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 678: 43


$\dfrac{15\sqrt{2}}{2} \text{ and } \dfrac{15\sqrt{2}}{2}$

Work Step by Step

The diagonal and the two sides of the square form a right triangle with the hypotenuse equal to $15.$ Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $ a=b $ and $ c=15 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ a^2+a^2=15^2 \\\\ 2a^2=225 \\\\ a^2=\dfrac{225}{2} \\\\ a=\sqrt{\dfrac{225}{2}} \\\\ a=\sqrt{\dfrac{225}{2}\cdot\dfrac{2}{2}} \\\\ a=\sqrt{\dfrac{225}{4}\cdot2} \\\\ a=\sqrt{\left( \dfrac{15}{2}\right)^2\cdot2} \\\\ a=\dfrac{15}{2}\sqrt{2} \\\\ a=\dfrac{15\sqrt{2}}{2} .\end{array} Hence, the missing sides measure $ \dfrac{15\sqrt{2}}{2} \text{ and } \dfrac{15\sqrt{2}}{2} $ units.
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