NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 1 — Figure it Out (Page 10)#
A perfect square is a number of the form $n \times n$. A perfect square always ends in 0, 1, 4, 5, 6 or 9, and its prime factors can be split into two identical groups. These two facts solve most of the questions below.
1. Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089 #
Solution
A perfect square can only end in 0, 1, 4, 5, 6 or 9. We look at the last digit of each number:
- 2032 ends in 2, so it is not a perfect square.
- 2048 ends in 8, so it is not a perfect square.
- 1027 ends in 7, so it is not a perfect square.
- 1089 ends in 9, so it may be a perfect square. We check it by factorising:
$1089 = 3 \times 3 \times 11 \times 11$
$1089 = (3 \times 11) \times (3 \times 11)$
$1089 = 33 \times 33$
So 1089 is a perfect square.
∴ 2032, 2048 and 1027 are not perfect squares.
2. Which one among 64², 108², 292², 36² has last digit 4? #
Solution
The last digit of a square depends only on the last digit of the number, so we square the last digit of each number:
- $64^2$: last digit 4, and $4 \times 4 = 16$, so $64^2$ ends in 6.
- $108^2$: last digit 8, and $8 \times 8 = 64$, so $108^2$ ends in 4.
- $292^2$: last digit 2, and $2 \times 2 = 4$, so $292^2$ ends in 4.
- $36^2$: last digit 6, and $6 \times 6 = 36$, so $36^2$ ends in 6.
∴ $108^2$ and $292^2$ have last digit 4.
3. Given 125² = 15625, what is the value of 126²? (i) 15625 + 126 (ii) 15625 + 26² (iii) 15625 + 253 (iv) 15625 + 251 (v) 15625 + 51² #
Solution
We write 126 as $125 + 1$ and expand the square:
$126^2 = (125 + 1)^2$
$126^2 = 125^2 + 2 \times 125 \times 1 + 1^2$
$126^2 = 15625 + 250 + 1$
$126^2 = 15625 + 251$
∴ The answer is (iv) 15625 + 251, that is $126^2 = 15876$.
4. Find the length of the side of a square whose area is 441 m². #
Solution
The area of a square is side × side, so the side is the square root of the area.
We factorise 441 into primes:
$441 = 3 \times 3 \times 7 \times 7$
$441 = (3 \times 7) \times (3 \times 7)$
$441 = 21 \times 21$
$\sqrt{441} = 21$
∴ The side of the square is 21 m.
5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10. #
Solution
A number divisible by 4, 9 and 10 is a multiple of their LCM.
$4 = 2 \times 2$
$9 = 3 \times 3$
$10 = 2 \times 5$
$LCM = 2 \times 2 \times 3 \times 3 \times 5 = 180$
In 180, the factors 2 and 3 come in pairs, but 5 is alone. To make a perfect square, we multiply by one more 5:
$180 \times 5 = 2 \times 2 \times 3 \times 3 \times 5 \times 5$
$180 \times 5 = (2 \times 3 \times 5) \times (2 \times 3 \times 5)$
$180 \times 5 = 30 \times 30$
$180 \times 5 = 900$
Check: $900 \div 4 = 225$, $900 \div 9 = 100$ and $900 \div 10 = 90$.
∴ The smallest square number divisible by 4, 9 and 10 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. #
Solution
We factorise 9408 into primes:
$9408 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7$
The 2s and the 7s come in pairs, but 3 is alone. So we multiply by 3:
$9408 \times 3 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7$
$9408 \times 3 = (2 \times 2 \times 2 \times 3 \times 7) \times (2 \times 2 \times 2 \times 3 \times 7)$
$9408 \times 3 = 168 \times 168$
$9408 \times 3 = 28224$
$\sqrt{28224} = 168$
∴ 9408 must be multiplied by 3. The product is 28224 and its square root is 168.
7. How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100 #
Between $n^2$ and $(n + 1)^2$ there are $2n$ numbers, because $(n + 1)^2 - n^2 = 2n + 1$, and we leave out the square at the end.
(i) 16 and 17
$16^2 = 256$
$17^2 = 289$
The numbers between them are 257, 258, …, 288.
Number of numbers $= 288 - 257 + 1 = 32$
This is $2 \times 16 = 32$.
∴ 32 numbers lie between $16^2$ and $17^2$.
(ii) 99 and 100
$99^2 = 9801$
$100^2 = 10000$
Number of numbers $= 2 \times 99 = 198$
Check: the numbers are 9802, 9803, …, 9999, and $9999 - 9802 + 1 = 198$.
∴ 198 numbers lie between $99^2$ and $100^2$.
8. In the following pattern, fill in the missing numbers: #
$1^2 + 2^2 + 2^2 = 3^2$
$2^2 + 3^2 + 6^2 = 7^2$
$3^2 + 4^2 + 12^2 = 13^2$
$4^2 + 5^2 + 20^2 = (\underline{\quad})^2$
$9^2 + 10^2 + (\underline{\quad})^2 = (\underline{\quad})^2$
Solution
In each line, the first two numbers are consecutive. The third number is their product, and the number on the right is one more than the product:
- $1 \times 2 = 2$, and $2 + 1 = 3$
- $2 \times 3 = 6$, and $6 + 1 = 7$
- $3 \times 4 = 12$, and $12 + 1 = 13$
For 4 and 5: $4 \times 5 = 20$, and $20 + 1 = 21$.
For 9 and 10: $9 \times 10 = 90$, and $90 + 1 = 91$.
Check:
$4^2 + 5^2 + 20^2 = 16 + 25 + 400$
$4^2 + 5^2 + 20^2 = 441$
$4^2 + 5^2 + 20^2 = 21^2$
$9^2 + 10^2 + 90^2 = 81 + 100 + 8100$
$9^2 + 10^2 + 90^2 = 8281$
$9^2 + 10^2 + 90^2 = 91^2$
∴ The completed lines are $4^2 + 5^2 + 20^2 = \mathbf{21}^2$ and $9^2 + 10^2 + \mathbf{90}^2 = \mathbf{91}^2$.
9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares. #
The picture is a square pattern of 9 rows with 9 tiles in each row. Every tile, whether upright or turned like a diamond, is a 5 × 5 grid of tiny squares.
Solution
Number of tiles $= 9 \times 9 = 81$
Number of tiny squares in one tile $= 5 \times 5 = 25$
Number of tiny squares $= 81 \times 25$
Number of tiny squares $= 2025$
We factorise 2025 into primes:
$2025 = 81 \times 25$
$2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5$
$2025 = 3^4 \times 5^2$
∴ There are 2025 tiny squares, and $2025 = 3^4 \times 5^2$.
Since the prime factors split into two identical groups, $(3 \times 3 \times 5) \times (3 \times 3 \times 5)$, the number 2025 is itself a perfect square: $2025 = 45^2$.