Using the prime factorization method, find which of the following numbers are perfect squares 5625

Question 3 Squares and Square Roots Exercise 3A

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Answer:

Given 5625,

A perfect square can always be expressed as a product of pairs of equal factors.

Now resolve 5625 into prime factors, we get

5625 = 225 X 25 = 9 X 25 X 25 = 3 X 3 X 5 X 5 X 5 X 5 = 3 X 5 X 5 X 3 X 5 X 5

= 75 X 75 = (75)2

Hence, 75 is the number whose square is 5625

∴ 5625 is a perfect square.

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Using the prime factorisation method, find which of the following numbers are perfect squares: i 441 ii 576 iii 11025 iv 1176 v 5625 vi 9075 vii 4225 viii 1089

Solution

(i) 441 = 3×3×7×7

=(3)2×(7)2=(21)2
∴ 441 is a perfect square.
(ii) 576= 2×2×2×2×2×2×3×

=(2)2×(2)2×(2)2×(3)2
=(2×2×2×3)2=(24)2
∴ 576 is a perfect square.
(iii) 11025= 3×3×5×5×7×7

=(3)2×(5)2×(7)2
=3×5×7)2=(105)2
(iv) 1176 = 2×2×2×3×7×7

=(2)2×2×3×(7)2 1176 is not a perfect square as it cannot be expressed as the product of pair of equal factors

(v) 5625 = 3×3×5×5×5×

=(3)2×(5)2×(5)2
=3×5×5)2=(75)2
∴ 5625 is a perfect square.
(vi) 9075 = 3×5×5×11×11

= 3×(5)2×(11)2
∴ 9075 is not a perfect square as it cannot be expressed as a product of pair of equal factors
(vii) 4225= 5×5×13×13

=(5)2×(13)2
=(5×13)2=(65)2
∴ 4225 is a perfect square.
(viii) 1089= 3×3×11×11

=(3)2×(11)2
=(3×11)2=(33)2
∴ 1089 is a perfect square.

Mathematics

Secondary School Mathematics VIII

Standard VIII