇒ ΔPAB = ΔPDC
Hence proved.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry InText Questions

Try This

Can you give any two axioms from your daily life. [Page No. 63]
Solution:
Example -1 : If we flip coin then there are only two chances that is head or tail.
Example – 2 : Things which are equal to the same thing are also equal to one another.
Example – 3 : The whole is greater than the part.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes InText Questions

Try This

Take a cube of edge T cm and cut it as we did in the previous activity and find total surface area and lateral surface area of cube. [Page No. 216]
Solution:

If we cut and open a cube of edgewe obtain a figure as shown above.
In the figure, A, B, C, D, E, F are squares of side
The faces A, C, D, F forms the lateral surfaces of the cube.
∴ Lateral surface area of the cube = 4l²
And all six faces form the cube.
∴ Total surface area of the cube = 6l²

Do This

Question 1.
Find the total surface area and lateral surface area of the cube with side 4 cm. By using the formulae deduced in above. Try this. [Page No. 216]1. Find the total surface area and lateral surface area of the cube with side 4 cm. By using the formulae deduced in above. Try this. [Page No. 216]

Solution:
Total surface area of a cube = 6l²
Given l = 4crn
∴ T.S.A. = 6 × 4² = 6 × 16 = 96 cm²
∴ L.S.A. = 4l² = 4 × 4² = 64cm²

Question 2.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area. [Page No. 216]
Solution:
Let the edge of a cube x units
Its surface area = 6l² = 6x² sq. units
If its edge is increased by 50%, then
new edge = x + 50% of x


Where x = increase/decrease

Try These

Question
a) Find the volume of a cube whose edge is ‘a’ units. [Page No. 217]

Solution:
V = edge³ = a³ cubic units.

b) Find the edge of a cube whose volume is 1000 cm³. [Page No. 217]
Solution:
V = edge³ = 1000 = 10 × 10 × 10 = 10³
∴ Edge =10 cm

Do These

Question 1.
Find the volume of a cuboid if l = 12 cm, b = 10 cm, h = 8 cm. [Page No. 218]
Solution:
Volume V = lbh = 12 × 10 × 8 = 960cm³

Question 2.
Find the volume of cube, if its edge is 10 cm [Page No. 218]
Solution:
Volume V = l³ = 10 × 10 × 10 = 1000cm³

Question 3.
Find the volume of isosceles right angled triangular prism. [Page No. 218]

Solution:
Volume = Area of base × height
= Area of isosceles triangle × height

Activity

Question
Take the square pyramid and cube containers of same base and with equal heights. [Page No. 218]

Fill the pyramid with a liquid and pour into the cube (prism) completely. How many times it takes to fill the cube? From this, what inference can you make?

Thus volume of pyramid = \(\frac { 1 }{ 3 }\) × of the volume of right prism
= \(\frac { 1 }{ 3 }\) × Area of the base × height.

Note : A right prism has bases perpendicu¬lar to the lateral edges and all lateral faces are rectangles.

Do These

Question 1.
Find the volume of a pyramid whose square base is 10 cm and height is 8 cm. (Page No. 219)
Solution:
Volume of a pyramid
= \(\frac { 1 }{ 3 }\) × Area of the base × height
= \(\frac { 1 }{ 3 }\) × 10 x 10 × 8 = \(\frac { 800 }{ 3 }\) cm³

Question 2.
The volume of a cube is 1200 cubic cm. Find the volume of square pyra¬mid of the same height. (Page No. 219)
Solution:
Volume of the square pyramid
= \(\frac { 1 }{ 3 }\) × volume of the square prism
= \(\frac { 1 }{ 3 }\) × 1200 = 400 cm³

Activity

Cut out a rectangular sheet of paper. Paste a thick string along the line as shown in the figure. Hold the string with your hands on either sides of the rectangle and rotate the rectangle sheet about the string as fast as you can.

Do you recognize the shape that the rotating rectangle is forming ?
Does it remind you the shape of a cylinder ? [Page No. 220]

Do This

Question 1.
Find C.S.A. of each of the following cylinders. [Page No. 221]
i) r = x cm; h = y cm
Solution:
CSA = 2πrh = 2πxy cm²

ii) d = 7 cm; h = 10 cm
Solution:

CSA = 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 10
= 220 cm²

iii) r = 3 cm; h = 14 cm
Solution:

CSA = 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 3 × 14
= 264 cm²

Question 2.
Find the total surface area of each of the following cylinder. [Page No. 222]
i)

