KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.4

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 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 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