112 pounds in a hundredweight
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Interesting one! If a rectangle has width w and height h, 1) the area a = wh 2) the perimeter p = 2w + 2h From 1), w=a/h Substituting into 2) p = 2a/h + 2h Multiplying by h pH = 2a + 2h2 Rearranging 2h2 - pH + 2a = 0 This is a quadratic equation in h. Factorising, using the standard quadratic formula, h = (p ± sqrt(p2 - 16a))/4 The two solutions to this will be the two dimensions.
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Let P be perimeter, A is area, L is length, and W is width.A = L x W, P = 2L + 2W; you know P and W, so L = (P - 2W)/2and A = W x (P - 2W)/2