AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.3

Question 1.
Draw the graph of each of the following linear equations.
i) 2y = – x + 1
ii) – x + y = 6
iii) 3x + 5y = 15
iv) \(\frac{x}{2}-\frac{y}{3}=3\)
Solution:
i) 2y = – x + 1
⇒ x + 2y = 1
ii) – x + y = 6
iii) 3x + 5y = 15
iv) \(\frac{x}{2}-\frac{y}{3}=3\)
⇒ \(\frac{3 x-2 y}{6}=3\)
⇒ 3x – 2y = 18

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 1
AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 2
AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 3

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 2.
Draw the graph of each of the following linear equations and answer the following questions.
i) y – x
ii) y = 2x
iii) y = – 2x
iv) y = 3x
v) y = – 3x
Solution:
i) y = x

x 1 2
y 1 2
(x, y) (1, 1) (2, 2)

ii) y = 2x

x 1 2
y 2 4

iii) y = – 2x

x 1 2
y -2 -4

iv) y = 3x

x 1 2
y 3 6

v) y = – 3x

x 1 2
y -3 -6

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 4

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

i) Are all these equations of the form y = mx; where m is a real number ?
Solution:
Yes. All the equations are of the form y = mx where m e R.

ii) Are all these graphs passing through the origin ?
Solution:
Yes. All these lines pass through the origin.

iii) What can you conclude about these graphs ?
Solution:
All lines of the form y = mx, pass through the origin.

Question 3.
Draw the graph of the equation 2x + 3y = 11. Find from the graph value of y when x = 1.
Solution:

x 1 4
y 3 1

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 5
From the graph; when x = 1 then y = 3.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 4.
Draw the graph of the equation y – x = 2. Find from the graph
i) the value of y when x = 4
ii) the value of x when y = – 3
Solution:
The given equation is y – x = 2 or – x + y = 2

x 0 -2
y 2 0

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 6
i) If x = 4 then y = 6 (∵ from the graph)
ii) When y = – 3 then x = – 5 (∵ from the graph)

Question 5.
Draw the graph of the equation 2x + 3y = 12. Find the solutions from the graph,
(i) Whose y-coordinate is 3 (OR) Whose y-coordinate is 2
(ii) Whose x-coordinate is – 3
Solution:
The given equation is 2x + 3y = 12

x 0 6
y 4 0

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 7
i) From the graph when y = 3 then 2x + 3(3) = 12 ⇒ 2x + 9 = 12 ⇒ 2x = 3 ⇒ x = \(\frac{3}{2}\) ; solution is (\(\frac{3}{2}\) , 3) (OR) When y = 2 then 2x + 3(2) = 12 ⇒ 2x + 6 = 12 ⇒ 2x = 6 ⇒ x = 3 solution is (3, 2),
ii) From the graph when x = – 3 then y = 6; solution is (- 3, 6)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 6.
Draw the graph of each of the equations given below and also find the coordinates of the points where the graph cuts the coordinate axes.
i) 6x – 3y = 12
Solution:
6x – 3y = 12

x 0 2
y -4 0

From the graph the line cuts the X-axis at (2, 0) and Y-axis at (0, – 4).

ii) -x + 4y = 8
Solution:

x 0 -8
y 2 0

From the graph the line cuts the X-axis at (- 8, 0) and Y -axis at (0, 2).

iii) 3x + 2y + 6 = 0
Solution:
3x + 2y + 6 = 0

x 0 -2
y -3 0

From the graph the line cuts the X-axis at (- 2, 0) and Y -axis at (0, -3).
AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 8

Question 7.
Rajiya and Preethi two students of class IX together collected ₹1000 for the Prime Minister Relief Fund for victims of natural calamities. Write a linear equation and
draw a graph to depict the statement. Clfp)
Solution:
Let Rajiya’s contribution to P.M.R.F be = ₹ x
Preethi’s contribution to P.M.R.F be = ₹ y
Then by problem x + y = 1000

x + y = 1000
x 200 300
y 800 700

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 9

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 8.
Gopaiah sowed w heat and paddy in two fields of total area 5000 sq. meters. Write a linear equation and draw a graph to represent the same.
Solution: Let the wheat be sowed in a land equal to x sq.m,
and the paddy be sowed in a land equal to y sq.m.
∴ By problem x + y = 5000

x + y = 5000
x 1000 2000
y 4000 3000

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 10

Question 9.
The force applied on a body of mass 6 kg. is directly proportional to the acceleration produced in the body. Write an equation to express this observation and draw the graph of the equation.
Solution:
Let the lorce = f; mass = 6 kg; acceleration = a
By problem f ∝ a or f = m . a ⇒ f = 6a

f = 6a
a 2 3
f 12 18

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 11

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 10.
A stone is falling from a mountain. The velocity of the stone is given by v = 9.8t.
Draw its graph and find the velocity of the stone 4 seconds after start.
Solution:
Given that, the velocity of the stone v = 9.8 t

v = 9.8t
v 49 98
t 5 10

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 12
The velocity after 4 seconds = v = 9.8 × 4 = 39.2 m/sec2.

Question 11.
In an election 60 % of voters cast their votes. Form an equation and draw the graph for this data. Find the following from the graph.
i) The total number of voters, if 1200 voters cast their votes,
ii) The number of votes cast, if the total number of voters are 800.
[Hint: If the number of voters who cast their votes be ‘x’ and the total number of voters be ‘y’ then x = 60 % of y.]
Solution:
Let the total number of votes be = y
Then the number of voters who cast their votes = x
By problem x = 60 % of y

x 1200 480
y 2000 800

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 13
i) From the graph when x = 1200, then y = 2000
ii) From the graph when y = 800 then x = 480

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 12.
When Rupa was born, her father was 25 years old. Form an equation and draw a graph for this data. From the graph find
i) The age of the father when Rupa is 25 years old.
ii) Rupa’s age when her father is 40 years old.
Solution:
Let her father’s age be = x years.
and Rupa’s age be = y years
By problem x – y = 25 years

x 40 50
y 15 25

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 14
i) From the graph age of the father when Rupa is 25 years is 50 years.
ii) Rupa’s age when her father is 40 years is 15 years.

Question 13.
An auto charges ₹15 for first kilometre and ₹ 8 each for subsequent kilometre. For a distance of x km. an amount of ₹y is paid. Write the linear equation representing this information and draw the graph. With the help of graph find the distance travelled if the fare paid is ₹55. How much would have to be paid for 7 kilometres?
Solution:
Charge for the first kilometre = ₹15
Charge for the subsequent kilometres = ₹ 8 per km.
Amount paid = ₹ y when the distance travelled is x km
∴ By problem y = 15 + 8x
∴ 8x – y + 15 = 0

x 2 1
y 31 23

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 15
i) When y = 55 then x = 5
ii) From the graph when x = 7 then y = 71

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 14.
A lending library has fixed charge for the first three days and an additional charges for each day thereafter. John paid ₹ 27 for a book kept for seven days. If the fixed charges be ₹ x and subsequent per day charges be ₹ y; then write the linear equation representing the above information and draw the graph of the same. From the graph if the fixed charge is ₹ 7, find the subsequent per day charge. And if the per day charge is ₹ 4, find the fixed charge, (charge is ₹7)
Solution:
John kept a book for 7 days. He paid ₹ 27
For first three days = ₹ x (fixed)
For the last four days = ₹ 4y (? y for a day)
By problem x + 4y = 27

x 3 11 7
y 6 4 5

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 16
When x = 7 then y = 5
When y = 4 then x = 11

Question 15.
The parking charges of a car in Hyderabad Railway station for first two hours is ₹ 50 and ₹10 for each subsequent hour. Write down an equation and draw the graph. Find the following charges from the graph.
i) For three hours ii) For six hours iii) How many hours did Rekha park her car if she paid ₹ 80 as parking charges ?
Solution: Let the total money paid be = ₹ y
Parking charges for the first two hours = ₹ 50
Parking charges for total x hours @ ₹10 per hour = 50 + (x – 2) 10
= 50 + 10x – 20
= 10x + 30
∴ By problem y = 10x + 30

x 3 5 6
y 60 80 90

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 17
i) For three hours = 50 + 10 × 1 = ₹ 60
[ ∵ From the graph we see the same]
ii) For six hours = 50 + 10 × 4 = 50 + 40 = ₹ 90
iii) Rekha parked her car for 5 hours.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 16.
Sameera was driving a car with uniform speed of 60 kmph. Draw distance – time graph. From the graph find the distance travelled by Sameera in
i) \(\frac { 1 }{ 2 }\) hours
ii) 2 hours
iii) 3\(\frac { 1 }{ 2 }\) hours
Solution:
Speed of the car = 60 kmph
Let the time taken be = x hours
Then the total distance travelled be = y hours
By problem 60x = y or 60x – y = 0

x 2 4 5
y 120 240 300

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 18
ii) Distance travelled in 1\(\frac { 1 }{ 2 }\) hours = 120 km
ii) Distance travelled in 2 hours = 120 km
iii) Distance travelled in 3\(\frac { 1 }{ 2 }\) hours = 210 km

