## KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2

Question 1.
If 4 is added to a number and the sum is multiplied 3, the result is 30. Find the number.
Let the number be ‘x’.
If 4 is added to it will be x + 4.
Sum is multiplied by 3 the result is 30.
∴ (x + 4)3 = 30
3x +12 = 30
3x = 30 – 12
3x = 18
x = $$\frac { 18 }{ 3 }$$
x = 6
∴The number is 6

Question 2.
Find three consecutive odd numbers whose sum is 219.
Let the odd number be ‘x’.
The next two consecutive numbers are x + 2 and x + 4
x + (x + 2) + (x + 4) = 219
3x + 6 = 219
3x = 219 – 6
3x = 213
x = $$\frac { 213 }{ 3 }$$
x = 71
x + 2 = 71 + 2 = 73
x + 4 = 71 + 4 = 75
Three consecutive odd numbers are 71, 73, 75

Question 3.
A number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.
Let the number be x
Number subtracted by 30 = 30 – x
14 subtracted by 3 times the number 3x – 14
∴30 – x = 3x – 14
30+ 14 = 3x + x
44 = 4x
x = $$\frac { 44 }{ 4 }$$
x =11
∴ The number is 11

Question 4.
If 5 is subtracted from three times a number the result is 16. Find the number.
Let the number be x, 5 is subtracted from 3 times the number the result is 16.
3x – 5 = 16
3x= 16 + 5
3x = 21
x = $$\frac { 21 }{ 3 }$$
x = 7
∴ The number is 7

Question 5.
Find two numbers such that one of them exceeds the other by 9 and their sum is 81.
Let the number is x. The other number is x + 9.
Their sum is 81
∴ x + (x + 9) = 81
2x = 81 – 9
2x = 72
x = $$\frac{72}{2}$$
x = 36
x + 9 = 36 + 9 = 45
∴ The number are 36 and 45

Question 6.
Prakruthi’s age is 6 time Sahil’s age. After 15 years prakruthi will be 3 times as old as Sahil. Find their age.
Let Sahil’s present age be x. Prakruthi’s present age is 6x, 15 years later Sahil age will be (x + 15) years and Prakruthi age will be (6x + 15) years.
Given that the Prakruthi age will be 3 times as old as Sahil.
∴6x +15 = 3(x + 15)
6x + 15 = 3x + 45
6x – 3x = 45 – 15
3x = 30
x = $$\frac{30}{3}$$ = 10
Sahils age = x = 10 years Prakruthis age = 6x = 6 x 10 = 60 years

Question 7.
Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will be twice that of his son. Find their present age.
Ahmed’s presents age be x
Fathers present age = 3x
12 years later Ahmed’s age = x + 12
and father’s age = 3x + 12
Given 3x + 12 = 2(x + 12)
3x+ 12 = 2x + 24
3x – 2x = 24 – 12
x = 12 years
∴ Ahmed’s age = 12 years
Fathers age = 3x = 3 × 12 = 36 years.

Question 8.
Sanju is 6 years older than his brother Nishu. If the sum of their ages is 28 years what are their present ages.
Let Nishu’s age be ‘x’ Sanju’s age = x + 6
Sum of their ages = 28 x + (x + 6) = 28
2x + 6 = 28
2x = 28 – 6
2x = 22
x = $$\frac{22}{2}$$
x = 11
Nishu’sage = x = 11 years
Sanju’ s age = x + 6 = 11 + 6 = 17 years

Question 9.
Viji is twice as old as his brother Deepu. If the difference of their ages is 11 years, find their present age.
Let Deep’s age be ‘x’, Viji’s age is 2x
Difference of their age = 11
2x – x = 11
x = 11
∴ Deepu’s age = x = 11 years
Viji’s age = 2x = 2 x 11 = 22 years

Question 10.
Mrs. Joseph is 27 years older than her daughter Bindu. After 8 years she will be twice as old as Bindu. Find their present ages.
Let Bindu’s present age be ‘x’ years.
Mrs. Joseph’s present age = x + 27 years.
After 8 years Bindu’s age = x + 8 and Mrs.
Josephs age = x + 27 + 8
= x + 35 years.
Given that x + 35 = 2(x + 8)
x + 35 = 2x + 16
35 – 16 = 2x – x
19 = x
x= 19
Bindu’s present age = 19 years
Mrs.Joseph’s age = x + 27 = 19 + 27 = 46 years

