Students can Download Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.4

1. Find the square root of the following numbers by factorization.

Question (i)

196

Answer:

196 = 2 × 2 × 7 × 7

= 2 × 7 × 2 7

196 = 14 × 14

√196 = 14

Question (ii)

256

Answer:

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

256 = 16 × 16

√256 = 16

Question (iii)

10404

Answer:

10404 = 2 × 2 × 3 × 3 × 17 × 17

= 2 × 3 × 17 × 2 × 17

10404 = 102 × 102

√10404 = 102

Question (iv)

1156

Answer:

1156 = 2 × 2 × 17 × 17

= 2 × 17 × 2 × 17

1156 = 34 × 34

√1156 = 34

Question (v)

13225

Answer:

13225 = 5 × 5 × 23 × 23

= 5 × 23 × 5 × 23

13225 = 115 × 115 7

√13225 = 115

2. Simplify

Question (i)

√100 + √36

Answer:

√100 + √36 = 10 + 6 = 16

Question (ii)

√1360 + 9

Answer:

√1360 + 9 = √1369

√1369 = 37 × 37

√1369 = 37

Question (iii)

√2704+√144+√289

Answer:

2704 = 2 × 2 × 2 × 2 × 13 × 13

= 2 × 2 × 13 × 2 × 2 × 13

2704 = 52 × 52

√2704 =52

√144 = 12 and √289 = 17

∴ √2704 + √144 + √289 = 52 + 12 + 17 = 81

Question (iv)

√225 – √25

Answer:

15 – 5=10

Question (v)

√1764 – √1444

Answer:

1764

1764 = 2 × 2 × 3 × 3 × 7 × 7

=2 × 3 × 7 × 2 × 3 × 7

1764 = 42 × 42

√l764 = 42

1444 = 2 × 2 × 19 × 19

= 2 × 19 × 2 × 19

1444 = 38 × 38

√1444 = 38

∴√l764 – √1444 = 42 – 38 = 4

Question (vi)

√169 × √361

Answer:

√169 × √361 = 13 × 19 = 247

Question 3.

A square yard has area 1764 m^{2}. From a corner of this yard, another square part of area 784 m^{2} is taken out for public utility. The remaining portion is divided into equal square parts. What is the perimeter of each of these equal parts?

Answer:

The area of the remaining portion= 1764 – 784 = 980 m²

It is divided into 5, equal square parts

∴ Each square part = \(\frac { 980 }{ 5 }\) = 196 m²

Area of each square part = 196 m²

(side)² = 196

side = √196

side = 14.

The length of each side of the square = 14m

Perimeter = 4 side = 4 × 14 = 56m

4. Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square.

Question (i)

847

Answer:

847 = 7 × 11 × 11

Here we observe that 7 appears only once.

∴ The required number of multiplied is 7.

Question (ii)

450

Answer:

450 = 2 × 5 × 5 × 3 × 3

Here we observe that 2 appears only once

∴ The required number is 2.

Question (iii)

1445

Answer: 1445 = 5 × 17 × 17

Here we observe that 5 appears only once

∴ The required number of multiplied is 5.

Question (iv)

1352

Answer:

1352 = 2 × 2 × 2 × 13 × 13

= 2 × 13 × 2 × 13 × 2

= 26 × 26 × 2

Here we observe that 2 appears only once in the product.

∴ The required number multiplier is 2

5. Find the largest perfect square factor of each of the following numbers.

Question (i)

48

Answer:

48 = 2 × 2 × 2 × 2 × 3

48= 16 × 3

∴ Largest perfect square factor is 16.

Question (ii)

11280

Answer:

11280 = 2 × 2 × 2 × 2 × 3 × 5 × 47

= 16 × 3 × 5 × 47

∴ Largest perfect square factor is 16.

Question (iii)

729

Answer:

729 = 3 × 3 × 3 × 3 × 3 × 3

= 27 × 27

= 729

∴ Largest perfect square factor is 729.

Question (iv)

1352

Answer: 1352 = 2 × 2 × 2 × 13 × 13

= 2 × 13 × 2 × 3 × 2

= 26 × 26 × 2

= 676 × 2

∴ The largest perfect square factor is 676