# 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?

hypotenuse $180.5$, legs $96$ and $88.25$ approx.
Let the known leg be ${c}_{0}$, the hypotenuse be $h$, the excess of $h$ over $2 c$ as $\delta$ and the unknown leg, $c$. We know that ${c}^{2} + {c}_{0}^{2} = {h}^{2}$(Pytagoras) also $h - 2 c = \delta$. Subtituting according to $h$ we get: ${c}^{2} + {c}_{0}^{2} = {\left(2 c + \delta\right)}^{2}$. Simplifiying, ${c}^{2} + 4 \delta c + {\delta}^{2} - {c}_{0}^{2} = 0$. Solving for $c$ we get.
$c = \frac{- 4 \delta \pm \sqrt{16 {\delta}^{2} - 4 \left({\delta}^{2} - {c}_{0}^{2}\right)}}{2}$
$c = \frac{2 \sqrt{4 {\delta}^{2} - {\delta}^{2} + {c}_{0}^{2}} - 4 \delta}{2} = \sqrt{3 {\delta}^{2} + {c}_{0}^{2}} - 2 \delta$