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

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 m²
It is divided into 5, equal square parts
∴ Each square part = 9805 = 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

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