Question 11.
After 16 years Leena will be three times as old as she is now. Find her present age.
Let Leena’s present age be ‘x’ years
16 years later she will be (x + 16) years
Given x + 16 = 3x
16 = 2x
x = $$\frac { 16 }{ 2 }$$
x = 8 years
Leena’ s present age = 8 years

Question 12.
A rectangle has a length which is 5 cm less than twice its breadth. If the length is decreased by 5 cm and breadth is increased by 2 cm the perimeter of the resulting rectangle will be 74 cm. Find the length and breadth of the original rectangle.
Let the breadth of the original rectangle be ‘b’ twice the breadth is 2b.
The length of the rectangle is 5 cm less than twice the breadth.
∴ Length = 2b – 5
If the length is decreased by 5 then length
is 2b – 5 – 5 = 2b – 10
If the breadth is increased by 2cm then breadth is b + 2 cm.
Perimeter of new rectangle = 2(length + breadth)
74 = 2(2b – 10 + b + 2)
74 = 2(3b – 8)
74 + 16 = 6b 90 = 6b
b = $$\frac{90}{6}$$
b = 15cm.
Breadth of the original rectangle = 15 cm
Length of the original rectangle = 2b – 5
= 2 × 15 – 5
= 30 – 5
= 25 cm

Question 13.
The length of a rectangular field is twice its breadth. If the perimeter of the field is 288m. Find the dimensions of the field.
Let breadth of the rectangular field bee ‘x’m
∴ Its length = 2x
Its perimeter = 288
2(1 + b) = 288
2(2x + x) = 288
2(3x) = 288
6x = 288
x = $$\frac { 288 }{ 6 }$$ = 48
Its length = 2x = 2 × 48 = 96m

Question 14.
Srishti’s salary is the same as 4 times Azar’s salary. If together they earn Rs 3750 a month find their individual salaries.
Let Axar’s salary be x. Sristi’s salary is 4x together they earn Rs. 3750
∴ x + 4x = 3750
5x = 3750
x = $$\frac{3750}{5}$$
x = 750
4x = 4 × 750 = 3000
∴ Azar’s salary is Rs.750 and Sristi’s salary is Rs.3000

## KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex 4.1

Students can Download Maths Chapter 4 Factorisation Ex 4.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 4 Factorisation Ex 4.1

1. Resolve into factors.

Question i.
x² + xy
x² + xy = x(x + y)

Question ii.
3x² – 6x
3x² – 6x = 3x (x – 2)

Question iii.
(1.6)a² – (0.8)a
(1.6)² – (0.8)a
= (0.8 x 2a²) – (0.8)a
= 0. 8a(2a- 1)

Question iv.
5 – 10m – 20n
5 – 10m -20n = 5(1 – 2m – 4n)

2. Froctorise:

Question i.
a² + ax + ab + bx
a² + ax + ab + bx a(a + x) + b(a + x)
(a + x)(a + b)

Question ii.
3ac + 7bc – 3ad – 7bd
3ac + 7bc – 3ad – 7bd
c(3a + 7b – d(3a + 7b) (3a + 7b) (c-d)

Question iii.
3xy – 6zy – 3xt + 6zt
3y (x – 2z) – 3t(x – 2z)
(x – 2z) (3y – 3t)

Question iv.
y3 + 3y² + 2y – 6 – xy + 3x
y² (y – 3) + 2(y – 3) – x(y – 3)
(y- 3) (y² + 2 – x)

3. Factorise:

Question i.
4a² – 25
4a² – 25
= (2a)² – 52
[a² – b²=(a+b)(a-b)]
= (2a + 5)(2a – 5)

Question ii.
$$x^{2}-\frac{9}{16}$$
Ans.

