Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 5 - Test - Page 264: 11b



Work Step by Step

$\triangle DEF$ is a 30-60-90 triangle. We are given the length of $FD$, which is $6\sqrt3$. The longest leg of the triangle, or the leg opposite the $60^{\circ}$ angle equals $(short leg)\sqrt3$. Since this side equals $6\sqrt3$, the shortest leg is equal to $6$. So $DE=6$. The hypotenuse is twice the length of the shortest leg. So the hypotenuse is $12$, or $EF=12$.
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