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 371: 42

Answer

(a) $\frac{P_{EHIL}}{P_{ABCD}} = \frac{\sqrt{2}}{2}$ (b) $\frac{A_{EHIL}}{A_{ABCD}} = \frac{4}{9}$

Work Step by Step

(a) Let each side of the square $ABCD$ be 3 units. We can find the length of the side $EH$ of the rectangle $EHIL$: $L_1 = \sqrt{(2)^2+(2)^2}$ $L_1 = \sqrt{4+4}$ $L_1 = \sqrt{8}$ $L_1 = 2\sqrt{2}$ We can find the length of the side $HI$ of the rectangle $EHIL$: $L_2 = \sqrt{(1)^2+(1)^2}$ $L_2 = \sqrt{1+1}$ $L_2 = \sqrt{2}$ We can find the ratio: $\frac{P_{EHIL}}{P_{ABCD}} = \frac{(2\times 2\sqrt{2})+(2 \times \sqrt{2})}{4\times 3} = \frac{6\sqrt{2}}{12} = \frac{\sqrt{2}}{2}$ (b) We can find the ratio: $\frac{A_{EHIL}}{A_{ABCD}} = \frac{(2\sqrt{2})(\sqrt{2})}{(3)^2} = \frac{4}{9}$
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