## Geometry: Common Core (15th Edition)

$x = 5\sqrt 2$
In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, the hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $x$, one of the legs: $10 = \sqrt 2(x)$ Divide each side of the equation by $\sqrt 2$ to solve for $x$: $x = \frac{10}{\sqrt 2}$ To simplify this fraction, we multiply both the numerator and denominator by the denominator: $x = \frac{10}{\sqrt 2} • \frac{\sqrt 2}{\sqrt 2}$ Multiply to simplify: $x = \frac{10\sqrt 2}{2}$ Divide the numerator and denominator by their greatest common factor, $2$: $x = 5\sqrt 2$
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