# Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set: 34

$4\sqrt{2} \text{ and } 4\sqrt{2}$

#### Work Step by Step

Since the given acute angle of the right triangle is $45^o,$ then the the legs ($a$ and $a$), measure $\dfrac{\sqrt{2}}{2}\cdot c,$ where $c$ is the hypotenuse. Hence, the legs measure \begin{array}{l}\require{cancel} \dfrac{\sqrt{2}}{2}\cdot c=a \\\\ \dfrac{\sqrt{2}}{2}\cdot 8=a \\\\ 4\sqrt{2}=a \\\\ a=4\sqrt{2} .\end{array} Hence, the missing sides have measure $4\sqrt{2} \text{ and } 4\sqrt{2}$ units.

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