Solution:
r = 7 cm; h = 10 cm
T.S.A. = 2πr (r + h)
=2 × \(\frac{22}{7}\) × 10 (7 + 10)
= 2 × \(\frac{22}{7}\) × 10 × 17
= 1068.5 cm²

ii)

h = 7cm; πr² = 250
πr² = 250 = 250
\(\frac{22}{7}\) × r² = 250 = 250
r² = 125 x \(\frac{7}{11}\)

Try These

Question 1.
If the radius of a cylinder is doubled keeping Its lateral surface area the same, then what is its height? [Page No. 225]
Solution:
Let the initial radius and height of the
cylinder be r and h.
Then L.S.A. = 2πrh
When r is doubled and the L.S.A. remains the same, then the height be hr By problem new L.S.A. = 2πrh
= 2π (2r) (h₁)
⇒ 2πrh = 4πrh₁
∴ \(\frac{2 \pi \mathrm{rh}}{4 \pi \mathrm{r}}=\frac{1}{2} \mathrm{~h}\)
Height becomes its half.

Question 2.
A hot water system (Geyser) consists of a cylindrical pipe of length 14 m and diameter 5 cm. Find the total radiating surface of hot water system. [Page No. 225]
Solution:
Radius (r) = \(\frac{\text { diameter }}{2}=\frac{5}{2}\) = 2.5 cm
Length of the pipe = height = 14 m Radiating surface = 2πrh
= 2 × \(\frac{22}{7}\) × 2.5 × 1400
= 22000 cm³

Activity

Question
Making a cone from a sector. [Page No. 227] Follow the instructions and do as shown in the figure.

i) Draw a circle on a thick paper Fig (a).
ii) Cut a sector AOB from it Fig (b).
iii) Fold the ends A, B nearer to each other slowly and join AB. Remember A, B should not overlap on each other. After joining A, B attach them with cello tape Fig (c).
iv) What kind of shape you have obtained? Is it a right cone?
While making a cone observe what hap-pened to the edges ‘OA’ and ‘OB’ and length of arc AB of the sector?

Try This

A sector with radius r and length of its arc / is cut from a circular sheet of paper. Fold it as a cone. How can you derive the formula of its curved surface area A = πrl. [Page No. 228]

C.S.A = πrl
Solution:
When a sector of radius ‘r’ and whose length of arc l is folded to form a cone. Radius ‘r’ becomes slant height ‘l’ and arc ‘l’ becomes perimeter of the base 2πr.

∴ Area of the sector = \(\frac{l r}{2}\) = Area of the cone
\(\frac{2 \pi \mathrm{r} l}{2}\) = Surface area of the cone
C.S.A = πrl

Do This

Question 1.
Cut a right angled triangle. Stick a string along its perpendicular side, as shown in fig. (T) hold both the sides of a string with your hands and rotate it with constant speed. What do you observe ? [Page No. 229]

Solution:
A right circular cone is observed.

Question 2.
Find the curved surface area and total surface area of the each following right circular cones.[Page No. 229]
Solution:

OP = 2 cm; OB = 3.5 cm
OP = h = 2 cm

OP = 3.5 cm; AB = 10 cm
r = \(\frac{\mathrm{AB}}{2}\) = 5cm; h = 3.5cm
r = OB = 3.5 cm
C.S.A. = πrl

T.S.A. = πr (r + l)
= \(\frac{22}{7}\) × 3.5(3.5 + 4.03)
= \(\frac{22}{7}\) × 3.5 × 7.53 = 82.83cm²

C.S.A. = πrl
\(\frac{22}{7}\) × 5 × 6.10
= 95.90cm²

T.S.A. = πr (r + l)
= \(\frac{22}{7}\) × 5 × (5 + 6.10)
= 174.42 cm²

Activity

Draw a circle on a thick paper and cut it neatly. Stick a string along its diam¬eter. Hold the both the ends of the string with hands and rotate with con¬stant speed and observe the figure so formed. [Paper 235]

Try This

Question 1.
Can you find the surface area of sphere in any other way ? (Page No. 235)
Solution:

Height of the pyramid is equal to r.
To derive the formula of the surface area of a sphere, we imagine a sphere with many pyramids inside of it until the base of all the pyramids cover the entire surface area of the sphere. In the figure below, only one of such pyra¬mid is shown.
Then, do a ratio of the area of the pyramid to the volume of the pyramid.
The area of the.pyramid is A.
The volume of the pyramid is V = (1/3) × A × r = (A × r)/3
So, the ratio of area to volume is A/V = A + (A × r) / 3 = (3 × A) / (A × r) = 3 / r