Question iii.
x4 – y4
x4 – y4
= (x²)² – (y²)²
Identity a² – b² = (a + b)(a – b)
= (x² + y²)(x² – y²)
= (x²+ y²)(x + y)(x – y)

Question iv.
$$\left(7 \frac{3}{10}\right)^{2}-\left(2 \frac{1}{10}\right)^{2}$$
Ans.

Question v.
(0.7)² – (0.3)²
(0.7)² – (0.3)²
Identity a² – b² = (a + b)(a – b)
= (0.7 + 0.3)(0.7 – 0.3)
= (1.0)(0.4)
= 0.4

Question vi.
(5a-2b)²-(2a-b)²
(5a-2b)² – (2a-b)²
Identity a² – b² = (a + b)(a – b)
= [5a – 2b + 2a – b]
= [5a – 2b – (2a – b)]
= (7a – 3b)[5a – 2b – 2a + b]
= (7a – 3b)(3a – b)

## KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.1

Students can Download Maths Chapter 9 Commercial Arithmetic Ex 9.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.1

Question 1.
In a school 30% students play chess 60% play carrom and the rest play other games. If the total number of students in the school is 900 find the exact number of students who play each game.
30% Students play chess. Total number of students is 900.
The number of players who play chess = 30% of 900.
$\frac { 300 }{ 100 }$ × 900 = 270
The number of players who play carrom board = 60% of 900
$\frac { 60}{ 100 }$ × 900 = 540
The number of players who play other games = 10% of 900
$\frac { 10 }{ 100 }$ × 900 = 90

Question 2.
In a school function Rs. 360 remained after spending 82% of the money. How much money was there in the beginning? verify your answer.
Let the money, in the beginning, be ₹ 100 The money spent is ₹ 82.
∴ The amount remaining is 100 – 82 = ₹ 18.
If the remaining money is ₹ 18 the money, in the beginning, is ₹ 100.
If the remaining money is ₹ 360 the money in the beginning is
$\frac { 100 }{ 18 }$ × 360 = 2000
Verification :
$\frac { 82}{ 100 }$ × 2000 = 1640
Money remained = 2000 – 1640 = ₹ 360.

Question 3.
Akshay’s income is 20% less than that of Ajay what percent is Ajay’s income more than that of Akshay?
Let Ajay’s income be Rs. 100.
Then Akshay’s income is (100 – 20) = Rs. 80.
If Akshay’s income is Rs. 80.
Ajay’s income is Rs. 100.
If Akshay’s income is Rs. 1 Ajay’s income is
$$\frac{100}{80}$$ × 1 = Rs. $$\frac { 100 }{ 80 }$$
Changing the scale to 100.
Ajay’s income is Rs. 100
Akshay’s income is
= $$\frac { 100 }{ 80 }$$ × 100 = 125
∴ Ajay’s income is 125 – 100 = 25% more than that of Akshay.

Question 4.
A daily wage employee spends 84% of his weekly earning. If he saves Rs. 384 find his weekly earning.
Let his weekly earning be ₹ 100.
His expenditure is ₹ 84.
Therefore his savings is (100 – 84) = ₹ 16
If the savings is ₹ 16 then the income is ₹ 100.
If the savings is ₹ 384 then the income is
$\frac { 100 }{ 16 }$ × 384 = ₹  2400
∴ His weekly earning is ₹ 2400.

Question 5.
A factory announces a bonus of 10% its employees. If an employee gets Rs. 10,780 find his actual salary.
Let the salary of the employee be ₹ 100.
He gets a bonus of ₹ 10.
He gets 100 + 10 = ₹ 110 including bonus.
∴ If the employee gets ₹ 10.780 his salary is
$\frac { 100 }{ 110 }$ × 10780 = 9800
∴ The salary of the employee is ₹ 9800

## KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex 1.3

Students can Download Maths Chapter 1 Playing with Numbers Ex 1.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.3

Question 1.
Find the quotient and remainder when each of the following number is divided by 13 : 8, 31, 44, 85, 1220.

Question 2.
Find the quotient and the remainder when each of the following numbers are divided by 304.
128, 636, 785, 1038, 2236, 8858

Question 3.
Find the least natural number larger than 100 which leaves the remainder 12 when divided by 19.
The least number greater than 100 divisible by 19 is 114.
Adding 12 to this number we get 114 + 12 = 126.
126 leaves 12 as the remainder when divided by 19.

