One leg of a right triangle is 96 inches. How do you find the hypotenuse and the other leg if the length of the hypotenuse exceeds 2 times the other leg by 4 inches?

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One leg of a right triangle is 96 inches. How do you find the hypotenuse and the other leg if the length of the hypotenuse exceeds 2 times the other leg by 4 inches?

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Answer 1 · 2016-05-13T14:43:30

hypotenuse $180.5$ $180.5$ , legs $96$ $96$ and $88.25$ $88.25$ approx.

Explanation:

Let the known leg be $c_{0}$ $c_0$ , the hypotenuse be $h$ $h$ , the excess of $h$ $h$ over $2 c$ $2c$ as $δ$ $delta$ and the unknown leg, $c$ $c$ . We know that $c^{2} + c_{0}^{2} = h^{2}$ $c^2+c_0^2=h^2$ (Pytagoras) also $h - 2 c = δ$ $h-2c = delta$ . Subtituting according to $h$ $h$ we get: $c^{2} + c_{0}^{2} = {(2 c + δ)}^{2}$ $c^2+c_0^2=(2c+delta)^2$ . Simplifiying, $c^{2} + 4 δ c + δ^{2} - c_{0}^{2} = 0$ $c^2+4delta c+delta^2-c_0^2=0$ . Solving for $c$ $c$ we get.
$c = \frac{- 4 δ \pm \sqrt{16 δ^{2} - 4 (δ^{2} - c_{0}^{2})}}{2}$ $c = (-4delta pm sqrt(16delta^2-4(delta^2-c_0^2)))/2$
Only positive solutions are allowed
$c = \frac{2 \sqrt{4 δ^{2} - δ^{2} + c_{0}^{2}} - 4 δ}{2} = \sqrt{3 δ^{2} + c_{0}^{2}} - 2 δ$ $c = (2sqrt(4delta^2-delta^2+c_0^2)-4delta)/2=sqrt(3delta^2+c_0^2)-2delta$

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One leg of a right triangle is 96 inches. How do you find the hypotenuse and the other leg if the length of the hypotenuse exceeds 2 times the other leg by 4 inches?

1 Answer

Explanation:

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