For a large number of pyramids, let say that n is such large number, the ratio of the surface area of the sphere to the volume of the sphere is the same as 3/r.
For n pyramids, the total area is n × A- Also for n pyramids, the total volume is n × V.
Therefore, ratio of total area to total volume is n × A/n × V = A/ V and we already saw before that A / V = 3 / r

Further more, n × A_pyramid = A_sphere (The total area of the bases of all pyramids or n pyramids is approximately equal to the surface area of the sphere).
n × V_pyramid = V_sphere (The total Volume of all pyramids or n pyramids is approximately equal to the volume of the sphere.
Putting observation # 1 and # 2 together, we get

Therefore, the total surface area of a sphere, call it S.A. is 4πr ²

Do These

Question 1.
A right circular cylinder just encloses a sphere of radius r (see figure). Find i) surface area of the sphere
ii) curved surface area of the cylinder
iii) ratio of the areas obtained in (i) and (ii) [Page No. 236]

[Page No. 236]
Solution:
i) Radius of the sphere = radius of the cylinder = r
∴Surface area of the sphere = 4πr²
ii) C.S.A. of cylinder = 2πr (2r) [∵ h = 2r]
= 4πr²
iii) Ratio of (i) and (ii) = 4πr² : 4πr² = 1:1

Question 2.
Find the surface area of each of the following figure.
i)

Surface area = 4πr²
C.S.A = 4 × \(\frac{22}{7}\) × 7 × 7
= 616cm²

ii)

C.S.A = 2πr² = 2 × \(\frac{22}{7}\) × 7 × 7 = 308cm²
Total Surface area = 3πr²
= 3 × \(\frac{22}{7}\) × 7 × 7 = = 462cm²

Do This

Question 1.
Find the volume of the sphere given below[Page No. 238]

Solution:
r = 3 cm
V = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) × 3 × 3 × 3
= 113.14cm³

d = 5.4 cm
r = \(\frac{\mathrm{d}}{2}=\frac{5.4}{2}\) = 2.7 cm
V = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) × 2.7 × 2.7 × 2.7 = 82.48cm³

Question 2.
Find the volume of sphere of radius 6.3 cm. [Page No. 238]
Solution:
r = 6.3 cm
V = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3 = 1047.81cm³

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions Exercise 13.2

Question 1.
Construct AABC in which BC = 7 cm,∠B = 75° and AB + AC =12 cm.
Solution:
A.

Steps:

• Draw a line segment BC = 7 cm.
• Erect ∠B = 75°
• MarkapointDon \(\overrightarrow{\mathrm{BX}}\) suchthat BD = AB + AC.
• Join D, C and draw the perpendicular bisector of \(\overline{\mathrm{CD}}\) meeting BD at A.
• Join A to C to form the ΔABC.

Question 2.
Construct ΔPQR in which QR = 8 cm, ∠B = 60° and AB – AC = 3.5 cm.
Read ∠Q = 60°and PQ – PR = 3.5 cm
Solution:
A.

Steps: I

• Draw QR = 8 cm.
• Construct ∠RQX = 300 at Q.
• Mark a point S on \(\overrightarrow{\mathrm{QX}}\) such that QS = PQ  –  PR = 3.5 cm.
• Join S, R and draw the perpendicular bisector to \(\overline{\mathrm{QR}}\) meeting \(\overrightarrow{\mathrm{QX}}\) at P.
• Join P, R to form the ΔPQR.

Question 3.
Construct ΔXYZ in which ∠Y = 30 °; ∠Z = 60 ° and XY + YZ + ZX = 10 cm.
Solution:
A.

Steps:

• Draw a line segment AB = XY + YZ + ZX = 10 cm.
• Construct ∠BAP = \(\frac { 1 }{ 2 }\) ∠Y at A and ∠ABQ = \(\frac { 1 }{ 2 }\) ∠Z at B meeting at X.
• Draw the perpendicular bisectors to XA and XB meeting \(\overline{\mathrm{AB}}\) at Y and Z respectively.
• Join X to Y and Z to form the ΔXYZ.

Question 4.
Construct a right triangle whose base is 7.5 cm and sum of its hypotenuse and otherside is 15 cm.
Solution:
A.

Steps:

• Draw BC = 7.5 cm.
• Construct ∠CBX = 90°
• Mark a point D on \(\overrightarrow{\mathrm{BX}}\) such that BD = 15 cm.
• Join C, D. ‘
• Draw the perpendicular bisectors of \(\overline{\mathrm{CD}}\) meeting BD at A.
• Join A, C to form the ΔABC.