Question 17.
The ratio of molecular weight of Hydrogen and Oxygen in water is 1 : 8. Set up an equation between Hydrogen and Oxygen and draw its graph. From the graph find the quantity of Hydrogen if Oxygen is 12 grams. And quantity of Oxygen if
Hydrogen is \(\frac { 3 }{ 2 }\) grams.
[Hint : If the quantities of hydrogen and oxygen ‘x’ and ‘y’ respectively, then
x : y = 1 : 8 ⇒ 8x = y]
Solution:
Let the quantity of Hydrogen = x grams
And the quantity of Oxygen = y grams
By problem 8x = y ⇒ 8x – y = 0

x 1 2 4 5
y 8 16 32 40

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 19
From the graph, the quantity of Hydrogen if Oxygen is 12 gm = \(\frac { 3 }{ 2 }\) g.
From the graph, the quantity of Oxygen if Hydrogen is \(\frac { 3 }{ 2 }\) g = 12 g.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 18.
In a mixture of 28 litres, the ratio of milk and water is 5 : 2. Set up the equation between the mixture and milk. Draw its graph. By observing the graph find the quantity of milk in the mixture.
[Hint: Ratio between mixture and milk = 5 + 2:5 = 7:5]
Solution:
Let the quantity of milk in the mixture be = x lit.
quantity of the mixture = y lit.
Ratio of the milk and water = 5:2
Sum of the terms of the ratio = 5 + 2 = 7
Quantity of milk x = \(\frac { 5 }{ 7 }\) y lit.
7x = 5y ⇒ 7x – 5y = 0

x 10 20 25
y 14 28 35

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 20
From the graph quantity of milk in the mixture = 20 lit.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3

Question 19.
In countries like U.S.A. and Canada temperature is measured in Fahrenheit whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius F = \(\frac { 9 }{ 5 }\) C + 32
i) Draw the graph of the above linear equation having Celsius on X-axis and Fahrenheit on Y-axis.
ii) If the temperature is 30°C, what is the temperature in Fahrenheit ?
iii) If the temperature is 95°F, what is the temperature in Celsius ?
iv) Is there a temperature that has numerically the same value in both Fahrenheit and Celsius ? If yes, find it.
Solution:
i) Given that F = \(\frac { 9 }{ 5 }\) C + 32

C 20 30 35 -40
F 68 86 95 -40

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.3 21
If C = 20, F = \(\frac { 9 }{ 5 }\) × 20 + 32 = 68
If c = 30, F = \(\frac { 9 }{ 5 }\) × 30 + 32 = 86
If C = 35, F = \(\frac { 9 }{ 5 }\) × 35 + 32 = 95
If C = -40, F = \(\frac { 9 }{ 5 }\) × (-40) + 32 = -40

ii) From the graph 30°C = 86°F
iii) 95°F = 35°C
iv) When C = – 40 then F = – 40

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.2

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

Question 1.
Find three different solutions of the each of the following equations.
i) 3x + 4y = 7
Solution:
Given equation is 3x + 4y = 7

Choice of  value of x Or y Simplification for y or x Solution
x = 0

 

3 x 0 + 4y = \(\frac{7}{4}\) (0, \(\frac{7}{4}\) )

 

y = 0

 

3x + 4(0) = 7 ⇒ x = \(\frac{7}{3}\) (\(\frac{7}{3}\) ,0)
x = 1

 

3(1) + 4y = 7

⇒ y = \(\frac{7-3}{-4}\) = 1

(1, 1)

Choice of x or y Simplification for y or x
Solution

ii) y = 6x
Solution: Given equation is y = 6x ⇒ 6x – y = 0

Choice of  value of x Or y Simplification for y or x Solution
x = 0 6(0) – y = 0 ⇒ y = 0 (0,0)
y = 0 6x – 0 = 0 ⇒ x = 0 (0,0)
x = 1 6(1) – y = 0 ⇒ y = 6 (1,6)
Y = 1 6x – 1 = 0 ⇒ 6x = 1 ⇒ x = \(\frac{1}{6}\) (\(\frac{1}{6}\),1)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

iii) 2x – y = 7
Solution:
Given equation is 2x – y = 7

Choice of  value of x Or y Simplification for y or x Solution
x = 0 2(0) – y = 7 ⇒ y = -7 (0, -7)
y = 0 2x – 0 = 7 ⇒ x = \(\frac{7}{2}\) (\(\frac{7}{2}\) , 0)
x = 1 2(1) – y = 7 ⇒ -y = 7 – 2  ⇒ y = -5 (1, -5)

iv) 13x – 12y = 25
Solution:
Given equation is 13x – 12y = 25

Choice of  value of x Or y Simplification for y or x Solution:
x = 0

 

13(0) – 12y = 25 ⇒ y = \(-\frac{25}{12}\) (0, \(-\frac{25}{12}\) )

 

y = 0

 

13x – 12(0) = 25 ⇒ y = \(\frac{25}{13}\) (\(\frac{25}{13}\) ,0)
x = 1

 

13(1) – 12y = 25
⇒ -12y = 25 – 13
y = \(\frac{12}{-12}\) = -1
(1, -1)

v) 10x + 11y = 21
Solution:
Given equation is 10x + 11y = 21

Choice of  value of x Or y Simplification for y or x Solution
x = 0 10(0) + 11y = 21 ⇒ y = \(\frac{21}{11}\) (0, \(\frac{21}{11}\))
y = 0 10x +11(0) = 21 ⇒ x = \(\frac{21}{10}\) (\(\frac{21}{10}\) , 0)
x = 1 10(1) + 11y = 21 ⇒ 11y = 21 – 10  ⇒ y = \(\frac{11}{11}\) = 1 (1, 1)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

vi) x + y = 0
Solution:
Given equation is x + y = 0

Choice of a value of x or y Simplification Solution
x = 0 0 + y = 0 ⇒ y = 0 (0, 0)
x = 1 1 + y = 0 ⇒ y = -1 (1,-1)
y = 1 x + 1 = 0 ⇒ x = -1 (- 1, 1)

Question 2.
If (0, a) and (b, 0) are the solutions of the following linear equations. Find a and b.
8x – y = 34
Solution:
Given that (0, a) and (b, 0) are the solutions of 8x – y = 34
∴ 8(0) – a = 34 and 8(b) – 0 = 34
⇒ a = -34 and b = \(\frac{34}{8}=\frac{17}{4}\)

ii) 3x = 7y – 21
Solution:
Given that (0, a) and (b, 0) are the solutions of 3x = 7y – 21
⇒ 3x – 7y = – 21
∴ 3(0) – 7(a) = -21 and 3(b) – 7(0) = -21
⇒ a = \(\frac{-21}{-7}\) and b = \(\frac{-21}{3}\)
⇒ a = 3 and b = -7

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

iii) 5x – 2y + 3 = 0
Given that (0, a) and (b, 0) are the solutions of 5x – 2y = – 3
i. e., 5(0) – 2a = – 3 and 5 (b) – 2(0) = – 3
⇒ -2a = -3 and 5b = -3
⇒ a = \(\frac{3}{2}\) and b = \(\frac{-3}{5}\)

Question 3.
Check which of the following is solution of the equation 2x – 5y = 10.
(i) (0, 2) (ii) (0,-2) (iii)(5, 0) (iv) (2√3, -√3) (v) (\(\frac{1}{2}\) , 2)
Solution:
i) (0, 2)
The given equation is 2x – 5y = 10
On substituting (0, 2), the L.H.S becomes
2(0)-5(2) = 0-10 = -10
R.H.S = 10
L.H.S ≠ R.H.S
∴ (0, 2) is not a solution.

ii) (0. – 2)
Substituting (0, – 2) in the L.H.S of 2x – 5y = 10, we get
2(0) – 5 (- 2) = 0 + 10 = 10 = R.H.S
∴ (0, – 2) is a solution.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

iii) (5, 0)
Substituting (5, 0) in the L.H.S of 2x – 5y = 10, we get
2(5)-5(0) = 10-0 = 10 = R.H.S
∴ (5, 0) is a solution.

iv) 2√3, -√3
On substituting (2√3, -√3 ), the L.H.S becomes
2(2√3) – 5(-√3) = 4√3 + 5√3 = 9√3 ≠ R.H.S
∴ (2√3, -√3) is not a solution.

v) (\(\frac{1}{2}\), 2)
Given equation is 2x – 5y = 10
Put x = \(\frac{1}{2}\) and y = 2 in the given equation.
Then 2(\(\frac{1}{2}\)) – 5(2) =10
1-10 = 10
-9 = 10 false
∴ (\(\frac{1}{2}\) , 2) is not a solution.

Question 4.
Find the value of k, if x = 2; y = 1 is a solution of the equation 2x + 3y = k. Find two more solutions of the resultant equation. £3 Each
Solution:
Given that x = 2 and y = 1 is a solution of 2x + 3y = k .
∴ 2(2) + 3(1) = k ⇒ 4 + 3 = k ⇒ k = .7 ’
∴ The equation becomes 2x + 3y = 7

x 0 1
y 2(0) + 3y = 7
3y = 7
y = \(\frac{7}{3}\)
2(1) + 3y = 7
3y = 7 – 2= y = \(\frac{5}{3}\)
(x, y) (0,\(\frac{7}{3}\)) (1, \(\frac{5}{3}\))

Two more solutions are (0, \(\frac{7}{3}\)) and (1, \(\frac{5}{3}\))

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

Question 5.
If x = 2 – α and y = 2 + α is a solution of the equation 3x – 2y + 6 = 0, find the value of ‘α’. Find three more solutions of the resultant equation.
Solution:
Given that x = 2 – α and y = 2 + α is a solution of 3x – 2y + 6 = 0
Thus 3 (2 – α) – 2 (2 + α) + 6 = 0
⇒ 6 – 3α – 4 – 2α + 6 = 0
⇒ -5α + 8 = 0
⇒ -5α = -8
∴ α = \(\frac{8}{5}\)
Three more solutions are

x 0 3x – 2y = -6
3x – 2(0) = -6
y = \(\frac{-6}{3}\) = -2
1
y 3x – 2y = -6
3(0) – 2y = -6
y = 3
0 3x – 2y = -6
3(1) – 2y = -6
y = \(\frac{9}{2}\)
Solutions (0, 3) (-2, 0) (1, \(\frac{9}{2}\))

