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 - Practice and Problem-Solving Exercises - Page 503: 10


$x = 15$ $y = 15$

Work Step by Step

In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, we know that the hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $x$, the length of a leg: $15 \sqrt 2 = \sqrt 2(x)$ Divide each side by $\sqrt 2$ to solve for $x$: $x = 15$ We know that in this type of triangle, the legs are equal to one another. Since one leg, $x$, measures $15$, the other leg, $y$, also equals $15$.
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