Question 5.
Construct a segment of a circle on a chord of length 5 cm containing the following angles i) 90° ii) 45° iii) 120°
Solution:
i) 90°
A.

Steps:

• Draw a rough sketch of ∠BAC = 90° and ∠BOC = 180°.
• Draw a line segment BC = 5 cm.
• Draw the perpendicular bisector of BC meeting \(\overline{\mathrm{BC}}\) at O
• Draw an arc of radius OB or OC with centre O.
• Mark any point A on the arc and join it with B and C.
• ∠BAC = 90°

ii) 45°

Steps:

• Draw a line segment BC = 5 cm.
• Construct ABOC such that BC = 5 cm, ∠B = 45° = ∠C.
• Draw a circle segment of radius OB or OC with centre ’O’.
• Mark any point A on the segment and join it with B and C.
• ∠BAC = 45°

iii) 120°

Steps:

• Draw a line segment AB = 5 cm. ,
• Construct ΔAOB in which ∠A = 30°; ∠B = 30°; AB = 5 cm.
• With ‘O’ as centre draw a circle segment.
• On the opposite side make any point C and join it with B and C.
• ∠ACB = 120°

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.3

Question 1.
The opposite angles of a parallelogram are (3x – 2)° and (x + 48)°. Find the measure of each angle of the parallelogram.
Solution:
Given that the opposite angles of a parallelogram are (3x – 2)° and (x + 48)°
Thus 3x – 2 = x + 48
(∵ opp. angles of a //gm are equal)
3x – x = 48 + 2
2x = 50
x = \(\frac{50}{2} \) = 25°
∴ The given angles are (3 x 25 – 2)° and (25 + 48) °
= (75 – 2)° and 73° = 73° and 73°
We know the consecutive angles are supplementary.
∴ The other two angles are (180°-73°) and (180°-73°)
= 107° and 107°
∴ The four angles are 73°, 107°, 73° and 107°.

Question 2.
Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.
Solution:
Let the smallest angle = x
Then its consecutive angle = 180 – x°
By problem (180 – x)° = (2x- 24)°
(∵ opp. angles are equal)
180 + 24 = 2x + x
3x = 204
x = \(\frac{204}{3} \) = 68°
∴ The angles are
68°; (2 x 68 – 24)°; 68°; (2 x 68 – 24)°
= 68°, 112°, 68°, 112°

Question 3.
In the given figure ABCD is a paral-lelogram and E is the mid point of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.

Solution:

Given that □ABCD is a parallelogram.
E is the midpoint of BC.
Let G be the midpoint of AD.
Join G, E.
Now in ΔAFD, GE is the line joining the midpoints G, E of two sides AD and FD.
∴GE // AF and GE = \(\frac{1}{2}\)AF
But GE = AB [ ∵ ABEG is a parallelo¬gram and AB, GE forms a pair of opp. sides]
\(\frac{1}{2}\) = AB ⇒ AF = 2AB
Hence Proved.

Question 4.
In the given figure ABCD is a paral¬lelogram. P, Q are the midpoints of sides AB and DC respectively. Show that
Solution:

□ABCD is a parallelogram.
P, Q are the mid points of AB and CD.
Join Q, P.
Now AB = CD (Opp. sides of a //gm)
\(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD
PB = QC
Also PB // QC.
Now in □PBCQ;
PB = QC; PB//QC
Hence □PBCQ is a parallelogram.

Question 5.
ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD//BA as shown in the figure. Show that i) ∠DAC = ∠BCA
ii) ABCD is a parallelogram.

Solution:
Given that AABC is isosceles; AB = AC
AD is bisector of ∠QAC

i) In ΔABC, AB = AC ⇒ ∠B = ∠ACB
(angles opp. to equal sides)
Also ∠QAC = ∠B + ∠ACB
∠QAC = ∠BCA + ∠BCA
(∵∠BCA = ∠B)
⇒ \(\frac{1}{2}\)∠QAC = \(\frac{1}{2}\) [2 ∠BCA]
⇒ ∠DAC = ∠BCA [ ∵ AD is bisector of ∠QAC]

ii) From (i) ∠DAC = ∠BCA
But these forms a pair of alt. int. angles for the pair of lines AD and BC; AC as a transversal.
∴ AD//BC
In □ABCD ; AB // DC; BC // AD
□ABCD is a parallelogram.

Question 6.
ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure). Show that 1) ΔAPB ≅ ΔCQD ii) AP = CQ.