Question 6.
If x = 1;y = 1 is a solution of the equation 3x + ay = 6, find the value of ‘a’.
Solution:
Given that x = 1; y = 1 is a solution of 3x + ay = 6.
Thus 3(1) + a(1) = 6
⇒ 3 + a = 6 ⇒ a = 6 – 3 = 3

Question 7.
Write five different linear equations in two variables and find three solutions for
each of them.
Solution:
i) Let the equations are 2x – 4y = 10

x 0 1 2
y 2(0) – 4y = 10
y = \(\frac{-5}{2}\)
2x – 4y = 10
2(1)  – 4y =10y = -2
2x – 4y = 10
2(2) – 4y = 10
y = –\(\frac{-3}{2}\))
Solutions (0, –\(\frac{-5}{2}\)) (1 , -2) (2, \(\frac{-3}{2}\))

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

ii) 5x + 6y = 15

x 0 5x + 6(0) = 15
x = 3
1
y 5(0) + 6y = 15
y = \(\frac{15}{6}\) = \(\frac{5}{2}\)
0 5(1) + 6y = 15
y = \(\frac{5}{3}\)
(x, y) (0, \(\frac{5}{2}\)) (3, 0 ) (1, \(\frac{5}{3}\))

iii) 3x-4y = 12

x 0 4 1
y -3 0 \(\frac{-9}{4}\)
(x, y) (0, -3) (4, 0 ) (1, \(\frac{-9}{4}\))

iv) 2x – 7y = 9

x 0 \(\frac{9}{2}\) 1
y \(\frac{-9}{7}\) 0 -1
(x, y) (0, \(\frac{-9}{7}\)) ( \(\frac{9}{2}\) , 0 ) (1, -1)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

v) 7x- 5y = 3

x 0 \(\frac{3}{7}\) 1
y \(\frac{-3}{5}\) 0 \(\frac{4}{5}\)
(x, y) (0, \(\frac{-3}{5}\) ) (\(\frac{3}{7}\), 0 ) (1, \(\frac{4}{5}\))

 

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.1

Question 1.
Express the following linear equations in the form of ax + by + c = 0 and indicate the values of a, b and c in each case.
i) 8x + 5y – 3 = 0
Solution:
8x + 5y – 3 = 0
⇒ 8x + 5y + (- 3) = 0
Here a = 8, b = 5 and c = – 3

ii) 28x – 35y = – 7
Solution:
28x – 35y = – 7
⇒ 28x + (- 35) y + 7 = 0
Here a = 28, b = – 35 and c = 7

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

iii) 93x = 12- 15y
Solution:
93x = 12 – 15y
⇒ 93x + 15y -12 = 0
⇒ 93x + 15y + (- 12) = 0
Here a = 93, b = 15 and c = – 12

iv) 2x = – 5y
Solution:
2x = – 5y
⇒ 2x + 5y = 0
Here a = 2, b = 5 and c = 0

v) \(\frac{x}{3}+\frac{y}{4}=7\)
Solution:
\(\frac{x}{3}+\frac{y}{4}=7\)
⇒ \(\frac{x}{3}+\frac{y}{4}-7=0\)
⇒\(\frac{4 x+3 y-84}{12}=0\)
⇒ 4x + 3y – 84 = 0
Here a = 4, b = 3 and c = – 84

vi) y = \(-\frac{3}{2} x\)
Solution:
y = \(-\frac{3}{2} x\)
⇒ 2y = -3x
⇒ 3x + 2y = 0
Here a = 3, b = 2 and c = 0

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

vii) 3x + 5y = 12
Solution:
3x + 5y = 12
⇒ 3x + 5y + (- 12) = 0
Here a = 3, b = 5 and c = – 12

Question 2.
Write each of the following in the form of ax + by + c = 0 and find the values of a, b and c.
i) 2x = 5
Solution:
2x – 5 = 0
a = 2
b = 0
c = -5

ii) y – 2 = 0
Solution:
y – 2 = 0
a = 0
b = 1
c = – 2

iii) \(\frac{y}{7}\) = 3
Solution:
\(\frac{y}{7}\) = 3
y = 21
y – 21 = 0
a = 0
b = 1
c = -21

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

iv) x = \(-\frac{14}{13}\)
x = \(-\frac{14}{3}\)
⇒ 13x = – 14
⇒ 13x + 14 = 0
a = 13
b = 0
c = 14

Question 3.
Express the following statements as a linear equation in two variables,
i)The sum of two numbers is 34.
Solution:
x + y = 34; x, y are any two numbers ⇒ x + y – 34 = 0

ii) The cost of a ball pen is ?5 less than half the cost of a fountain pen.
Solution:
Let the cost of a fountain pen = x
Let the cost of ball pen = y
Then y = x – 5 or x – y – 5 = 0

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

iii) Bhargavi got 10 more marks than double of the marks of Sindhu. |l M)
Solution:
Let Sindhu’s marks = x
Bhargavi’s marks = y
Then by problem y = 2x + 10 or 2x – y + 10 = 0

iv) The cost of a pencil is ₹2 and one ball point pen costs ₹15. Sheela pays ₹100 for the pencils and pens she purchased.
Solution:
Giver: that cost of a pencil = ₹2
Cost of a ball point pen = ₹15
Let the number of pencils purchased = x
Let the number of pens purchased = y
Then the total cost of x – pencils = 2x
Then the total cost of y – pens = 15y
By problem 2x + 15y = 100

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.3

Question 1.
Plot the following points on the Cartesian plane whose x, y co-ordinates are given.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 1
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 1 (i)
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 2

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 2.
Are the positions of (5, -8) and (-8, 5) Is same ? JustIfy your answer.
Solution:
The positions of (5, -8) and (-8, 5) are not same. They are two distinct points. (5, -8) lies at a distance of 5 units from Y – axis and 8 units from X – axis on down side of the origin. So it lies in Q4. Where as (- 8, 5) lies in Q2. The point is at a distance of 8 units from Y – axis on left side of the origin and 5 units from X – axis.

Question 3.
What can you say about the position of the points (1, 2), (1, 3), (1, – 4), (1, 0) and (1, 8), Locate on a graph sheet.
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 3
All the given points lie on a line parallel to Y-axis at a distance of 1 cm.

Question 4.
What can you say about the position of the points (5, 4), (8, 4), (3, 4), (0, 4), (-4, 4), (-2,4)? Locate the points on a graph sheet and justify your answer.
Solution:
The points (5, 4), (8, 4), (3, 4), (0, 4), (- 4, 4) and (- 2, 4) all lie on a line parallel to X-axis at a distance of 4-units from it.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 4

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 5.
Plot the points (0, 3), (4, 3), (3, 4) (4, 0) in graph sheet. Join the points with straight lines to make a rectangle. Find the area of the rectangle.
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 5
From the graph, area of the rectangle =12 square units (OR)
Length = 4 units ; breadth = 3 units
A = l/b = 4 × 3 = 12 sq. units.

Question 6.
Plot the points (2, 3), (6, 3) and (4, 7) in a graph sheet. Join them to make it a triangle. Find the area of the triangle.
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 6
From the graph
Base of the triangle = 4 units
Height of the triangle = 4 units
∴ Area of the triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) × 4 × 4 = 8sq. units.

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 7.
Plot at least six points in a graph sheet, each having the sum of its co-ordinates equal to 5. [Hint: (- 2, 7), (1, 4)………….]
Solution:
Given that (x co-ordinate) + (y co-ordinate) = 5
Let the points be
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 7

Question 8.
Look at the graph. Write the co-ordinates of the points
A, B, C, D, E, F. G. H, 1, J, K, L, M, N, O, P and Q.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 8
Solution:
A(- 3, 4) ; B(0, 5) : C (3, 4) ; D (2, 4) ; E (2, 0) ;
F (3, 0) ; G (3, – 1) ; H (0, – 1) ; I (- 3, – 1) ;
J (- 3, 0) ; K (- 2, 0) ; L (- 2, 4) ; M (- 1, 0) ;
N (-1, 3); O (0, 0) ; P (1. 3) and Q (1, 0)

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 9.
In a graph sheet plot each pair of points, join them by line segments.
i) (2, 5), (4, 7)
ii) (-3, 5) (-1,7)
iii) (-3, -4), (2, -4)
iv) (-3, -5) (2, -5)
v) (4, -2), (4, -3)
vi) (-2, 4), (-2, 3)
vii) (-2, 1), (-2, 0)
Now join the following pairs of points by straight line segments, in the same graph.
viii) (-3, 5), (-3, 4)
ix) (2, 5), (2, -4)
x) (2, -4), (4, -2)
xi) (2, -4), (4, -3)
xii) (4, -2), (4, 7)
xiii) (4, 7), (-1, 7)
xiv) (-3, 2), (2, 2)
Now you will get a surprise figure. What is it ?
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 9

AP Board 9th Class Maths Solutions

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.2

Question 1.
Write the quadrant in which the following points lie.
i) (- 2, 3)
ii) (5, – 3)
iii) (4, 2)
iv) (- 7, – 6)
v) (0, 8)
vi) (3, 0)
vii) (-4,0)
viii) (0, – 6)
Solution:
i) (- 2, 3) – Q2 (second quadrant)
ii) (5, – 3) – Q4 (fourth quadrant)
iii) (4, 2) – Q1 (Iirst quadrant)
iv) (- 7, – 6) – Q3 (third quadrant)
v) (0, 8) – on Y-axis
vi) (3, 0) – on X-axis
vii) (-4,0) – on X’ – axis
viii) (0, – 6) – on Y’: axis

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 2.
Write the abscissae and ordinates of the following points.
i)(4,-8)
ii)(-5,3)
iii)(0,0)
iv)(5, 0)
v)(0, -8)
Note: Plural of abscissa is abscissae.
Solution:

Point abscissa ordinate
i) (4, – 8) 4 -8
ii)  (-5, 3) -5 3
iii) (0,0) 0 0
iv) (5,0) 5 0
v) (0,-8) . 0 -8

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 3.
Which of the following points lie on the axes ? Also name the axis.
i) (-5,-8)
ii) (0, 13)
iii) (4, – 2)
iv) (- 2,0)
v) (0, – 8)
vi) (7,0)
vii) (0,0)
Solution:
The points (ii) (0,13) ; (v) (0,- 8) lie on Y – axis.
The points (iv) (- 2, 0), (vi) (7, 0) lie on X – axis.
The point (vii) (0, 0) lie on both X – axis and Y – axis.
The points (i) (- 5, – 8); (iii) (4, – 2) do not lie on any axis.

