Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 349: 21

Answer

144$ unit^{2}$

Work Step by Step

Given a figure PQST Lets draw perpendicular from T on base PQ at M and from S on QR at N Then TSNM is a rectangle and TS = MN = 12 therefore PM = 6 =NR By Pythagoras theorem $\triangle$TMP $TM^{2}$ = $10^{2}$ - $6^{2}$ = 100 - 36 = 64 TM = 8 units similarly NS = 8 units Area of parallelogram PQST = PQ*TM =12 * 8 = 96 $unit^{2}$ Area of triangle SQR = $\frac{1}{2}$ * b * h = $\frac{1}{2}$ * 12 * 8 =48 $unit^{2}$ therefore total shaded area = 96 + 48 = 144$unit^{2}$
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