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Q. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. |
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Answer» Answer: Let the sides of the first and second square be X and Y . Area of the first square = (X)² Area of the second square = (Y)² According to question, (X)² + (Y)² = 468 m² ——(1). Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y According to question, 4X – 4Y = 24 ——–(2) From equation (2) we get, 4X – 4Y = 24, 4(X-Y) = 24 X – Y = 24/4 , X – Y = 6 X = 6+Y ———(3) Putting the value of X in equation (1) (X)² + (Y)² = 468, (6+Y)² + (Y)² = 468 (6)² + (Y)² + 2 × 6 × Y + (Y)² = 468 36 + Y² + 12Y + Y² = 468 2Y² + 12Y – 468 +36 = 0 2Y² + 12Y -432 = 0 2( Y² + 6Y – 216) = 0 Y² + 6Y – 216 = 0 Y² + 18Y – 12Y -216 = 0 Y(Y+18) – 12(Y+18) = 0 (Y+18) (Y-12) = 0 (Y+18) = 0 Or (Y-12) = 0 Y = -18 OR Y = 12 Putting Y = 12 in EQUATION (3) X = 6+Y = 6+12 = 18 Side of first square = X = 18 m Side of second square = Y = 12 m |
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