Question 4.
Write the following based on the graph.
i) The ordinate of L
ii) The ordinate of Q
iii) The point denoted by (- 2,-2)
iv) The point denoted by (5, – 4)
v) The abscissa of N
vi) The abscissa of M
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 1
Solution:
i) – 7
ii) 7
iii) The point ’R’
iv) The point P
v) 4
vi) – 3

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 5.
State true or false and write correct statement.
i) In the Cartesian plane the horizon-tal line is called Y – axis, if) in the Cartesian plane, the vertical line is called Y – axis.
iii) The point which lies on both the axes is called origin.
iv) The point (2, – 3) lies in the third quadrant.
v) (-5, -8) lies in the fourth quadrant.
vi) The point (- x, – y) lies in the first quadrant where x < 0; y < 0.
Solution:
i) False
Correct statement: In the Cartesian plane the horizontal line is called X – axis.
ii) True
iii) True
iv) False
Correct statement: The point (2, -3) lies in the fourth quadrant.
v) False
Correct statement: (- 5, – 8) lies in the third quadrant.
vi) True

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 6.
Plot the following ordered pairs on a graph sheet. What do you observe ?
i) (1, 0), (3, 0), (- 2, 0), (- 5, 0), (0, 0), (5, 0), (- 6, 0)
ii) (0, 1), (0, 3), (0, – 2), (0, – 5), (0, 0), (0,5), (0,-6)
Solution:
i) All points lie on X – axis,
ii) All points lie on Y – axis.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 2

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.1

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.1

Question 1.
In a locality, there is a main road along North – South direction. The map is given below. With the help of the picture answer the following questions.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.1
i) What is the 3rd object on the left side in street no. 3 while going in east direction ?
ii) Find the name of the 2nd house which is on right side of street 2 while going in east direction.
iii) Locate the position of Mr. K’s house.
iv) How do you describe the position of the post office ?
v) How do you describe the location of the hospital ?
Solution:
i) Water tank
ii) Mr. J’s house
iii) In street No. 2, 3rd house on right side.
iv) In street No. 4, the first house on right side.
v) In street No. 4, the last house on left side.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4

Question 1.
In the given triangles, find out ∠x, ∠y and ∠z
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 1
Solution:
In fig(i)
x° = 50° + 60°
(∵ exterior angle is equal to sum of the opposite interior angles)
∴ x= 110°

In fig (ii)
z° = 60° + 70°
(∵ exterior angle is equal to sum of the opposite interior angles)
∴ z = 130°

In fig (iii)
y° = 35° + 45° = 80°
(∵ exterior angle is equal to sum of the opposite interior angles)

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

Question 2.
In the given figure AS // BT; ∠4 = ∠5, \(\overline{\mathbf{S B}}\) bisects ∠AST. Find the measure of ∠1.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 2
Solution:
Given AS // BT
∠4 = ∠5 and SB bisects ∠AST.
∴ By problem
∠2 = ∠3 …………..(1)
For the lines AS // BT
∠2 = ∠5 ( ∵alternate interior angles)
∴ In ΔBST
∠3 = ∠5 = ∠4
Hence ΔBST is equilateral triangle and each of its angle is equal to 60°.
∴∠3 = ∠2 = 60° [by eq. (1)]
Now ∠1 + ∠2 + ∠3 = 180°
∠1 + 60° + 60° = 180°
[ ∵ angles at a point on a line]
∴∠1 = 180° – 120° = 60°

Question 3.
In the given figure AB // CD; BC // DE then find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 3
Solution:
Given that AB // CD and BC // DE.
∴ 3x = 105° (∵ alternate interior angles for AB // CD)
x = \(\frac { 105° }{ 3 }\) = 35°
Also BC // DE
∴∠D = 105°
(∵ alternate interior angles)
Now in ΔCDE
24° + 105° + y = 180°
(∵ angle sum property)
∴ y = 180° – 129° = 51°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

Question 4.
In the given figure BE ⊥ DA and CD ⊥ DA then prove that m∠1 = m∠3.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 4
Solution:
Given that CD ⊥ DA and BE ⊥ DA.
⇒ Two lines CD and BE are perpendicular to the same line DA.
⇒ CD // BE (or)
∠D =∠E ⇒ CD // BE
(∵ corresponding angles for CD and BE and DA are transversal)
Now m∠1 = m∠3
(∵alternate interior angles for the lines CD // BE ; DB are transversal)
Hence proved.

Question 5.
Find the values of x, y for which lines AD and BC become parallel.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 5
Solution:
For the lines AD and BC to be parallel x – y = 30° (corresponding angles) ……… (1)
2x = 5y ………….(2)
(∵ alternate interior angles)
Solving (1) & (2)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 6
y = \(\frac{60}{3}\) = 20°
Substituting y = 20° in eq. (1)
x – 20° = 30°
⇒ x = 50°
∴ x = 50° and y = 20°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

Question 6.
Find the values of x and y in the figure.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 7
Solution:
From the figure y + 140° = 180°
(∵ linear pair of angles)
∴ y = 180° – 140° = 40°
And x° = 30° + y°
(∵ exterior angle = sum of the opposite interior angles)
x° = 30° + 40° = 70°

Question 7.
In the given figure segments shown by arrow heads are parallel. Find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 8
Solution:
From the figure
x° = 30° (∵ alternate interior angles)
y° = 45° + x° (∵ exterior angles of a triangle = sum of opp. interior angles)
y = 45° + 30° – 75°

Question 8.
In the given figure sides QP and RQ of ∠PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 9
Solution:
Given that ∠SPR = 135° and ∠PQT =110°
From the figure
∠SPR + ∠RPQ = 180°
∠PQT + ∠PQR = 180°
[∵ linear pair of angles]
⇒ ∠RPQ = 180° – ∠SPR
= 180° – 135° = 45°
⇒ ∠PQR = 180° – ∠PQT
= 180°-110° = 70°
Now in APQR
∠RPQ + ∠PQR + ∠PRQ = 180°
[∵ angle sum property]
∴ 45° + ’70° + ∠PRQ = 180°
∴ ∠PRQ = 180°-115° = 65°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

Question 9.
In the given figure ∠X = 62° ; ∠XYZ = 54°. In ΔXYZ. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respec-tively find ∠OZY and ∠YOZ.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 10
Solution:
Given that ∠X = 62° and ∠Y = 54°
YO arid ZO are bisectors of ∠Y and ∠Z.
In ΔXYZ
∠X + ∠XYZ + ∠XZY = 180° .
62° + 54° + ∠XZY = 180°
=> ∠XZY = 180°- 116° = 64°
Also in Δ𝜏OYZ
∠OYZ = 1/2 ∠XYZ = 1/2 x 54° = 27°
(∵ YO is bisector of ∠XYZ)
∠OZY = 1/2 ∠XZY = 1/2 x 64° = 32
(∵ OZ is bisector of ∠XYZ)
And ∠OYZ + ∠OZY + ∠YOZ = 180°
(∵ angle sum property, ΔOYZ)
⇒ 27 + 32° + ∠YOZ = 180°
⇒ ∠YOZ = 180° – 59° = 121°

Question 10.
In the given figure if AB // DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 11
Solution:
Given that AB // DE, ∠CDE = 53°;
∠BAC = 35°
Now ∠E = 35°
( ∵ alternate interior angles)
Now in ∆CDE
∠C + ∠D + ∠E = 180°
(∵angle sum property, ACDE)
∴ ∠DCE + 53° + 35° = 180°
⇒ ∠DCE = 180° – 88° = 92°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

Question 11.
In the given figure if line segments PQ and RS intersect at point T, such that∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 12
Solution:
Given that ∠PRT = 40°; ∠RPT = 95°;
∠TSQ = 75°
In ∆PRT ∠P + ∠R + ∠PTR = 180°
(∵angle sum property)
95° + 40° + ∠PTR = 180°
⇒ ∠PTR = 180° – 135° = 45°
Now ∠PTR = ∠STQ
(∵ vertically opposite angles)
In ΔSTQ ∠S + ∠Q + ∠STQ = 180°
(∵ angle sum property)
75° + ∠SQT + 45° = 180°
∴ ∠SQT = 180° – 120° = 60°

Question 12.
In the given figure, ABC is a triangle in which ∠B = 50° and ∠C = 70°. Sides AB and AC are produced. If ∠ is the measure of angle between the bisec¬tors of the exterior angles so formed, then find ‘z’.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 13
Solution:
Given that ∠B = 50°; ∠C = 70°
Angle between bisectors of exterior angles B and C is ∠.
From the figure
50° + 2x = 180°
70° + 2y = 180°
(∵ linear pair of angles)
∴ 2x= 180°-50°
2x= 130°
x = \(\frac{130}{2}\)
= 65°

