In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.4 pdf, free KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.4 pdf download. Now you will get step by step solution to each question.

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

## Karnataka Board 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 an area 1764 m2. From a corner of this yard, another square part of area 784 m2 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 m^{2}

It is divided into 5, equal square parts

∴ Each square part = 9805 = 196 m^{2}

Area of each square part = 196 m^{2}

(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

**All Chapter KSEEB Solutions For Class 8 maths**

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