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 505: 38


$\sqrt {11}$

Work Step by Step

We can find the other leg by using the Pythagorean theorem, which states that $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. Let's plug in what we know into the Pythagorean theorem: $5^2 + b^2 = 6^2$ Evaluate the exponents: $25 + b^2 = 36$ Subtract $25$ from each side of the equation to move constants to the right side of the equation: $b^2 = 11$ Take the positive square root to solve for $b$: $b = \sqrt {11}$ The other leg would be $\sqrt {11}$.
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