Show that each of the following numbers is a perfect square. Also find the number whose square is the given number in each case :
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761.

To do :

We have to show that each of the given number is a perfect square and find the numbers whose squares are the given numbers.

Solution:

Perfect Square: A perfect square has each distinct prime factor occurring an even number of times.

(i) Prime factorisation of 1156 $=2\times2\times17\times17$

$=(2)^2\times(17)^2$

$=(2\times17)^2$

$=(34)^2$

1156 has distinct prime factors occurring an even number of times.

Therefore, 1156 is a perfect square and it is a square of 34. 

(ii) Prime factorisation of 2025 $=3\times3\times3\times3\times5\times5$

$=(3)^2\times(3)^2\times(5)^2$

$=(3\times3\times5)^2$

$=(45)^2$

2025 has distinct prime factors occurring an even number of times.

Therefore, 2025 is a perfect square and it is a square of 45. 

(iii) Prime factorisation of 14641 $=11\times11\times11\times11$

$=(11)^2\times(11)^2$

$=(11\times11)^2$

$=(121)^2$

14641 has distinct prime factors occurring an even number of times.

Therefore, 14641 is a perfect square and it is a square of 121. 

(iv) Prime factorisation of 4761 $=3\times3\times23\times23$

$=(3)^2\times(23)^2$

$=(3\times23)^2$

$=(69)^2$

4761 has distinct prime factors occurring an even number of times.

Therefore, 4761 is a perfect square and it is a square of 69. 

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Updated on: 10-Oct-2022

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