Solution:
Given that □ABCD is a parallelogram.
BD is a diagonal.
AP ⊥ BD and CQ ⊥ BD
i) In ΔAPB and ΔCQD
AB = CD ( ∵ Opp. sides of //gm ABCD)
∠APB = ∠CQD (each 90°)
∠PBA = ∠QDC (alt. int. angles for the lines AB and DC)
∴ ΔAPB ≅ ΔCQD (AAS congruence)

ii) From (1) ΔAPB ≅ ΔCQD
⇒ AP = CQ (CPCT)

Question 7.
In Δs ABC and Δs DEF, AB = DC and AB//DE; BC = EF and BC//EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that
i) ABED is a parallelogram
ii) BCFE is a parallelogram
iii) AC = DF
iv) ΔABC = ΔDEF

Solution:

Given that in ΔABC and ΔDEF
AB = DE and AB // DE
BC = EF and BC//EF.
i) In □ABED AB//ED and AB = ED
Hence □ABED is a parallelogram.

ii) In □BCFE; BC = EF and BC//EF
Hence □BCFE is a parallelogram.

iii) ACFD is a parallelogram (In a paral-lelogram opposite sides are equal).
So, AC = DF.

iv) Consider ΔABC = ΔDEF
AB = DE (given);
AC = DF (proved)
BC = EF (given)
∴ ΔABC ≅ ΔDEF (SSS congruency rule).

Question 8.
ABCD is a parallelogram. AC and BD are the diagonals intersect at ‘O’. P and Q are the points of trisection of the diagonal BD. Prove that CQ//AP and also AC bisects PQ.

Solution:
Given □ABCD is a parallelogram;
BD is a diagonal.
P, Q are the points of trisection of BD.
In ΔAPB and ΔCQD
AB = CD (•.• Opp. sides of //gm ABCD)
BP = DQ (given)
∠ABP = ∠CDQ (alt. int. angles for the lines AB//DC, BD as a transversal)
ΔAPB = ΔCQD (SAS congruence)
Similarly in ΔAQD and ΔCPB
AD = BC (opp. sides of //gm ABCD)
DQ = BP (given)
∠ADQ = ∠CBP (all int. angles for the lines AD//BC, BD as a transversal)
ΔAQD ≅ ΔCPB
Now in □APCQ
AP = CQ (CPCT of AAPB, ACQD)
AQ = CP (CPCT of AAQD and ACPB)
∴ □APCQ is a parallelogram.
∴ CQ//AP (opp. sides of//gm APCQ)
Also AC bisects PQ. [ ∵ diagonals of //gm APCQ]

Question 9.
ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution:

Given that ABCD is a square.
E, F, G, H are the mid points of AB, BC, CD and DA.
Also AE = BF = CG = DH
In ΔABC; E, F are the mid points of sides AB and BC.
∴ EF//AC and EF = \(\frac{1}{2}\) AC
Similarly GH//AC and GH = AC
GF//BD and GF = \(\frac{1}{2}\) BD
HE//BD and HE = \(\frac{1}{2}\) BD

But AC = BD (∵ diagonals of a square)
∴ EF = FG = GH = HE
Hence EFGH is a rhombus.
Also AC ⊥ BD
(∵ diagonals of a rhombus)
∴ In //gm OIEJ [ ∵ 0I // EJ; IE // OJ]
We have ∠IOJ = ∠E
[ ∵ Opp. angles of a //gm]
∴ ∠E – 90°
Hence in quad. EFGH; all sides are equal and one angle is 90°.
∴ EFGH is a square.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation InText Questions

Think, Discuss and Write

Question
Which of the following expressions are polynomials ? Which are not ? Give reasons. [Page No. 28]
Solution:
i) 4x² + 5x – 2 is a polynomial.
ii) y² – 8 is a polynomial.
iii) 5 is a constant polynomial.
iv) \(2 x^{2}+\frac{3}{x}-5\) is not a polynomial as x is in denominator.
v) √3x² + 5y is a polynomial.
vi) \(\frac{1}{x+1}\) is not a polynomial as the variable x is in denominator.
vii) √x is not a polynomial as its exponent is not an integer.
viii) 3xyz is a polynomial.

Do These

Question
Write two polynomials with variable ‘x’. [Page No. 29]
Solution:
5x² + 2x – 8 and 3x² – 2x + 6.

Question
Write three polynomials with variable ‘y’.
Solution:
y³ – y² + y ; 2y² + 7y – 9 + 3y³; y⁴ – y + 6 + 2y².

Question
Is the polynomial 2x² + 3xy + 5y² in one variable ?
Solution:
No. It is in two variables x and y.

