# A right triangle has one leg that is 8 inches shorter than the other leg, the hypotenuse is 30 inches long, how do you find the length of each leg?

Jun 3, 2017

The legs are $24.83 \mathmr{and} 16.83$ inches long.

#### Explanation:

The lengths of the two short sides (the legs of the right-angle) are:

$x \mathmr{and} \left(x - 8\right)$ inches

The length of the hypotenuse is $30$ inches

Write an equation using Pythagoras' Theorem:

${x}^{2} + {\left(x - 8\right)}^{2} = {30}^{2}$

${x}^{2} + {x}^{2} - 16 x + 64 = 900 \text{ } \leftarrow$ make =0

$2 {x}^{2} - 16 x - 836 = 0 \text{ } \leftarrow \div 2$

${x}^{2} - 8 x - 418 = 0 \text{ } \leftarrow$ does not factorise

Solve for $x$ by completing the square method:

${x}^{2} - 8 x + 16 = 418 + 16$

${\left(x - 4\right)}^{2} = 434$

$x - 4 = \pm \sqrt{434} \text{ } \leftarrow$ ignore the negative root.

$x = \sqrt{434} + 4$

$x = 24.83$

The legs are $24.83 \mathmr{and} 16.83$ inches long.

Check:

$\sqrt{{24.83}^{2} + {16.83}^{2}} = 30$