Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - 8-2 Special Right Triangles - Lesson Check - Page 503: 3


$x = 4 \sqrt 2$

Work Step by Step

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 $x$, the length of a leg: $8 = \sqrt 2(x)$ Divide each side by $\sqrt 2$ to solve for $x$: $x = \frac{8}{\sqrt 2}$ To simplify this fraction, we need to get rid of radicals in the denominator. We do this by multiplying both the numerator and denominator by the denominator: $x = \frac{8 \sqrt 2}{\sqrt 4}$ Take the square root of the denominator: $x = \frac{8 \sqrt 2}{2}$ Divide both the numerator and denominator by their greatest common factor to simplify: $x = 4 \sqrt 2$
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