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Question 3 Squares and Square Roots Exercise 3A Next
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.
Was This helpful? > 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 |