Let the side of the first square be 'a' m and that of the second be ′A′ m. Area of the first square = a2 sq m. Area of the second square = A2 sq m. Their perimeters would be 4a and 4A respectively. Given 4A - 4a = 24 A - a = 6 ......(1) A2 + a2 = 468 ......(2) From (1), A = a + 6 Substituting for A in (2), we get
(a + 6)2 + a2 = 468 a2 + 12a + 36 + a2 = 468 2a2 + 12a + 36 = 468 a2 + 6a + 18 = 234 a2 + 6a - 216 = 0 a2 + 18a - 12a - 216 = 0 a(a + 18) - 12(a + 18) = 0 (a - 12)(a + 18) = 0 a = 12, - 18 So, the side of the first square is 12 m. and the side of the second square is 18 m. |