Question
Write the formulae of area and volume of different solid shapes. Find out the variables and constants in them. [Page No. 29]
Solution:

Question 1.
Write the degree of each of the following polynomials. [Page No. 30]
Solution:
i) 7x³ + 5x² + 2x – 6 – degree 3
ii) 7 – x + 3x² – degree 2
iii) 5p – √3 – degree 1
iv) 2 – degree 0
v) – 5 xy² – degree 3

Question 2.
Write the co-efficient of x² in each of the following. [Page No. 30]
Solution:
i) 15 – 3x + 2x² : co-efficient of x² is 2
ii) 1 -x² : co-efficient of x² is -1
iii) πx² – 3x + 5 : co-efficient of x² is π
iv) √2x² + 5x – 1 : co-efficient of x² is √2

Think, Discuss and Write

Question
How many terms a cubic (degree 3) polynomial with one variable can have? Give examples. [Page No. 31]
Solution:
A cubic polynomial can have atmost 4 terms.
E.g.: 5x³ + 3x² – 8x + 4; x³ – 8

Try These

Question 1.
Write a polynomial with 2 terms in variable x. [Page No. 31]
Solution:
2x + 3x²

Question 2.
How can you write a polynomial with 15 terms in variable ‘x’. [Page No. 31]
Solution:
a₁₄p¹⁴ + a₁₃p¹³ + a₁₂p¹²+ …………….+ a₁p + a₀

Do This

Question
Find the value of each of the follow ing polynomials for the indicated value of variables. [Page No. 33]
(i) p(x) = 4x² – 3x + 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
4(1)² – 3(1) + 7 = 8

ii) q(y) = 2y³ – 4y + √11 at y = 1.
Solution:
The value of q(y) at y = 1 is
2(1)³ – 4(1) + √11 = -2 + √11

iii) r(t) = 4t⁴ + 3t³ – t² + 6 at t = p, t ∈ R.
Solution:
The value of r(t) at t = p is
4p⁴ + 3p³ – p² + 6

iv) s(z) = z³ – 1 at z – 1.
Solution:
The value of s(z) at z = 1 is 1³ – 1 = 0

v) p(x) = 3x² + 5x – 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
3(1)² + 5(1) — 7 = 1.

vi) q(z) = 5z³ – 4z + √2 at 7. = 2.
Solution:
The value of q(z) at z = 2 is
5(2)³ – 4(2) + √2 = 40 – 8 + √2
= 32 + √2

Try These

Question
Find zeroes of the following polyno¬mials. [Page No. 34]

1. 2x-3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)
∴ x = \(\frac{3}{2}\) is the zero of 2x – 3.

2. x² – 5x + 6
Solution:
x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x (x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x – 2 = 0 or x – 3 = 0
⇒ x = 2 or x = 3
∴ x = 2 or 3 are the zeroes of x² – 5x + 6.

3. x + 5
Solution:
x + 5 = 0
x = – 5
∴ x = – 5

Do This

Fill in the bianks : [Page No. 35]

Solution:

Think, and Discuss

Question 1.
x² + 1 has no zeroes. Why ? [Page No. 36]
Solution:
x² + 1 = 0 ⇒ x² = -1
No real number exists such that whose root is – 1.
∴ x² + 1 has no zeroes.

Question 2.
Can you tell the number of zeroes of a polynomial of degree ‘n’ will have? [Page No. 36]
Solution:
A polynomial of degree n will have n- zeroes.

Do These

Question 1.
Divide 3y³ + 2y² + y by ‘y’ and write division fact. [Page No. 38]
Solution:
(3y³ + 2y² + y) ÷ y = \(\frac{3 y^{3}}{y}+\frac{2 y^{2}}{y}+\frac{y}{y}\)
= 3y² + 2y + 1
Division fact = (3y² + 2y + 1) y
= 3y³ + 2y² + y

Question 2.
Divide 4p² + 2p + 2 by ‘2p’ and write division fact.
Solution:
4p² + 2p ÷ 2 = \(\frac{4 p^{2}}{2 p}+\frac{2 p}{2 p}+\frac{2}{2 p}\)
= 2p + 1 + \(\frac{1}{\mathrm{P}}\)
Division fact:
(2p + 1 + \(\frac{1}{\mathrm{P}}\)).2p = 4p² + 2p + 2

Try These

Show that (x – 1) is a factor of xⁿ – 1. [Page No. 45]
Solution:
Let p(x) = xⁿ – 1
Then p(1) = 1ⁿ – 1 = 1 – 1 = 0
As p(1) = 0, (x – 1) is a factor of p(x).

