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 2.5 times the other leg by 4 inches?

Jan 15, 2016

Use Pythagoras to establish $x = 40$ and $h = 104$

Explanation:

Let $x$ be the other leg
then the hypotenuse $h = \frac{5}{2} x + 4$
And we are told the the first leg $y = 96$

We can use Pythagoras' equation ${x}^{2} + {y}^{2} = {h}^{2}$
${x}^{2} + {96}^{2} = {\left(\frac{5}{2} x + 4\right)}^{2}$
${x}^{2} + 9216 = 25 {x}^{2} / 4 + 20 x + 16$

Reordering gives us
${x}^{2} - 25 {x}^{2} / 4 - 20 x + 9200 = 0$

Multiply throughout by $- 4$
$21 {x}^{2} + 80 x - 36800 = 0$

Using the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(80\right) \pm \sqrt{6400 + 3091200}}{- 42}$
$x = \frac{- 80 \pm 1760}{42}$

so $x = 40$ or $x = - \frac{1840}{42}$

We can ignore the negative answer as we re dealing with a real triangle, so the other leg $= 40$

The hypotenuse $h = 5 \cdot \frac{40}{2} + 4 = 104$