Question 4.
What is the least natural number you have to add to 1024 to get a multiple of 181?
The Multiples of 181 are 181, 362, 543, 724, 905, 1086.
The nearest multiple of 181 to 1024 is 1.086.
Hence 62 (1086 – 1024) must be added to 1024 to get a multiple of 181.

## KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Ex 7.2

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## Karnataka Board Class 8 Maths Chapter 7 Rational Numbers Ex 7.2

Question 1.
Write down ten rational numbers which are equivalent to $$\frac { 5 }{ 7 }$$ and the denominator not exceeding 80.
Multiply both numerator and denominator by 2, 3, 4………
$$\frac{10}{14}, \frac{15}{21}, \frac{20}{28}, \frac{35}{35}, \frac{30}{42}, \frac{35}{49}, \frac{40}{56}, \frac{45}{63}, \frac{50}{70}, \frac{55}{77}$$

Question 2.
Write down 15 rational numbers which are equivalent to $$\frac { 11 }{ 5 }$$ and the numerator not exceeding 180.
$$\begin{array}{l}{\frac{22}{10}, \frac{33}{15}, \frac{44}{20}, \frac{55}{25}, \frac{66}{30}, \frac{77}{35}, \frac{88}{40}, \frac{99}{45}} \\ {\frac{110}{50}, \frac{121}{55}, \frac{132}{60}, \frac{143}{65}, \frac{154}{70}, \frac{165}{75}, \frac{176}{80}}\end{array}$$

Question 3.
Write down 10 positive rational numbers such that the sum of the numerator and the denominator of each is 11. Write them in decreasing order.

Question 4.
Write down ten positive rational numbers such that numerator – denominator for each of them is -2. Write to them in increasing order.
Numerator – denominator = – 2
therefore the denominator is greater than the numerator by 2.
$$\frac{1}{3}, \frac{2}{4}, \frac{3}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{9}, \frac{7}{9}, \frac{8}{10}, \frac{9}{11}, \frac{10}{12}$$

Question 5.
Is $$\frac { 3 }{ -2 }$$ a rational number? If so, how do you write it in the form conforming to the definition of a rational number (that is, the denominator as positive integer)?
$$\frac{3}{-2}$$ is a rational number because the denominator is negative.
It can be written as $$\frac{3}{-2}$$ since $$\frac{3}{-2}$$ is same as $$\frac{3}{-2}$$

Question 6.
Earlier you have studied decimals 0.9, 0.8, can you’ write these as rational numbers?
$$0.9=\frac{9}{10} \text { and } 0.8=\frac{8}{10}=\frac{4}{5}$$

## KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Ex 7.1

Students can Download Maths Chapter 7 Rational Numbers Ex 7.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 7 Rational Numbers Ex 7.1

1. Identify the property in the following statements.

Question (i)
2 + (3 + 4) = (2 + 3) + 4

Question (ii)
2 × 8 = 8 × 2
Commutative property of multiplication

Question (iii)
8 × (6 + 5) = (8 × 6) + (8 × 5)
Distributive property of multiplication over addition in integers.

Question 2.
Find the additive inverse of the following integers.
6                                      -6
9                                      -9
123                                -123
-76                                   76
-85                                   85
1000                             -1000

3. Find the integer m in the following:

Question (i)
m + 6 = 8
m = 8 – 6

Question (ii)
m + 25 = 15
m =15 – 25
m = -10

Question (iii)
m – 40 = -26
m = – 26 + 40
m = 14

Question (iv)
m + 28 = – 49
m = – 49 – 28
m = – 77

Question 4.
Write the following in increasing order: 21, -8, -26, 85, 33, -333, -210, 0, 2011
-333 < -210 < -26 < -8 < 0 < 21 < 33 < 85 < 2011.
-333, -210, -26, -8, 0, 21, 33, 85, 2011

Question 5.
Write the following in decreasing order: 85, 210, -58, 2011, -1024, 528, 364, -10000,12
2011 > 528 > 364 > 210 > 85 > 12 > – 58 > -1024 < -10,000
2011, 528, 364, 210, 85, 12, -58, -1024, -10,000.

## KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Ex 1.2

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## Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.2

Question 1.
In the following find the digits represented by the letters.

Here : 3 + 4 = 7 ∴ B = 4

Here 6 + A = 1 ∴ A = 5
6 + 5 = 11 is written in units place and 1 is carried.
∴ 1 + 1 + 2 = B
∴ B = 4

Here A × A is A.
Hence A = 0, 1, 5, or 6 If A = 1 we get 21 × 1 = 21 but the product is 12A.
If A = 5 then 25 × 5 = 125.
∴ A = 5

Here A + A = A
Hence A = 0

Answer: Here A × A = A
Hence A = 0, 1, 5 or 6
If A = 1, then 11 × 11 = 121
∴ A = 1 and B = 2

3A × A
36 × 6
216
∴ A = 6 & B = 1

Question 2.
In the adjacent sum A, B, C are consecutive digits. In the third row A, B, C appear in some order. Find A, B, C.

∴ A = 3, B = 4 and C = 5

## KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1

1. Solve the following:

Question i.
x + 3 = 11
x + 3 = 11
x = 11 – 3
x = 8

Question ii.
y – 9 = 21
y – 9 = 21
y = 21 + 9
y = 30

Question iii.
10 = z + 3
10 = z + 3 10-3 =z
7 = z or z = 7

Question iv.
$$\frac{3}{11}+x=\frac{9}{11}$$

Question v.
10x = 30
10x = 30
x = $$\frac { 30 }{ 10 }$$
x = 3

Question vi.
$$\frac{s}{7}=4$$
$$\frac{s}{7}=4$$
S = 4 × 7
S = 28

Question vii.
$$\frac{3 x}{6}=10$$
$$\frac{3 x}{6}=10$$
3x = 10 × 6
3x = 60
x = $$\frac { 60 }{ 3 }$$
x = 20

Question viii.
$$1.6=\frac{x}{1.5}$$
$$1.6=\frac{x}{1.5}$$
1.6 × 1.5 = x
2.40 = x
x = 2.4

Question ix.
8x – 8 = 48
8x – 8 = 48
8x = 48 + 8
8x = 56
x = $$\frac { 56 }{ 8 }$$
x = 7

Question x.
$$\frac{x}{3}+1=\frac{7}{15}$$

Question xi.
$$\frac{x}{5}=12$$
$$\frac{x}{5}=12$$
x = 12 × 5
x = 60

Question xii.
$$\frac{3 x}{5}=15$$

Question xiii.
3(x + 6) = 24
3(x + 6) = 24
3x+ 18 = 24
3x = 24 – 18
3x = 6
x = $$\frac { 6 }{ 3 }$$ = 2

Question xiv.
$$\frac{x}{4}-8=1$$

Question xv.
3(x+2) – 2(x—1) = 7
3(x + 2) – 2(x – 1) = 7
3x + 6 – 2x + 2 = 7
x + 8 = 7 .
x = 7 – 8
x = -1

2. Solve the equations :

Question i.
5x = 3x + 24
5x = 3x + 24
2x = 24
x = $$\frac { 24 }{ 2 }$$
x = 12

Question ii.
8t + 5 = 2t – 31
8t + 5 = 2t – 31
6t = -36
t = $$\frac { -36 }{ 6 }$$
t = -6

Question iii.
7x -10 = 4x + 11
7x – 10 = 4x + 11
7x – 4x = 11 + 10
3x = 21
x = $$\frac { 21 }{ 3}$$
x = 7

Question iv.
4z + 3 = 6 + 2z
4z + 3 = 6 + 2z
4z – 2z = 6 – 3
2z = 3

Question v.
2x – 1 = 14 – x
2x – 1 = 14 – x
2x + x = 14 + 1
3x = 15
x = $$\frac { 15 }{ 3 }$$
x = 5

