Class 8

Students can Download Maths Chapter 10 Exponents Ex 10.2 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 10 Exponents Ex 10.2

Question 1.
Simplify
i. 3¹ × 3² × 3³ × 3⁴ × 3^s × 3⁶
ii. 2² × 3³ × 2⁴ × 3⁵ × 3⁶
Answer:
i. 3^1+2+3+4+5+6
₌ 3²¹
[a^m x aⁿ = a^m+n]

ii. 2² × 3³ × 2⁴ × 3⁵ × 3⁶ [a^m x aⁿ = a^m+n]
= 2²⁺⁴ × 3³⁺⁵⁺⁶
= 2⁶ × 3¹⁴

Question 2.
How many zeros are there in 10⁴ x 10³ x 10² x 10?
Answer:
10⁴ × 10³ × 10² × 10
= 10^4+3+2+1
= 10¹⁰
There are 10 zeros.

Question 3.
Which is larger (5³ x 5⁴ x 5^s x 5⁶) or (5⁷ x 5⁸)
Answer:
5³ x 5⁴ x 5⁵ x 5⁶
= 5^3+4+5+6
= 5¹⁸ 5⁷ x 5⁸
= 5⁷⁺⁸
= 5⁸
∴ 5³ x 5⁴ x 5^s x 5⁶ is larger than 5⁷ x 5⁸

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.3 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 15 Quadrilaterals Ex 15.3

Question 1.
The adjacent angles of a parallelogram are in the ratio 2 : 1. Find the measures of all the angles.
Answer:
The adjacent angles are in the ratio 2 : 1.
Let the angles be 2x and x
2x + x = 180°
[adjacent angles of a parallelogram are supplementary]
3x = 180° .
x =  = 60°
2x = 2 × 60° = 120°
∴The angles of the parallelogram are 60°, 120°. 60° and 120°

Question 2.
A field is in the form of a parallelogram whose perimeter is 450m and one of its sides is larger than the other by 75m. Find the lengths of all the sides.
Answer:
Let one side be x, then the other side is x + 75

Perimeter = 450
x + (x + 75) + x + (x + 75) = 450
4x + 150 = 450
4x = 450 – 150
4x = 300
x = \(\frac { 300 }{ 4 }\) = 75
x = 75m
x + 75 = 75 + 75 = 150 m
∴ The four sides are 75m, 150m, 75m, 150m.

Question 3.
In the figure ABCD is a parallelogram. The diagonals AC and BD intersect at 0; and ∠DAC = 40°, ∠CAB = 35° and ∠DOC = 110° calculate the measure ∠ABO, ∠ADC, ∠ACB and ∠CBD
Answer:

∠AOB = ∠DOC = 110°
[Vertically opposite angles]
In ΔAOB,
∠AOB + ∠OAB + ∠OBA = 180°
110 + 55 +∠OBA = 180°
∠OBA = 180 – 165
∠ABO = 15°
∠ADC + ∠DAB = 180°
[adjacent angles of a parallelogram , are supplementary]
∠ADC + 95 = 180°
∠ADC = 180 – 95
∠ ADC = 85°
∠ACB = ∠DAC = 40°
[AD 11BC alternate angle]
∠ACB = 40°
∠ABC = ∠ ADC = 85°
[Opposite angles pf a parallelogram]
[∠ABC = ∠ABO + ∠CBO]
15 + ∠CBO = 85°
∠CBO = 85 – 15
∠CBO = 70
∴ ∠CBO = ∠CBD = 70°
∴ ∠CBD = 70°

Question 4.
In a parallelogram ABCD, the side DC is produced to E and ∠BCE = 105° calculate ∠A, ∠E, ∠C, and∠D.