2y= 180°-70°
2y= 110°
x = \(\frac{110°}{2}\)
= 55°

Now in ΔBOC
x + y + ∠ = 180° (∵ angle sum property)
65° + 55° + ∠ = 180°
z = 180° -120° = 60°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

Question 13.
In the given figure if PQ ⊥ PS; PQ // SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 14
Solution:
Given that PQ ⊥ PS ; PQ // SR
∠SQR = 28°, ∠QRT = 65°
From the figure
∠QSR = x° (∵ alt. int. angles for the lines PQ // SR)
Also 65° = x + 28° (∵ ext. angles = sum of the opp. interior angles)
∴ x° = 65° – 28° = 37°
And x° + y° = 90°
[ ∵ PQ ⊥ PS and PQ // SR. ⇒ ∠P = ∠S]
37° + y = 90°
∴ y = 90° – 37° = 53°

Question 14.
In the given figure ΔABC side AC has been produced to D. ∠BCD = 125° and ∠A: ∠B = 2:3, find the measure of ∠A and ∠B

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 15
Solution:
Given that ∠BCD = 125°
∠A : ∠B = 2 : 3
Sum of the terms of the ratio
∠A : ∠B = 2 + 3 = 5
We know that ∠A + ∠B = ∠BCD
(∵ exterior angles of triangle is equal to sum of its opp. interior angles)
∴ ∠A = \(\frac{2}{5}\) x 125° = 50°
∠B = \(\frac{3}{5}\) x 125° = 75°

Question 15.
In the given figure, it is given that, BC // DE, ∠BAC = 35° and ∠BCE = 102°. Find the measure of 0 ∠BCA i0 ∠ADE and iii) ∠CED.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 16
Solution:
Given that BC // DE ; ∠BAC = 35°;
∠BCE = 102°

i) From the figure
102° + ∠BCA = 180°
(∵ linear pair of angles)
∴ ∠BCA = 180° – 102° = 78°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

ii) ∠ADE + ∠CBD = 180°
(∵ interior angles on the same side of the transversal)
∠ADE + (78° + 35°) = 180°
(∵ ∠CBD = ∠BAC + ∠BCA)
∴ ∠ADE = 180° – 113° = 67°

iii) From the figure .
∠CED = ∠BCA = 78°
(∵ corresponding angles)

Question 16.
In the given figure, it is given that AB = AC; ∠BAC = 36°; ∠ADB = 45° and ∠AEC = 40°. Find i) ∠ABC
i) ∠ACB iii) ∠DAB iv) ∠EAC.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 17
Solution:
Given that AB = AC; ∠BAC = 36°,
∠ADB = 45°, ∠AEC = 40°
(i) & (ii)
In ∆ABC ; AB = AC
⇒ ∠ABC = ∠ ACB
And 36° + ∠ABC + ∠ACB = 180°
(∵ angle sum property)
∴ ∠ABC = \(\frac{180^{\circ}-36^{\circ}}{2}=\frac{144^{\circ}}{2}=72^{\circ}\)
∠ACB = 72°

iii) From the figure
∠ABD + ∠ABC = 180°
∠ABD = 180° – 72° = 1086
In ΔABD
∠DAB + ∠ABD + ∠D = 180°
∠DAB + 108° + 45° = 180°
∠DAB = 180° – 153° = 27°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4

iv) In ΔADE
∠D + ∠A + ∠E = 180°
45° + ∠A + 40° = 180°
⇒ ∠A = 180° -85° = 95°
But ∠A = ∠DAB + 36° + ∠EAC
95° = 27°, + 36° + ∠EAC
∴ ∠EAC = 95° – 63° = 32°

Question 17.
Using information given in the figure, calculate the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 18
Solution:
From the figure In ∆ACB
34° + 62° + ∠ACB = 180°
(∵ angle sum property)
∴ ∠ACB = 180° – 96° = 84° .
And x + ∠ACB = 180°
(∵ linear pair of angles) .
∴ x + 84° = 180°
x = 180°-84° = 96°
(OR)
x = 34° + 62° = 96°
( ∵ x is exterior angle, ∆ABC)
y = 24° + x°
= 24° + 96° = 120°
(∵ y is exterior angle, ∆DCE)

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.3

Question 1.
It is given that l // m; to prove ∠1 is supplement to ∠8. Write reasons for the Statements.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 1

Solution:

Statement Reasons
i) l //m ∠1 + ∠8 = 180° (exterior angles on the same side of the transversal)
ii) ∠1 = ∠5 corresponding angles
iii) ∠5 + ∠8 = 180° linear pair of angles
iv) ∠1 + ∠8 = 180° exterior angles on the same side of the transversal.
v) ∠1 is supplement is ∠8 exterior angles on the same side of the transversal.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 2.
In the given figure AB || CD; CD || EF and y: z = 3:7 find x.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 2
Solution:
Given that AB//CD; CD//EF.
⇒ AB // EF
Also y : z = 3 : 7
From the figure x + y = 180 …………. (1)
[∵ interior angles on the same side of the transversal]
Also y + z = 180 ………….. (2)
Sum of the terms of the ratio y : z
= 3 + 7 = 10
∴ y = \(\frac{3}{10}\) x 180° = 54°
y = \(\frac{7}{10}\) x 180° = 126°
From (1) and (2)
x + y = y + z
⇒ x = z = 126°

Question 3.
In the given figure AB//CD; EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 3
Solution:
Given that EF ⊥ CD; ∠GED = 126°
i. e., ∠FED = 90° and
∠GEF = ∠GED – ∠FED
∠GEF = 126° – 90° = 36°
In ∆GFE
∠GEF + ∠FGE + ∠EFG = 180°
36 + ∠FGE + 90° = 180°
∠FGE = 180° – 126° = 54°
∠AGE = ∠GFE + ∠GEF
(exterior angle in ∆GFE)
= 90°+ 36°= 126°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 4.
In the given figure PQ//ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. [Hint : Draw a line parallel to ST through point R.]
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 4
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 5
Given PQ // ST
Draw a lipe ‘l’ parallel to ST through R.
From the figure
a + 110° = 180° and c + 130 = 180°
[ ∵ Interior angles on the same side of the transversal]
∴ a = 180° -110° = 70°
c = 180° – 130° = 50°
Also a + b + c = 180° (angles at a point on a line)
70° + b + 50° = 180°
b = 180° – 120° = 60°
∴ ∠QRS = 60°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 5.
In the given figure m // n. A, B are any two points on in and n respectively. Let C be an interior point between the lines m and n. Find ∠ACB.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 6
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 7
Draw a line ‘l’ parallel to m and n through C.
From the figure
x = a [ ∵ alternate interior angles for l, m]
y = b [ ∵ alternate interior angles for l, n]
∴ z = a + b = x + y

Question 6.
Find the values of a and b, given that p // q and r // s.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 8
Solution:
Given that p // q and r // s.
∴ From the figure
2a = 80° (∵ corresponding angles)
a = \(\frac { 80° }{ 2 }\) = 40°
Also 80° + b = 180° ( ∵ interior angles on the same side of the transversal)
∴ b = 180° – 80° = 100°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 7.
If in the figure a // b and c // d, then name the angles that are congruent to (i) ∠1 and (ii) ∠2.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 9
Solution:
Given that a // b and c // d.
∠1 = ∠3 (∵ vertically opposite angles)
∠1 = ∠5 (∵ corresponding angles)
∠1 = ∠9 (∵ corresponding angles)
Also ∠1 = ∠3 = ∠5 = ∠7 ;
∠1 = ∠11 = ∠9 = ∠13 = ∠15
Similarly ∠2 = ∠4 = ∠6 = ∠8
Also ∠2 = ∠10 = ∠12 = ∠14 = ∠16

Question 8.
In the figure the arrow head segments are parallel, find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 10
Solution:
From the figure
y = 59° ( ∵ alternate interior angles)
x = 60° ( ∵ corresponding angles)

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 9.
In the figure the arrow head segments are parallel then find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 11
Solution:
From the figure 35° + 105° + y = 180°
∴ y = 180° – 140°
= 40°
∴ x = 40° (∵ x, y are corresponding angles)

Question 10.
Find the values of x and y from the figure.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 12
Solution:
From the figure 120° + x = 180°
(∵ exterior angles on the same side of the transversal)
∴ x = 180° – 120°
x = 60°
Also x = (3y + 6)
(∵ corresponding angles)
3y + 6 = 60°
3y = 60° – 6° = 54°
y = \(\frac { 54 }{ 3 }\) = 18°
∴ x = 60°; y = 18°

Question 11.
From the figure find x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 13
Solution:
From the figure
52° + 90° + (3y + 5)° = 180°
(∵ interior angles of a triangle)
∴ 3y + 147 = 180°
⇒ 3y = 33°
⇒ y = \(\frac { 33 }{ 3 }\) = 11°
Also x + 65° + 52° = 180°
(∵ interior angles on the same side of the transversal)
∴ x = 180° -117° = 63°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 12.
Draw figures for the following statement.
“If the two arms of one angle are respectively perpendicular to the two arms of another angle then the two angles are either equal or supplementary
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 14
AO ⊥ PQ, OB ⊥ QR
Angles are supplementary.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 15
AO ⊥ PQ, OB ⊥ QR
Angles are equal.