Do These

Question
Factorise the following. [Page No. 46]

1. 6x² + 19x + 15
Solution:
6x² + 19x + 15 = 6x² + 10x + 9x + 15
= 2x (3x + 5) + 3 (3x + 5)
= (3x + 5) (2x + 3)

2. 10m² – 31m – 132
Solution:
10m² – 31m – 132
= 10m² – 55m + 24m – 132
= 5m (2m- 11) + 12 (2m- 11)
= (2m – 11) (5m + 12)

3. 12x² + 11x + 2
Solution:
12x² + 11x + 2
= 12x² + 8x + 3x + 2
= 4x (3x + 2) + 1 (3x + 2)
= (3x + 2) (4x + 1)

Try This

Question
Try to draw the geometrical figures for other identities. [Page No. 49]
i) (x + y)² ≡ x² + 2xy + y²

Step – 1 : Area of fig. I = x x = x²
Step – 2 : Area of fig. II = x y = xy
Step – 3 : Area of fig. III = x y = xy
Step – 4 : Area of fig. IV = y y = y²

Area of big square = sum of the areas of figures I, II, III and IV
∴ (x + y) (x + y) = x² + xy + xy + y²
(x + y)² = x² + 2xy + y²

ii) (x + y) (x – y) ≡ x² – y²

Step -1: Area of fig! I = x (x – y) = x² – xy
Step – 2: Area of fig. II = (x – y) y = xy – y²
Area of big rectangle = sum of areas of figures I & II
(x + y) (x – y) = x² – xy + xy – y²
= x² – y²
∴ (x + y) (x-y) = x2-y2

iii) (x + a) (x + b) ≡ x² + (a + b) x + ab

Step – 1 : Area of fig. I = x²
Step – 2 : Area of fig. II = ax
Step – 3 : Area of fig. Ill = bx
Step – 4 : Area of fig. IV = ab
∴ Area of big rectangle = Sum of areas of four small figures.
∴ (x + a) (x + b) = x² + ax + bx + ab
(x + a) (x + b) = x² + (a + b) x + ab

Do These

Question
Find the following product using appropriate identities. [Page No. 49]
i) (x + 5) (x + 5)
Solution:
(x + 5) (X + 5) = (x + 5)²
= x² + 2(x) (5) + 5²
= x² + 10x + 25

ii) (p – 3) (p + 3)
Solution:
(p – 3) (p + 3)
= p² – 3²
= p² – 9

iii) (y – 1) (y – 1)
Solution:
(y – 1) (y – 1)
= (y – 1)²
= y² – 2y + 1

iv) (t + 2) (t + 4)
Solution:
(t + 2) (t + 4)
= t² + t(2 + 4) + 2 x 4
= t² + 6t + 8

v) 102 x 98
Solution:
102 x 98 = (100 + 2) (100 -2)
= 100² – 2²
= 10000 – 4
= 9996

Do These

Question
Factorise the following using appro-priate identities. [Page No. 50]
i) 49a² + 70ab + 25b²
Solution:
49a² + 70ab + 25b²
= (7a)² + 2 (7a) (5b) + (5b)²
= (7a + 5b)²
= (7a + 5b)(7a + 5b)

ii) \(\frac{9}{16} x^{2}-\frac{y^{2}}{9}\)
Solution:

iii) t² – 2t + 1
= (t)² – 2(t) (1) + (1)²
= (t – 1)² = (t – 1) (t – 1)

iv) x² + 3x + 2
Solution:
x² + 3x + 2 = x2 + (2 + 1) x + (2 x 1)
(x + 2) (x + 1)

Do These

Question i)
Write (p + 2q + r)² in expanded form. [Page No. 52]
Solution:
(p + 2q + r)² = (p)² + (2q)² + (r)²
+ 2 (P) (2q) + 2 (2q) (r) + 2(r) (p)
= p² + 4q² + r² + 4pq + 4qr + 2rp

Question ii)
Expand (4x – 2y – 3z)² using identity. [Page No. 52]
Solution:
(4x – 2y – 3z)² = (4x)² + (- 2y)² + (- 3z)² + 2 (4x) (- 2y) + 2 (- 2y) (- 3z) + 2 (- 3z) (4x)
= 16x² + 4y² + 9z² – 16xy + 12yz – 24zx.