Answer:
∠DCB + ∠DCE = 180° [Linear pair] ∠DCB + 1050 = 180°
∠DCB = 180 – 105
∠DCB = 75°
∠A = ∠DCB = 75°
∠A = ∠C = 75°
[Opposite angles of a parallelogram]
∠A +∠B = 180°
[adjacent angles of a parallelogram]
75° + ∠B = 180°
∠B = 180 + 75
∠B = 105°
∠B = ∠D = 105°
[Opposite angles of a parallelogram]

Question 5.
In a parallelogram KLMN, ∠K = 60 °. Find the measures of all the angles.
Answer:

∠K = ∠M = 60° [Opposite angles of parallelogram]
∠K + ∠N = 180° [adjacent angles of a parallelogram]
60 + ∠N = 180°
∠N = 180 – 60
∠N = 120°
∠N = ∠L = 120°
[Opposite angles of a parallelograom]
∴ Angles of a paralelogram are 60°, 120°, 60° & 120°

Students can Download Maths Chapter 11 Congruency of Triangles Ex 11.2 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 11 Congruency of Triangles Ex 11.2

Question 1.
In the adjoining figure PQRS is a rectangle. Identify the congruent triangles formed by the diagonals.

Answer:
Δ PSR ≅ Δ PQR, Δ POQ ≅ Δ SOR
Δ SPQ ≅ Δ SRQ, Δ POS ≅ Δ QOR

Question 2.
In the figure ABCD is a square. M, N, O & P are the midpoints of sides AB, BC, CD and DA respectively. Identify the congurent triangles.

Answer:
Δ APM ≅ Δ BMD ≅ Δ CNO ≅ Δ DOP

Question 3.
In a triangle ABC, AB = AC. Points G on AB and D on AC are such that AE = AD. Prove that triangles BCD and CBE are congruent.

Answer:
In the figure AB = AC (data)
∴∠ABC = ∠ACB
[Base angles of an isosceles triangle]
AB = AC(data)
AE = AD(data)
AB – AE = AC – AD (Axiom 3)
BE = DC
In Δ BCD and Δ CBE
∠BCD = ∠EBC
BE = DC (Proved)
BC = BC (Common side)
∴ Δ BCD= Δ CBE [SAS postulate]

Question 4.
In the adjoining figure the sides BA and CA have been produced such that BA = AD and CA = AE prove that DE || BC (hint : use the concept of alternate angles.)

Answer:
Δ AED and Δ ABC
AE = AC (data)
AD = AB (data)
∠EAD = ∠BAC
[Vertically opposite angles]
∴ Δ AED = ABC
[SAS postulate]
∴ ∠AED = ∠ACB [Congruent triangle property]
But these are alternate angles.
∴ DE || BC
Consequences of SAS postulate.

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.2 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 15 Quadrilaterals Ex 15.2

Question 1.
In a trapezium PQRS, PQ || RS and ∠E= 70° and Q = 80° calculate the measure of ∠S and ∠R.
Answer:

∠P + ∠S = 180°
[adjacent angles correspondingto parallel sides 70° + ∠S = 180°]
∠S = 180 – 70
∠S = 110°
∠R + ∠Q = 180°
[adjacent angles corresponding to parallel sides ∠R + 80° = 180°]
∠R = 180 – 80
∠R = 100°

Question 2.
In trapezium ABCD, AB || CD. It is given that AD is not parallel to BC. Is ∆ABC ≅ ∆ADC ? Give reasons.
Answer:
∆ ABC and ∆ ADC are not congruent. It is not possible to show that they are congruent to each other by using any of the four congruence postulates.

Question 3.
In the figure, PQRS is an isosceles trapezium. ∠SRP = 30° and ∠POS = 40°. Calculate the angles ∠RPO and∠RSO.

