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: 119

Answer

Perimeter = 18 $\sqrt 5$ feet Area = 100 Square Feet

Work Step by Step

$ P = 2l + 2w $ Substitute $ l = \sqrt 125 $ and $w = 2\sqrt 20 $ $P = 2 \sqrt 125 + 2 ( 2\sqrt 20)$ $ = 2 \sqrt 125 + 4\sqrt 20$ Split $ \sqrt 125 $ and $\sqrt 20$ into two factors such that one is a perfect square. $ = 2 \sqrt( 25 \times 5) + 4 \sqrt( 4 \times 5) $ $ = 2 \sqrt 25 \times \sqrt 5 + 4 \sqrt 4 \times \sqrt 5 $ $ = 2 \times 5 \times \sqrt 5 + 4 \times 2 \times \sqrt 5 $ $ = 10\sqrt 5 + 8 \sqrt 5 $ Apply Distributive Property $= (10 + 8) \sqrt 5$ $P = 18 \sqrt 5$ feet Area A $ = l \times w$ $ = \sqrt 125 \times 2 \sqrt 20$ $=\sqrt (25 \times 5) \times 2 \sqrt (4 \times 5)$ $=\sqrt 25 \times \sqrt 5 \times 2\times \sqrt 4\times \sqrt 5$ $= 5 \times \sqrt 5 \times 2 \times 2 \times \sqrt 5$ $= 20 \sqrt5 \sqrt 5$ $= 20 \times 5$ Area A$ = 100 $ Square feet
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