Question iii)
Factorise 4a² + b² + c² – 4ab + 2bc – 4ca
using identity. [Page No. 52]
Solution:
4a² + b² + c² – 4ab + 2bc – 4ca
= (2a)² + (- b)² + (- c)² + 2(2a) (- b) + 2 (- b) (- c) + 2(- c) (2a)
= (2a – b – c)² = (2a – b – c) (2a – b – c)

Try These

Question
How can you find (x – y)³ without actual multiplication ? Verify with actual multiplication. [Page No. 52]
Solution:
(x – y)³ = x³ – 3x²y + 3xy² – 3y³ from identity.
By actual multiplication
(x – y)³ = (x – y)² (x – y)
= (x² – 2xy + y²) (x – y)
= x³ – 2x²y + xy² – x²y + 2xy² – y³
= x³ – 3x² y + 3xy² – y³
Both are equal.

Do These

Question 1.
Expand (x + 1)³ using an identity. [Page No. 54]
Solution:
(x + 1)³ = (x)³ + (1)³ + 3 (x) (1) (x + 1)
= x³ + 1 + 3x (x + 1)
= x³ + 1 + 3x² + 3x = x³ + 3x² + 3x + 1

Question 2.
Compute (3m – 2n)³. [Page No. 54]
Solution:
(3m-2n)³
=(3m)³ – 3 (3m)² (2n) + 3 (3m) (2n)² – (2n)³
= 27m³ – 54m²n + 36mn² – 8n³

Question 3.
Factorise a³ – 3a²b + 3ab² – b³. [Page No. 54]
Solution:
a³ – 3a²b + 3ab² – b³
= (a)³ – 3 (a)² (b) + 3 (a) (b)² – (b)³
= (a – b)³
= (a – b) (a – b) (a – b)

Do These

Question 1.
Find the product (a – b – c) (a² + b² + c² – ab + be – ca) without actual multi-plication. [Page No. 55]
Solution:
The problem is incorrect.

Question 2.
Factorise 27a³ + b³ + 8c³ – 18abc using identity. [Page No. 55]
Solution:
27a³ + b³ + 8c³ – 18abc
= (3a)³ + (b)³ + (2c)³ – 3(3a) (b) (2c)
= (3a + b + 2c) (9a² + b² + 4c² – 3ab – 2be – 6ca)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals Exercise 8.2

Question 1.
In the given figure ABCD is a parallelogram. ABEF is a rectangle. Show that ΔAFD ≅ ΔBEC

Solution:
Given that □ABCD is a parallelogram.
□ABEF is a rectangle.
In ΔAFD and ΔBEC
AF = BE ( ∵ opp. sides of rectangle □ABEF)
AD = BC (∵ opp. sides of //gm □ABCD)
DF = CE (∵ AB = DC = DE + EC , AB = EF = DE + DF)
∴ ΔAFD ≅ ΔBEC (SSS congruence)

Question 2.
Show that the diagonals of a rhombus divide it into four congruent triangles.
Solution:

□ABCD is a rhombus.
Let AC and BD meet at O’.
In ΔAOB and ΔCOD
∠OAB = ∠OCD (alt.int. angles)
AB = CD (def. of rhombus)
∠OBA = ∠ODC ………………….(1) (alt. int. angles)
∴ ΔAOB ≅ ΔCOD (ASA congruence)
Thus AO = OC (CPCT)
Also ΔAOD ≅ ΔCOD …………..(2)
[ ∵ AO = OC; AD = CD; OD = OD SSS congruence]
Similarly we can prove
ΔAOD ≅ ΔCOB ……………. (3)
From (1), (2) and (3) we have
ΔAOB ≅ ΔBOC ≅ ΔCOD ≅ ΔAOD
∴ Diagonals of a rhombus divide it into four congruent triangles.

Question 3.
In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O. Prove that ∠COD = \(\frac{1}{2}\) (∠A + ∠B) .
(OR)
In a quadrilateral ABCD, the bisectors of ∠A and ∠B are intersects at ‘O’ then prove that ∠AOB = \(\frac{1}{2}\) (∠C + ∠D)
Solution:

In a quadrilateral □ABCD
∠A + ∠B + ∠C + ∠D = 360°
(angle sum property)
∠C + ∠D = 360° – (∠A + ∠B)
\(\frac{1}{2}\) (∠C + ∠D) = 180 – \(\frac{1}{2}\) (∠A + ∠B) ………….. (1)
(∵ dividing both sides by 2) .
But in ΔCOD
\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D + ∠COD = 180°
\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D = 180° – ∠COD
∴\(\frac{1}{2}\)(∠C +∠D) = 180° -∠COD………….(2)
From (1) and (2);
180° – ∠COD = 180° – \(\frac{1}{2}\) (∠A + ∠B)
∴ ∠COD = \(\frac{1}{2}\) (∠A + ∠B)
Hence proved.