Question 13.
In the given figure, if AB // CD; ∠APQ = 50° and ∠PRD = 127°, find x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 16
Solution:
Given that AB // CD.
∠PRD = 127°
From the figure x = 50°
(∵ alternate interior angles)
Also y + 50 = 127°
(∵ alternate interior angles)
∴ y = 127-50 = 77°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 14.
In the given figure PQ and RS are two mirrors placed parallel to each other.
An incident ray \(\overline{\mathrm{AB}}\) strikes the mirror PQ at B, the reflected ray moves along the path \(\overline{\mathrm{BC}}\) and strikes the mirror RS at C and again reflected back along CD. Prove that AB // CD. [Hint : Perpendiculars drawn to parallel lines are also parallel]
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 17
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 18
Draw the normals at B and C.
then ∠x = ∠y (angle of incidence angle of reflection are equal)
∠y = ∠w (alternate interior angles)
∠w = ∠z (angles of reflection and incidence)
∴ x + y = y + z (these are alternate interior angles to \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{CD}}\))
Hence AB // CD.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 15.
In the figures given below AB // CD. EF is the transversal intersecting AB and CD at G and H respectively. Find the values of x and y. Give reasons.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 19
Solution:
For fig(i)
3x = y (∵ alternate interior angles)
2x + y = 180° (∵ linear pair of angles)
∴ 2x + 3x = 180°
5x= 180°
x = \(\frac { 180 }{ 5 }\) = 36°
and y = 3x = 3 x 36 = 108°

For fig (ii)
2x + 15 = 3x- 20°
(∵ corresponding angles)
2x-3x = -20-15
– x = – 35
x = 35°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

For fig (iii)
(4x – 23) + 3x = 180° ,
( ∵ interior angles on the same side of the transversal)
7x- 23 = 180°
7x = 203
x = \(\frac { 203 }{ 7 }\) = 29°

Question 16.
In the given figure AB // CD, ‘t’ is a transversal intersecting E and F re-spectively. If ∠2 : ∠1 = 5 : 4, find the measure of each marked angles.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 20
Solution:
Given that AB // CD and ∠2 : ∠1 = 5 : 4 ∠1 + ∠2 = 180° (-. linear pair of angles) Sum of the terms of the ratio ∠2 :∠1 = 5 + 4 = 9
∴ ∠1 = \(\frac { 4 }{ 9 }\) x 180° = 80°
∠2= \(\frac { 5 }{9 }\) x 180° = 100°
Also ∠1, ∠3, ∠5, ∠7 are all equal to 80°. Similarly ∠2, ∠4, ∠6, ∠8 are all equal to 100°.

Question 17.
In the given figure AB//CD. Find the values of x, y and z.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 21
Solution:
Given that AB // CD.
From the figure (2x + 3x) + 80° = 180°
(∵ interior angles on the same side of the transversal)
∴ 5x = 180° – 80°
x = \(\frac{100}{5}\) = 20°
Now 3x = y (∵ alternate interior angles)
y = 3 x 20° = 60° .
And y + z = 180°
(∵ linear pair of angles)
∴ z = 180°-60° = lg0°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 18.
In the given figure AB // CD. Find the values of x, y and z.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 22
Solution:
Given that AB // CD.
From the figure x° + 70° + x° = 180°
(∵ The angles at a point on the line)
∴ 2x = 180° – 70°
x = \(\frac { 110° }{ 2 }\) = 55-
Also 90° + x° + y° = 180°
[∵ interior angles of a triangle]
90° + 55° + y = 180°
y = 180° – 145° = 35°
And x° + z° = 180°
[∵ interior angles on the same side of a transversal]
55° + z = 180°
z = 180°-55° = 125°

Question 19.
In each of the following figures AB // CD. Find the values of x in each case.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 23
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 24
In each case draw a line ‘l’ parallel to AB and CD through F.
fig (i)
a + 104° = 180° ⇒ a = 180° – 104° = 76°
b+ 116° = 180° ⇒ -b = 180°- 116° = 64°
[∵ interior angles on the same side]
∴ a + b = x = 76° + 64° = 140°

fig-(ii)
a = 35°, b = 65° [∵ alt. int. angles]
x = a + b = 35° + 65° = 100°

fig- (iii)
a + 35° = 180° ⇒ a = 145°
b + 75° = 180° ⇒ b = 105°
[ ∵ interior angles on the same side]
∴ x = a + b = 145° + 105° = 250°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.2

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 1.
In the given figure three lines \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{EF}}\) intersecting at ‘O’. Find the values of x, y and z, it is being given that x : y : z = 2 : 3 : 5
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 1
Solution:
From the figure
2x + 2y + 2z = 360°
⇒ x + y + z = 180°
x : y : z = 2 : 3 : 5
Sum of the terms of the ratio
= 2 + 3 + 5 = 10
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 2

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 2.
Find the value of x in the following figures.
i)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 3
Solution:
From the figure
3x + 18= 180° – 93° (∵ linear pair )
3x + 18 = 87
3x = 87- 18 = 69
∴ x = \(\frac{69}{3}\) = 23

ii)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 4
Solution:
From the figure
(x – 24)° + 29° + 296° = 360“
(∵ complete angle)
x + 301° = 360°
∴ x = 360° – 301° = 59°

iii)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 5
Solution:
From the figure
2 + 3x = 62
(∵ vertically opposite angles)
3x = 62 – 2
∴ 3x = 60° ⇒ x = \(\frac{60}{3}\)
∴ x = 20°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

iv)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 6
Solution:
From the figure
40 + (6x + 2) = 90°
(∵ complementary angles)
6x = 90° – 42°
6x = 48
x = \(\frac{48}{6}\) = 8°

Question 3.
In the given figure lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) intersect at ’O’. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 7
Solution:
Given that ∠AOC +∠BOE = 70°
∠BOD = 40°
∠AOC = 40°
(∵ ∠AOC, ∠BOD are vertically opposite angles)
∴ 40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30°
Also ∠AOC + ∠COE +∠BOE = 180°
( ∵ AB is a line)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 180° -70° = 110°
∴ Reflex ∠COE = 110°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 4.
In the given figure lines \(\overline{\mathrm{XY}}\) and \(\overline{\mathrm{MN}}\) . intersect at O. If ∠POY = 90° and a : b = 2:3, find c.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 8
Solution:
Given that XY and MN are lines.
∠POY = 90°
a : b = 2 : 3
From the figure a + b = 90°
Sum of the terms of the ratio a : b
= 2 + 3 = 5
∴ b = \(\frac{3}{5}\) x 90° = 54°
From the figure b + c = 180°
(∵ linear pair of angles)
54° + c = 180°
c = 180°-54° = 126°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 5.
In the given figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 9
Solution:
Given that ∠PQR = ∠PRQ
From the figure
∠PQR + ∠PQS = 180° ………….. (1)
∠PRQ + ∠PRT = 180° …………..(2)
From (1) and (2)
∠PQR + ∠PQS = ∠PRQ + ∠PRT
But ∠PQR = ∠PRQ
So ∠PQS = ∠PRT
Hence proved.

Question 6.
In the given figure, if x + y = w + z, then prove that AOB is a line.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 10
Solution:
Given that x + y = w + z = k say
From the figure
x + y + z + w = 360°
(∵ Angle around a point)
Also x + y = z + w
∴ x + y = z + w = \(\frac{360^{\circ}}{2}\)
∴ x + y = z + w = 180°

OR

k + k = 360°
2k = 360°
k = \(\frac{360^{\circ}}{2}\)

(i.e.) (x,y) and (z, w) are pairs of adjacent angles whose sum is 180°.
In other words (x, y) and (z, w) are linear pair of angles ⇒ AOB is a line.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 7.
In the given figure \(\overline{\mathrm{PQ}}\) is a line. Ray \(\overline{\mathrm{OR}}\) is perpendicular to line \(\overline{\mathrm{PQ}}\).\(\overline{\mathrm{OS}}\) os is another ray lying between rays \(\overline{\mathrm{OP}}\) and \(\overline{\mathrm{OR}}\) Prove that
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 11
Solution:
Given : OR ⊥ PQ ⇒ ∠ROQ = 90°
To prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)
Solution:
Proof: From the figure
∠ROS = ∠QOS – ∠QOR ……………(1)
∠ROS = ∠ROP – ∠POS ……………..(2)
Adding (1) and (2)
∠ROS + ∠ROS = ∠QOS – ∠QOR +∠ROP – ∠POS [ ∵ ∠QOR = ∠ROP = 90° given]
⇒ 2∠ROS = ∠QOS – ∠POS
⇒ ∠ROS = \(\frac{1}{2}\) [∠QOS – ∠POS]
Hence proved.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 8.
It is given that ∠XYZ = 64° and XY is produced to point P. A ray \(\overline{\mathrm{YQ}}\) bisects ∠ZYP. Dräw a figure from the given Information. Find ∠XYQ and reflex ∠QYP.
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 12
∠XYQ = 32°
∠QYP = 32°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.1

Question 1.
In the given figure, name:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 1
i) Any six points
Solution:
A, B, C, D, P, Q, M, N etc.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

ii) Any five line segments
Solution:
\(\overline{\mathrm{AX}}, \overline{\mathrm{XM}}, \overline{\mathrm{MP}}, \overline{\mathrm{PB}}, \overline{\mathrm{MN}}, \overline{\mathrm{PQ}}, \overline{\mathrm{AB}} \ldots \ldots\) etc.

iii) Any four rays
Solution:
\(\overline{\mathrm{MA}}, \overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{NC}}, \overline{\mathrm{QD}} \ldots \ldots\) etc.

iv) Any four lines
Solution:
\(\overline{\mathrm{MA}}, \overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{NC}}, \overline{\mathrm{QD}} \ldots \ldots\)

v) Any four collinear points
Solution:
A, X, M, P and B are collinear points on the line \(\overline{\mathrm{AB}}\).

