# Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set - Page 677: 29

$5\sqrt{2} \text{ and } 5$

#### Work Step by Step

Since the acute angle of the right triangle is $45^o,$ then the right triangle is isosceles. Hence, the legs both measure $5$ units. Let the right triangle have $a$ and $b$ as the legs and $c$ as the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $a=5$ and $b=5 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 5^2+5^2=c^2 \\\\ 25+25=c^2 \\\\ 50=c^2 \\\\ c=\sqrt{50} \\\\ c=\sqrt{25\cdot2} \\\\ c=\sqrt{(5)^2\cdot2} \\\\ c=5\sqrt{2} .\end{array} Hence, the missing sides have measure $5\sqrt{2} \text{ and } 5$ units.

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