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1.Sum of the areas of two squares is 468 m2 and difference between theirperimeters is 24m. Then find the sides of the squares. |
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Answer» Let the sides of squares be a m & b m long such that a > b. By 1st condition, a^2 + b^2 = 468 ... (1) By 2nd condition,4a-4b = 24a - b = 6.a = b + 6. Putting this value of a in equation (1),(b+6)^2 + b^2 = 468b^2 + 12b + 36 + b^2 = 468b^2 + 6b + 18 = 234b^2 + 6b - 216 = 0.b^2 + 18b - 12b - 216 = 0.b (b+18) - 12 (b+18) = 0(b+18)(b-12) = 0.b = -18 or b = 12.But b = -18 is unacceptable so b = 12.a = 18.Lengths of sides of given squares are 12 m & 18 |
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