SOLUTION: The diagonal of a rectangle measures 8sqrt(2) inches. If the width is 8 inches less than the length, then find the dimensions of the rectangle.
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-> SOLUTION: The diagonal of a rectangle measures 8sqrt(2) inches. If the width is 8 inches less than the length, then find the dimensions of the rectangle.
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Question 1104923: The diagonal of a rectangle measures 8sqrt(2) inches. If the width is 8 inches less than the length, then find the dimensions of the rectangle. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! use the Pythagorean Theorem to solve
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let x be the length of the rectangle
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x^2 + (x-8)^2 = (8sqrt(2))^2
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x^2 +x^2-16x+64 = 64*2
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2x^2 -16x +64 = 128
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divide both sides of = by 2
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x^2 -8x +32 = 64
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x^2 -8x -32 = 0
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use quadratic formula
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x = (-(-8) + sqrt((-8)^2 -4 * 1 * (-32)))/2(1) = 4 + 4 * sqrt(3) = 4(1 + sqrt(3))
x = 4(1 - sqrt(3))
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we reject the negative value for x
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length of rectangle = 4(1 + sqrt(3)) is approximately 10.93 inches
width is 10.93 - 8 = 2.93 inches
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check the answer
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(10.93)^2 + (2.93)^2 = (8*sqrt(2))^2
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119.46 + 8.58 = 128
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128.04 approximately equals 128
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