Question vi.
6x + 1 = 3(x -1) + 7
6x + 1 = 3(x —1) + 7
6x + l =3x-3 + 7
6x + l = 3x + 4
6x – 3x = 4 – 1
3x = 3
x = $$\frac { 3 }{ 3 }$$
x = 1

Question vii.
$$\frac{2 x}{5}-\frac{3}{2}=\frac{x}{2}+1$$

Question viii.
$$\frac{x-3}{5}-2=\frac{2 x}{5}$$

Question ix.
3(x + 1) = 12 + 4(x – 1)
3(x + 1)= 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x + 3 = 4x + 8
3x – 4x = 8 – 3
-x = 5
x = -5

Question x.
2x – 5 = 3(x – 5)
2x – 5 = 3(x – 5)
2x – 5 = 3x – 15
2x – 3x = -15 + 5
-x = -10
x= 10

Question xi.
6(1 – 4x) + 7(2 + 5x) = 53
6( 1 – 4x) + 7(2 + 5x) = 53
6 – 24x+ 14 + 35x = 53
35x – 24x + 6 + 14 = 53
11x+ 20 = 53
11x = 53 – 20
x = $$\frac { 33 }{ 11 }$$
x = 3

Question xii.
3(x + 6) + 2(x + 3) = 64
3(x + 6) + 2(x + 3) = 64
3x + 18 + 2x + 6 = 64
3x + 2x + 18 + 6 = 64
5x + 24 = 64
5x = 64 – 24
5x = 40
x = $$\frac { 40 }{ 5 }$$
∴ x = 8

Question xiii.
$$\frac{2 m}{3}+8=\frac{m}{2}-1$$

Question xiv.
$$\frac{3}{4}(x-1)=(x-3)$$
$$\frac{3}{4}(x-1)=(x-3)$$
3(x – 1) = 4(x – 3)
3x – 3 = 4x – 12
3x – 4x = -12 + 3
-x = -9
x = 9

## KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3

Question 1.
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find the angles of the triangle.
In triangle ABC, BC is produced on either sides.
Let ∠ABC = 140° and ∠ACE = 136°
∠ABD +∠ABC = 180° [ Linear pair]
104 + ∠ABC = 180°
∠ABC = 180 – 104
∠ ABC = 76°

∠ACB + ∠ACE = 180° [Linearpair]
∠ACB + 136° = 180°
∠ACB = 180 – 136 ∠ACB = 44°
∠ABD + ∠ACB + ∠BAC = 180°
[Sum of the angles of a triangle 180° ]
76 + 44 + ∠BAC = 180°
120 + ∠BAC = 180°
∠BAC = 180 – 120 ∠BAC = 60°
Three angles are 76°, 44° & 60°

Question 2.
Sides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles ∠ACD, ∠BAE, and∠CBF.
Show that ∠ACD + ∠BAE + ∠CBF = 360°.

ABC + ∠CBF = 180° [Linear pair]………..(i)
∠ACB + ∠ACD = 180° [Linear pair]……(ii)
∠BAC + ∠BAE = 180° [Linear pair]…….(iii)
∠ABC + ∠CBF + ∠ACB + ∠ACD+
∠BAC + ∠BAE = 180 + 180 + 180
∠ABC + ∠ACB + ∠BAC +
∠CBF + ∠ACD + ∠BAE = 540°
180 + ∠CBF + ∠ACD + ∠BAE = 540°
[Sum of the angles of a triangle 180°]
∠CBF + ∠ACD + ∠BAE = 360°

Question 3
Compute the value of x in each of the following figures
i.

AB = AC
∴ ∠ABC = ∠ACB = 50°
∠ACB + ∠ACD = 180° [Linear pair]
50 + x = 180°
x = 180 – 50 = 130°

ii.

∠ABC + ∠CBF = 180° [Linear pair].
106°+∠ABC = 180°
∠ABC = 180 – 106 = 74°
∠EAC = ∠ABC + ∠ACB
130° =74 +x
130 – 74 = x
∴x = 56°

iii.

∠QPR = ∠TPU = 65°
[Vertically opposite angles]
∠PRS = ∠PQR + ∠QPR [Exterior angle=Sum of interior opposire angle]
100 = 65 + x
100 – 65 = x
35 = x
∴ x = 35°.
iv.