Question 2.
Observe the following figures and identify the type of angles in them.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 2
Solution:
∠A – reflex angle
∠B – right angle
∠C – acute angle

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

Question 3.
State whether the following state¬ments are true or false
i) A ray has no end point.
ii) Line \(\overline{\mathrm{AB}}\) is the same as line \(\overline{\mathrm{BA}}\)
iii) A ray \(\overline{\mathrm{AB}}\) is same as the ray \(\overline{\mathrm{BA}}\)
iv) A line has a definite length.
v) A plane, has length and breadth but no thickness.
vii) Two lines may intersect in two points.
viii) Two intersecting lines cannot both be parallel to the same line.
Solution:
i) A ray has no end point. – False
ii) Line \(\overline{\mathrm{AB}}\) is the same as line \(\overline{\mathrm{BA}}\) – True
iii) A ray \(\overline{\mathrm{AB}}\) is same as the ray \(\overline{\mathrm{BA}}\) – False
iv) A line has a definite length. – False
v) A plane, has length and breadth but no thickness. – True
vi) Two distinct points always determine a unique line. – True
vii) Two lines may intersect in two points. – False
viii) Two intersecting lines cannot both be parallel to the same line. – True

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

Question 4.
What is the angle between two hands of a clock when the lime in the clock
is
a) 9 ‘o clock
b) 6 ‘o clock
c) 7 ; 00 p.m.
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 3
a) 12 hours = = 360°
1 hour = \(\frac{360^{\circ}}{12}\) = 30°
∴Angle between hands when the time is 9 o clock = 3 x 30 = 90

b)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 4
Angle between hands = 6 x 30° = 180°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

c)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 5
Angle between hands = 7 x 30° = 210°

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry Exercise 3.1

Question 1.
Answer the following:
i) How many dimensions a solid has ?
Solution:
A solid has three dimensions namely length, breadth and height or depth.

ii) How many books are there in Euclid’s Elements ?
Solution:
There are 13 volumes in Euclid’s elements.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

iii) Write the number of faces of a cube and cuboid.
Solution:
Cube : 6 faces
Cuboid : 6 faces

iv) What is the sum of interior angles of a triangle ?
Solution:
The sum of interior angles of a triangle is 180°.

v) Write three undefined terms of geometry.
Solution:
Point, line and plane are three undefined terms in geometry.

Question 2.
State whether the following statements are true or false. Also give reasons for your answers.
a) Only one line can pass through a given point
b) All right angles are equal
c) Circles with same radii are equal
d) A finite line can be extended on its both sides endlessly to get a straight line
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 1
e) From figure AB > AC
Solution:
a) Only one line can pass through a given point – False.
Reason : (Since, infinitely many lines can pass through a given point)
b) All right angles are equal – True.
c) Circles with same radii are equal – True.
d) A finite line can be extended on its both sides endlessly to get a straight line – True.
e) From figure AB > AC – True.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 3.
In the figure given below, show that the length AH > AB + BC + CD.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 2
Solution:
Given a line \(\stackrel{\leftrightarrow}{\mathrm{AH}}\)
To prove AH > AB + BC + CD
From the figure AB + BC + CD = AD
AD is a part of whole AH.
From Euclid’s axiom whole is greater than part.
∴ AH > AD
⇒ AH > AB + BC + CD

Question 4.
If a point Q lies between two points P and R such PQ = QR, prove that PQ = \(\frac{1}{2}\)PR.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 3
Let PR be a given line.
Given that PQ = QR
i. e., Q is a point on PR.
⇒ PQ + QR = PR
⇒ PQ + PQ = PR [∵ PQ = QR]
⇒ 2PQ = PR
⇒ PQ = \(\frac{1}{2}\) PR
Hence proved.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 5.
Draw an equilateral triangle whose sides are 5.2 cm
Soluton:
Step – 1 : Draw a line segment AB of length 5.2 cm. *
Step – 2 : Draw an arc of radius 5.2 cm with centre A.
Step – 3 : Draw an arc of radius 5.2 cm with centre B.
Step – 4 : Two arcs intersect at C; join C to A and B.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 4
Δ ABC is the required triangle.

Question 6.
What is a conjecture? Give an example for it.
Solution:
Mathematical statements which are neither proved nor disproved are called conjectures. Mathematical discoveries often start out as conjectures. This may be an educated guess based on observations.
Eg : Every even number greater than 4 can be written as sum of two primes. This example is called Gold Bach Conjecture

Question 7.
Mark two points P and Q. Draw a line through P and Q. Now how many lines are parallel to PQ, can you draw ?
Solution:
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 5
Infinitely many lines parallel to PQ can be drawn.

Question 8.
In the figure given below, a line n falls on lines / and m such that the sum of the interior angles 1 and 2 is less than 180°, then what can you say about lines l and m ?
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 6
Solution:
Given : l, m and n are lines, n is a transversal.
∠1 < 90°
∠2 < 90°
If the lines l and m are produced on the side where angles 1 and 2 are formed, they intersect at one point.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 9.
In the figure given below, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4 write the rela-tion between ∠1 and ∠2 using Euclid’s postulate.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 7
Solution :
Given : ∠1 = ∠3
∠3 = ∠4
∠2 = ∠4
∴∠1 = ∠2

∵Both ∠1 and ∠2 are equal to ∠4. (By Euclid’s axiom things which are equal to same things are equal to one another).

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 10.
In the figure given below, we have BX = \(\frac{1}{2}\) AB, BY= \(\frac{1}{2}\) BC and AB = BC. Show that BX = BY.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 8
Solution:
Given : BX = \(\frac{1}{2}\) AB
BY = \(\frac{1}{2}\)BC
AB = BC
To prove : BX = BY
Proof: Given AB = BC [ ∵ By Euclid’s axiom things which are halves of the same things are equal to one another]
\(\frac{1}{2}\) AB = \(\frac{1}{2}\) BC
BX = BY
Hence proved.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.5

Question 1.
Use suitable identities to find the following products.
i) (x + 5) (x + 2)
Solution:
(x + 5) (x + 2)
= x2 + (5 + 2)x + 5 x 2
[ ∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= x2 + 7x + 10

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

ii) (x – 5) (x – 5)
Solution:
(x – 5) (x – 5)
= (x – 5)2 = x2 – 2(x) (5) + 52
[ ∵(x – y)2 = x2 – 2xy + y2]
= x2 – 10x + 25

iii) (3x + 2) (3x – 2)
Solution:
(3x + 2) (3x – 2) = (3x)2 – (2)2
[∵ (x + y) (x – y) =x2 – y2]
= 9x2 – 4

iv) \(\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{2}-\frac{1}{x^{2}}\right)\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 1(i)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

v) (1 + x) (1 + x)
Solution:
(1 + x) (1 + x)
= (1 + x)2 = 12 + 2 (1) (x) + x2
[∵(x + y)2 = x2 + 2xy + y2]
= 1 + 2x + x2

Question 2.
Evaluate the following products with¬out actual multiplication.
i) 101 x 99
Solution:
101 x 99
= (100 + 1) (100 – 1)
= 1002 – 12
= 10000 – 1
= 9999

ii) 999 x 999
Solution:
999 x 999
= 9992
= (1000 – 1)2
= 10002 – 2 x (1000) x 1 + 12
= 1000000-2000 + 1
= 998001

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

iii) \(50 \frac{1}{2} \times 49 \frac{1}{2}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 1

iv) 501 x 501
Solution:
501 x 501
= (500 + 1) (500 + 1)
= (500 + 1)2
= 5002 + 2 x (500) x 1 + 12
= 250000 + 1000 + 1 = 251001

v) 30.5 x 29.5 = (30 + 0.5) (30 – 0.5)
= 302 – (0.5)2
= 900 – 0.25
= 899.75

Question 3.
Factorise the following using appro-priate identities.
i) 16x2 + 24xy + 9y2
Solution:
16x2 + 24xy + 9y2
= (4x)2 + 2 (4x) (3y) + (3y)2
= (4x + 3y)2 = (4x + 3y) (4x + 3y)
[ ∵ (x + y)2 = x2 + 2xy + y2]

ii) 4y2 – 4y + 1
Solution:
4y2 – 4y + 1
= (2y)2 – 2 (2y) (1) + (1)2
[ ∵ (x -y)2 = x2 – 2xy + y2]
= (2y -1)2 = (2y – 1) (2y-1)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

iii) \(4 x^{2}-\frac{y^{2}}{25}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 2

iv) 18a2 – 50
Solution:
18a2 – 50 = 2 (9a2 – 25)
= 2[(3a)2 – (5)2]
[ ∵ x2 – y2 = (x + y) (x – y)]
= 2 (3a + 5) (3a – 5)

v) x2 + 5x + 6
Solution:
x2 + 5x + 6 = x2 + (3 + 2) x + 3 x 2
[ ∵ (x + a) (x + b) = x2 + (a + b) x + a . b]
= (x + 3) (x + 2)

vi) 3p2 – 24p + 36
Solution:
3p2 – 24p + 36
= 3[p2 – 8p + 12]
= 3[p2 + (- 6 – 2)p + (- 6) (- 2)]
[ ∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= 3 (p – 6) (p – 2)

Question 4.
Expand each of the following, using suitable identities.
i) (x + 2y + 4z)2
(x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)
[ ∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

ii) (2a – 3b)3
Solution:
(2a – 3b)3 = (2a)3 – 3 (2a)2 (3b) + 3 (2a) (3b)2 – (3b)3
[ ∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 8a3 – 3(4a2) (3b) + 3 (2a) (9b2) – 27b3
= 8a3 – 36a2b + 54ab2-27b3
(or)
∵ (a – b)3 = a3 – b3– 3ab (a – b)]
= (2a)3 – (3b)3 – 3(2a) (3b) (2a – 3b)
= 8a3 – 27b3 – 18ab (2a – 3b)

iii) (- 2a + 5b – 3c)2
Solution:
(- 2a + 5b – 3c)2
= (- 2a)2 + (5b)2 + (- 3c)2 + 2 (- 2a) (5b) + 2 (5b) (- 3c) + 2 (- 3c) (- 2a)
= 4a2 + 25b2 + 9c2 – 20ab – 30bc + 12ca
[ ∵ (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz +2za]

iv) \(\left[\frac{a}{4}-\frac{b}{2}+1\right]^{2}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 3

v) (p + 1)3
Solution:
(p + 1)3
= (P)3 + 3 (p)2 (1) + 3 (p) (1)2 + (1)3
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3]
= p3 + 3p2 + 3p + 1

