Sum of the areas of two squares is 260 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the sides of two squares be 'a' m and 'b' m. As we know that area of square = (side)2 and perimeter of square = 4 x side It is given that sum of the areas of two squares is 260 m2. ⇒ a2 + b2 = 260 .......... (1) And, the difference of their perimeters is 24 m. ⇒ 4a - 4b = 24 ⇒ 4(a - b) = 24 ⇒ a - b = ⇒ a - b = 6 ⇒ a = 6 + b .......... (2) Substituting the value of a in equation (1), we get ⇒ (6 + b)2 + b2 = 260 ⇒ 36 + b2 + 12b + b2 = 260 ⇒ 36 + 2b2 + 12b - 260 = 0 ⇒ 2b2 + 12b - 224 = 0 ⇒ 2(b2 + 6b - 112) = 0 ⇒ b2 + 6b - 112 = 0 ⇒ b2 + 14b - 8b - 112 = 0 ⇒ b(b + 14) - 8(b + 14) = 0 ⇒ (b + 14)(b - 8) = 0 ⇒ (b + 14) = 0 or (b - 8) = 0 ⇒ b = -14 or b = 8 Since, side of square cannot be negative. Side of one square = 8 m Side of other square = 6 + 8 = 14 m Thus, side of squares = 8 m and 14 m.

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Mathematics

Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Quadratic Equations

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