College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 50: 120

Answer

Perimeter P $ = 24 \sqrt 5 $ feet Area A = 160 Square feet

Work Step by Step

$ P = 2l + 2w $ Substitute $ l = 4\sqrt 20 $ and $w = \sqrt80 $ $P = 2 \times 4\sqrt 20 + 2 \times \sqrt 80$ $ = 8\sqrt 20 + 2 \sqrt 80$ Split $\sqrt 20$ and $\sqrt 80$ into two factors such that one is a perfect square. $ = 8 \sqrt (4 \times 5) + 2 \sqrt (16 \times 5)$ $= 8 \sqrt 4 \sqrt 5 + 2 \sqrt 16 \sqrt 5$ $= 8 \times 2 \times \sqrt 5 + 2 \times 4 \times \sqrt 5$ $ = 16\sqrt 5 + 8 \sqrt 5$ $ = 24\sqrt 5$ P $ = 24\sqrt 5$ feet Area = $l \times w$ $= 4\sqrt 20 \times \sqrt 80$ $= 4 \sqrt (4 \times 5) \times \sqrt (16 \times 5)$ $ = 4 \times\sqrt 4 \times\sqrt 5 \times\sqrt 16 \times\sqrt 5$ $ = 4 \times 2 \times \sqrt 5 \times 4 \times \sqrt 5 $ $ = 32 \times \sqrt 5 \times \sqrt 5 $ $ = 32 \times 5 $ Area = 160 Square feet
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