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

vi) \(\left(x-\frac{2}{3} y\right)^{3}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 4

Question 5.
Factorise
i) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
Solution:
25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
= (5x)2 + (- 4y)2 + (- 2z)2 + 2(5x) (- 4y) + 2 (- 4y) (- 2z) + 2 (- 2z) (5x)
= (5x – 4y – 2z)2 = (- 5x + 4y +, 2z)2

ii) 9a2 + 4b2 + 16c2 + 12ab – 16bc – 24ca
Solution:
9a2 + 4b2 + 16c2 + 12ab – 16bc -24ca
= (3a)2 + (2b)2 + (- 4c)2+ 2 (3a) (2b) + 2 (2b) (- 4c) + 2(- 4c) (3a)
= (3a + 2b – 4c)2

Question 6.
If a + b + c = 9 and ab + be + ca = 26, find a2 + b2 + c2.
Solution:
Given that a + b + c = 9
Squaring on both sides,
(a + b + c)2 = 92
⇒ a2+ b2 + c2+ 2 (ab + be + ca) = 81 ⇒ a2 + b2 + c2 = 81 – 2 (ab + be + ca)
(by problem)
= 81 – 2 x 26
= 81 – 52 = 29

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 7.
Evaluate the following by using suit¬able identities. m EachgM)
i) (99)3
Solution:
(99)2 = (100 – 1)3
= 1003 – 3 (100)2 (1) + 3 (100) (1)2 – 13
[ ∵ (x – y)3 = x3 – 3x2y + 3xy2 + y3]
= 1000000 – 30000 + 300 – 1
= 970299

ii) (102)3
Solution:
(102)3 = (100 + 2)3
= 1003 + 3 (100)2 (2) + 3 (100) (2)2 + 23
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3]
= 1000000 + 60000 + 1200 + 8
= 1061208

iii) (998)3
Solution:
(998)3 =(1000 – 2)3
[ ∵ (x – y)3 = x3 – 3x2y + 3xy2 – y3] = 10003– 3(1000)2(2) + 3(1000)(2)2– 23
= 1000000000 – 6000000 + 12000 – 8
= 994011992

iv) (1001)3
Solution:
(1001)3 = (1000 + 1)3 .
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3] = 10003 + 3(1000)2(1) + 3(1000) (1)2 + 13
= 1000000000 + 3000000 + 3000 + 1
= 1003003001

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 8.
Factorise each of the following.
i) 8a3 + b3 + 12a2 b + 6ab2
Solution:
8a3 + b3 + 12a2 b + 6ab2
= (2a)3 + (b)3 + 3 (2a)2 (b) + 3 (2a) (b)2
= (2a + b)3

ii) 8a3 – b3 – 12a2 b + 6ab2
Solution:
8a3 – b3 – 12a2 b + 6ab2
= (2a)3 – (b)3 – 3 (2a)2 (b) + 3 (2a) (b)2
= (2a – b)3

iii) 1 – 64a3 -12a + 48a2
Solution:
1 – 64a3 – 12a + 48a2
= (1)3 – (4a)3 – 3(1)2 (4a) + 3(1) (4a)2
= (1 – 4a)3

iv) \(8 p^{3}-\frac{12}{5} p^{2}+\frac{6}{25} p-\frac{1}{125}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 5

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 9.
Verify i) x3 + y3 = (x + y) (x2 – xy + y2);
ii) x3 – y3 = (x – y) (x2 + xy + y2)
Using some non-zero positive integers and check by actual multiplication. Can you
call these as identities ?
i) x3 + y3 = (x + y) (x2 – xy + y2)
Solution:
Given x3 + y3 = (x + y) (x2 – xy + y2)
L.H.S = x3 + y3
R.H.S = (x + y) (x2 – xy + y2)
= x (x2 – xy + y2) + y (x2 – xy + y2)
= x3 -x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S
∴ L.H.S = R.H.S

Take x = 3, y = 2
L.H.S = 33 + 23 = 27 + 8 = 35
R.H.S = (3 + 2) (32 – 3 x 2 + 22)
= 5 x (9 – 6 + 4)
= 5 x 7 = 35
∴ L.H.S = R.H.S

ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
Given that x3 – y3 = (x – y) (x2 + xy + y2)
L.H.S = x3 – y3
R.H.S = (x – y) (x2 + xy + y2)
= x (x2 + xy + y2) – y (x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3= L.H.S

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

L.H.S = 33 – 23 = 27 – 8 = 19
R.H.S = (3 – 2) (32 + 3 x 2 + 22)
= 1 x (9 + 6 + 4)
= 1 x 19 = 19
∴ L.H.S = R.H.S
We can call the above two expressions as identities

Question 10.
Factorise by using the above results (identities).
i) 27a3 + 64b3
Solution:
27a3+ 64b3 = (3a)3 + (4b)3
= (3a + 4b) {(3a)2 – (3a) (4b) + (4b)2}
= (3a + 4b) (9a2 – 12ab + 16b2)

ii) 343y3 – 1000
Solution:
343y3 – 1000 = (7y)3 – (10)3
= (7y – 10) [(7y)2 + (7y) (10) + (10)2]
= (7y – 10) (49y2 + 70y + 100)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 11.
Factorise 27x3 + y3 + z3 – 9xyz using identity.
Solution:
Given 27x3 + y3 + z3 – 9xyz
= (3x)3 + (y)3 + (z)3 – 3 (3x) (y) (z)
= (3x + y + z)
[(3x)2 + y2 + z2 – (3x) (y) – (y) (z) – (z) (3x)]
[ ∵ (x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2– xy – yz – zx)
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)

Question 12.
Verify that x3+ y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2 ]
(OR)
Verify that
p3 + q3 + r3 – 3pqr = 1/2 (p + q + r)
[(p – q)2 + (q – r)2 + (r – p)2]
Solution:
Given x3+ y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2 ]
R-H.S = 1/2 (x + y + z) [(x – y)2 + (y – z)2+ (z – x)2]
= 1/2 (x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2xz]
= 1/2 (x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= 1/2 (x + y + z) (2) [x2 + y2 + z2 – xy – yz – zx]
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= L.H.S
Hence proved.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz
Solution:
Given x + y + z = 0
To prove x3 + y3 + z3 = 3xyz
We have an identity
(x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz
Substituting x + y + z = 0in the above equation, we get
0 x (x2 + y2 + z2 -xy-yz-zx)
= x3 + y3 + z3 – 3xyz
⇒ x3 + y3 + z3 – 3xyz = 0
⇒ x3 + y3 + z3 = 3xyz

Question 14.
Without actual calculating the cubes, find the value of each of the following.
i) (- 10)3 + 73 + 33
Solution:
Given (-10)3 + 73 + 33
Sum of the bases = -10 + 7 + 3 = = 0
∴ (- 10)3 + 73 + 33
= 3 (- 10) x (7) x 3
= -630
[ ∵ x + y + z = 0 then x3 + y3 + z3 = 3xyz]

ii) (28)3 + (- 15)3 + (- 13)3
Solution:
Given (28)3 + (- 15)3+ (- 13)3
Sum of the bases = 28 + (- 15) + (- 13) = 0
∴ (28)3 + (- 15)3 + (- 13)3
= 3 x 28 x (- 15) x (- 13)
= 16380

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

iii) \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}\) read it as \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(\frac{-5}{6}\right)^{3}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 6

iv) (0.2)3 – (0.3)3 + (0.1)3
Solution:
Given that (0.2)3 – (0.3)3 + (0.1)3
= (0.2)3 + (- 0.3)3 + (0.1)3
Sum of the bases = 0.2 – 0.3 + 0.1 = 0
∴ (0.2)3 + (-0.3)3 + (0.1)3
= 3 x (0.2) (- 0.3) (0.1)
= -0.018

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 15.
Give possible expressions for the length and breadth of the rectangle whose area is given by
i) 4a2 + 4a – 3
Given that area = 4a2 + 4a – 3
= 4a2 + 6a – 2a – 3
= 2a (2a + 3) – 1 (2a + 3)
= (2a – 1) (2a + 3)
∴ Length = (2a + 3); breadth = (2a – 1).

ii) 25a2 – 35a + 12
Solution:
Given that area = 25a2 – 35a +12
= 25a2 – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
∴ (5a – 4) (5a – 3) are the length and breadth.

Question 16.
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?
i) 3x3 – 12x
Solution:
Volume = 3x3 – 12x
= 3x (x2 – 4)
= 3x (x + 2) (x – 2) are the dimensions.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

ii) 12y2 + 8y – 20
Solution:
Given that volume = 12y2 + 8y – 20
= 4 (3y2 + 2y – 5)
= 4 [3y2 + 5y – 3y – 5]
= 4 [y (3y + 5) – 1 (3y + 5)]
= 4 (3y + 5) (y – 1)
Hence 4, (3y + 5) and (y – 1) are the dimensions.

Question 17.
Show that if 2 (a2 + b2 ) = (a + b)2 then a = b
Solution:
Given that 2 (a2 + b2 ) = (a + b)2
To prove a = b
As 2 (a2 + b2 ) = (a + b)2
We have
2a2 + 2b2 = a2 + 2ab + b2
2a2 – a2 + 2b2 – b2 = 2ab
a2 + b2 = 2ab
This is possible only when a = b
∴ a = b