Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Section 5.5 - Special Right Triangles - Exercises - Page 247: 13

Answer

So $HL= 6$ $HK = 12$ $MK= 6$

Work Step by Step

In $\triangle HKL$ $\angle HKL = 30^{\circ}$ So $ \angle LHK =60^{\circ}$ now By Theorem 5.5.2/ (30-60-90 Theorem) $LK= 6\sqrt 3 $ given So $HL= 6$ $HK = 12$ $MK= HK/2 = 12/2= 6$
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