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The sum of areas of two squares is 468m2 If the difference of their perimeters is 24cm, find the sides of the two squares.
Ans: Let the side of the larger square be x.
Let the side of the smaller square be y. APQ x2+y2 = 468
Cond. II 4x-4y = 24
⇒ x - y = 6
⇒ x = 6 + y
x2 + y2 = 468
⇒ (6+y)2 +y2 = 468
on solving we get y = 12
⇒ x = (12+6) = 18 m
∴ sides are 18m & 12m.
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