∠BAE + ∠BAC = 180° [Linear pair]
120° + ∠BAC = 180°
∠BAC = 180 – 120
∠BAC = 60°
∠ACD = ∠BAC + ∠ABC
[Exterior angle = Sum of interior opposuite angle]
112° = 60+x
112 – 60 = x
52 = x
x = 52°
v.

In Δ ABC, BA = BC (data)
∴ ∠BAC = ∠BCA = 20°
[Base angles of an isosceles traingle]
∠ABD = ∠BAC + ∠BCA
[Exterior angle=Sum of interior opposite angle]
x = 20 + 20
x=40°

Question 4.
In figure QT ⊥ PR, ∠TOR = 40° and ∠SPR = 30° find ∠TRS and ∠PSQ.

In ΔTQR,
∠TQR + ∠QTR + ∠TRQ = 180°
40 + 90 + ∠TRQ = 180°
130 + ∠TRQ = 180
∠ TRQ = 180 – 130
∠ TRQ = 50°
∠ TRS = 50° [∠PRS is same as ∠TRS ]
In Δ PRS, RS is produced to Q
∴Exterior∠PSQ = ∠SPR + ∠PRS = 30 + 50
∠PSQ = 80°

Question 5.
An exterior angle of a triangle is 120° and one of the interior opposite angle is 30 °. Find the other angles of the triangle.

In Δ ABC, BC is produced to D. Let
∠ACD = 120° and ∠ABC = 30°
Exterior ∠ACD = ∠BAC + ∠ABC
120 = ∠BAC + 30
120 – 30 = ∠BAC
90° = ∠BAC
∴∠BAC = 90
∠ACB +∠ACD = 180° [Linear pair]
∠ACB + 120 = 180°
∠ACB = 180 – 120
∠ACB = 60°
Other two angles are 90° & 60°

## KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Ex 16.2

Students can Download Maths Chapter 16 Mensuration Ex 16.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 16 Mensuration Ex 16.2

Question 1.
Find the total surface area and volume of a cube whose length is 12 cm.
l = 12 cm
T.S.A of cube = 6 l² = 6 x 12² = 6 x 144 = 864 cm²
Volume of cube = P = (12)3 = 1728 cm3

Question 2.
Find the volume of a cube whose surface area is 486 cm².
T.S.A of a cube = 486 cm²
612 = 486
l² = $$\frac{486}{6}$$
l² = 81
∴ l = √81 = 9 cm
Volume = l3 = 93= 729 cm3

Question 3.
A tank, which is cuboidal in shape has a volume 6.4m3. The length and breadth of the base are 2m and 1.6m respectively. Find the depth of the tank.
V = 6.4 m3
l = 2 m
b = 1.6 m
h = ?
Volume of cuboid = 6.4 m3
l × b × h = 6.4
2 × 1.6 × h = 6.4
3.2h = 6.4
h = $$\frac{6.4}{3.2}$$
h = 2m
h = 2m .
∴ The depth of the tank is 2 m.

Question 4.
How many m3 of soil has to be excavated from a rectangular well 28cm deep and whose base dimensions are 10cm and 8m. Also, find the cost of plastering its vertical walls at the rate of Rs. 15/m2.
l = 10 m
b = 8
h = 28 m
Volume of soil = volume of cuboid
= l × b × h = 10 × 8 × 28 = 2240 m3
2240 m3 of soil has to be excavated.
Area to be plastered = L.S.A of cuboid
= 2h(l + b)
= 2 × 28(10 + 8)
= 56 × 18
= 1008 m2
The cost of plastering 1m² = Rs. 15
∴ The cost of plastering = 15 × 1008 = Rs. 15,120

Question 5.
A solid cubical box of fine wood costs Rs. 256 at the rate of Rs. 500/m3. Find its volume and length of each side.
$$\frac{256}{500}=\frac{128}{250}=\frac{64}{125} \mathrm{m}^{3}$$
∴ Volume of the wood = $$\frac{64}{125} \mathrm{m}^{3}$$ = 0.512 m3
$$l=\sqrt[3]{0.512}$$