Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set - Page 677: 31


$7 \text{ and } 7\sqrt{3}$

Work Step by Step

Since the given acute angle of the right triangle is $30^o,$ then the side opposite this angle measures $ 7 $ (half the given hypotenuse, $14.$) 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=7 $ and $ c=14 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 7^2+b^2=14^2 \\\\ 49+b^2=196 \\\\ b^2=196-49 \\\\ b^2=147 \\\\ b=\sqrt{147} \\\\ b=\sqrt{49\cdot3} \\\\ b=\sqrt{(7)^2\cdot3} \\\\ b=7\sqrt{3} .\end{array} Hence, the missing sides have measure $ 7 \text{ and } 7\sqrt{3} $ units.
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