A Square and a Cube l Exercise 1.1 l Class 8 l NCERT Solutions l Ganita Prakash

NCERT Solutions for Class 8 Maths – Ganita Prakash

Exercise 1.1

1. Which of the following numbers are not perfect squares?

Given numbers: 2032, 2048, 1027, 1089

We know:

33² = 1089

So, 1089 is a perfect square.

Therefore, the numbers which are not perfect squares are:

2032, 2048, 1027

2. Which one among 64², 108², 292², 36² has last digit 4?

To find the last digit of a square, we only check the last digit of the number.

64² → last digit of 4² = 16 → last digit = 6
108² → last digit of 8² = 64 → last digit = 4
292² → last digit of 2² = 4 → last digit = 4
36² → last digit of 6² = 36 → last digit = 6

Therefore, the numbers having last digit 4 are:

108² and 292²

3. Given 125² = 15625, what is the value of 126²?

We use the identity:

(a + 1)² = a² + 2a + 1

Here, a = 125

126² = (125 + 1)²
= 125² + 2 × 125 + 1
= 15625 + 250 + 1
= 15625 + 251
= 15876

Therefore,

126² = 15876

Correct option: (iv) 15625 + 251

4. Find the length of the side of a square whose area is 441 m².

Formula:

Area of square = side × side = side²

Given area = 441 m²
So, side = √441 = 21

Therefore, the length of the side is 21 m.

5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.

First, find the prime factorisation:

4 = 2²
9 = 3²
10 = 2 × 5

LCM of 4, 9 and 10 = 2² × 3² × 5 = 180

Now, for a perfect square, each prime must appear an even number of times.

180 = 2² × 3² × 5

Here, power of 5 is odd, so multiply by 5:

180 × 5 = 2² × 3² × 5² = 900

Therefore, the smallest square number is 900.

6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.

Prime factorisation of 9408:

9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
= 2⁶ × 3 × 7²

For a perfect square, all powers must be even.

Power of 2 is 6 → even
Power of 3 is 1 → odd
Power of 7 is 2 → even

So, we must multiply by 3.

Required product = 9408 × 3 = 28224

Now,
28224 = 2⁶ × 3² × 7²

√28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168

Therefore:

Smallest number to be multiplied = 3
Square root of the product = 168

7. How many numbers lie between the squares of the following numbers?

We know:

Number of integers between n² and (n + 1)² = 2n

(i) 16 and 17

Numbers between 16² and 17²
= 2 × 16
= 32

So, 32 numbers lie between 16² and 17².

(ii) 99 and 100

Numbers between 99² and 100²
= 2 × 99
= 198

So, 198 numbers lie between 99² and 100².

8. In the following pattern, fill in the missing numbers:

1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = (21)²
9² + 10² + (90)² = (91)²

Therefore, the missing numbers are:

9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.

From the picture:

There are 81 big squares in total.
Each big square contains 25 tiny squares.

So, total number of tiny squares = 81 × 25 = 2025

Prime factorisation of 2025:

2025 = 45 × 45
= (9 × 5) × (9 × 5)
= 3⁴ × 5²