Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 369: 2

Answer

46 in

Work Step by Step

Lets give the name to the intersecting point be E In parallelogram opposite sides are equal So AD = BC = 13 in AB = CD AD = AE +ED 13in = AE + 7 in AE= 6 in By using pythagoras theorem in triangle ABE $AB^{2}$ =$AE^{2}$ + $ED^{2}$ $AB^{2}$ = $6^{2}$ + $8^{2}$ $AB^{2}$ = 36 + 64 AB = $\sqrt 100$ AB =10 in The perimeter of parallelogram ABCD =AB+BC+CD+DA 10+13+10+13 =46 in
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