Answer:
PQ || RS
∠PRS = ∠RPQ = 30° [Alternate angles]
∠RPQ = 30°
∠RSQ = ∠PQS = 40° [Alternate angles]
∠RSQ = 40°

Students can Download Maths Chapter 10 Exponents Ex 10.1 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 10 Exponents Ex 10.1

Question 1.
Express the following numbers in the exponential form.

i. 1728
ii. \(\frac { 1 }{ 512 }\)
iii. 0.000169
Answer:
i. 1728 = 12 x 12 x 12
1728 = 12³

ii. 

iii. 0.000169 = (0.013)^-2

Question 2.
Write the following numbers using base 10 and exponents.
i. 12345
ii. 1010.0101
iii. 0.1020304
Answer:

Question 3.
Find the value of (-0.2)^-4
Answer:

= -5 x -5 x -5 x -5
= 625

Students can Download Maths Chapter 12 Construction of Triangles Ex 12.12 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 12 Construction of Triangles Ex 12.12

Question 1.
Construct triangle ABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Answer:

Question 2.
Construct triangle ABC in which BC = 5cm, AB – AC = 2.8 cm and ∠B = 40°.
Answer:

Question 3.
Construct triangle ABC in which BC = 6 cm AB – AC = 3.1 cm and <B = 30°.
Answer:

Students can Download Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7 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 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.7

Question 1.
Find the cube root by prime factorisation,
i) 1728
ii) 3375
iii) 10648
iv) 46656
v) 15625
Answer:
i) 1728 = 2 × 2 × 2 × 6 × 6 × 6 = 12 × 12 × 12
\(\sqrt[3]{1728}=12\)

ii) 3375 = 5 × 5 × 5 × 3 × 3 × 3 = 15 × 15 × 15
\(\sqrt[3]{3375}=15\)

iii) 10648 = 2 × 2 × 2 × 11 × 11 × 11
= 2 × 11 × 2 × 11 × 2 × 11
= 22 × 22 × 22
10648 = 22³
∴ \(\sqrt[3]{10648}=22\)

iv) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= 4 × 4 × 4 × 9 × 9 × 9
= 4 × 9 × 9 × 4 × 9 × 4
46656 = 36 × 36 × 36 = 36³
∴ \(\sqrt[3]{46656}=36\)

v) 15625 = 5 × 5 × 5 × 5 × 5 × 5
= 25 × 25 × 25
15625 = 25³
∴ \(\sqrt[3]{15625}=25\)

2. Find the cube root of the following by looking at the last digit and using estimation.

Question (i)
91125
Answer:
Unit digit of 91125 in 5. Therefore the units digit in its cube root is 5.
Let us split 91125 as 91 and 125. We find that 43 = 64 < 91 < 125 = 5³
Hence 40³ = 64000 <91125 < 125000 = 50³
∴Cube root of 91125 lies between 40 and 50 and units digit is 5 the only such number is 45.
∴ \(\sqrt[3]{91125}=45\)

Question (ii)
166375
Answer:
Units digit of 166375 is 5. Therefore the units digit of its cube root is 5.
Let us split 166375 as 166 and 375.
5³ = 125 < 166 < 216 = 6³
Hence 50³ = 125000, < 166375 < 216000 = 60³
∴ \(\sqrt[3]{166375}\) lies between 50 and 60. .
Since the units digit is 5 the only such number in 55.
∴ \(\sqrt[3]{166375}=55\)

Question (iii)
704969
Answer:
The units digit of 704969 is 9. Therefore the units digits of its cube root are 9.
Let us split 704969 as 704 and 969
8³ = 512 < 704 < 729 – 9³
Hence = 80³= 512000 < 704969 < 72900 = 90³
∴ \(\sqrt[3]{704969}\) lies between 80 and 90.
Since the units digits is 9 the only such number is 89.
∴ \(\sqrt[3]{704969}=89\)

3. Find the nearest integer to the cube root of each of the following.
(i) 331776
(ii) 46656
(iii) 373248

Question (i)
331776
Answer:
60³ = 216000 < 331776 < 343000 = 70³
Hence \(\sqrt[3]{331776}\) lies between 60 and 70.
We do not know whether 331776 in a perfect cube or not.
However we may sharpen the bound.
68³ = 314432, 69³ = 328509
Hence \(\sqrt[3]{331776}\) lies between 69 and 70
331776 – 328509 = 3267
343000 – 331776 = 11224
331776 in nearer to 69³
∴ The closest integer to \(\sqrt[3]{331776}\) is 69.

Question (ii)
46656
Answer:
30³ = 2700 < 46656 < 64000 – 40₃
\(\sqrt[3]{46656}\) lies between 30 and 40 we do not know whether 46656 in a perfect cube or not. However we may sharper the bound
35³ = 42, 875, 36³ = 46656
∴ \(\sqrt[3]{46656}=36\)

Question (iii)
373248
Answer:
70³ = 343000 < 373248 < 512000 – 80³
\(\sqrt[3]{373248}\) lies between 70 and 80. We do not know whether 373248 is a perfect cube or not.
However we may sharpen the bound. 71³ = 357911, 72³ = 373248
∴ \(\sqrt[3]{373248}=72\)

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.1 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 15 Quadrilaterals Ex 15.1

Question 1.
Two angles of a quadrilateral are 70° and 130° and the other two angles are equal. Find the measure of these two angles.
Answer:
Let the equal angles be x
70 + 130 + x + x = 360°
[Sum of the angles of a quadrilateral 360°]
200 + 2x = 360°
2x = 160
x = 
x = 80°
Each of the other two angles is 80°

Question 2.
In the figure. Suppose ∠P and ∠Q are supplementary angles and ∠R = 125 °. Find the measure of∠S.

Answer:
∠P + ∠Q +∠R + ∠s = 360°
[Sum of the angles of a quadrilateral]
180 + 125° + ∠S = 360°
[∠P and ∠Q are supplementary]
305 + ∠S = 360°
∠ S = 360 – 305
∠S = 55°

Question 3.
Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth angles are 90°. Find the measure of the other three angles.
Answer:
Three angles are in the ratio 2 : 3 : 5 Let the angles be 2x, 3x and 5x
[Sum of the angles of a quadrilateral]
2x + 3x + 5x + 90° = 360°
10x + 90 = 360
10x = 360 – 90
10x = 270°
x =  = 27°
∴ 2x = 2 × 27 = 54°
3x = 3 × 27 = 81°
5x = 5 × 27 = 135°
Three angle are 54°, 81° & 135°

Question 4.
In the adjoining figure, ABCD is a quadrilateral such that ∠D + ∠C = 100°. The bisectors of ∠A and ∠B meet at P.
Determine ∠APB.

Answer:
∠A + ∠B + ∠C + ∠D = 360°
[Sum of the angles of a quadrilateral]
∠A + ∠B = 360 – 100
∠A + ∠B = 260
∠A +   ∠B = × 260°
[Multiplying by  ]
∠PAB + ∠PBA = 130°
[AP and BP are bisectors of ∠A and ∠B]
In ∆ APB. ∠APB +∠PAB +∠PB A = 180°
∠PAB + ∠PBA + ∠APB = 180°
[Sum of the angles of a triangle]
∠APB + 130 = 180°
∠APB = 180 – 130
∠ APB = 50°

Students can Download Maths Chapter 12 Construction of Triangles Ex 12.11 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 12 Construction of Triangles Ex 12.11

Question 1.
Construct triangle ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠Q = 60°.
Answer:

Question 2.
Construct triangle ABC in which AB + AC = 5.6 cm BC = 4.5 cm and ∠B = 45°.
Answer:

Question 3.
Construct triangle PQR in which PQ + PR = 6.5 cm QR = 5.4 cm and ∠Q = 40°.
Answer:

Students can Download Maths Chapter 12 Construction of Triangles Ex 12.10 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 12 Construction of Triangles Ex 12.10

Question 1.
Construct a triangle PQR in which PQ = 5.5 cm PR = 6.2 cm and the length of the perpendicular from P to QR is 4 cm
Answer:

Question 2.
Construct a triangle MNP in which MN = 4.5 cm, MP = 5.2 cm and the length of the perpendicular from M on NP is 3.8 cm.
Answer: