Elementary Geometry for College Students (6th Edition)

(a) $\frac{P_{EGIK}}{P_{ABCD}} = \frac{\sqrt{5}}{3}$ (b) $\frac{A_{EGIK}}{A_{ABCD}} = \frac{5}{9}$
(a) Let each side of the square $ABCD$ be 3 units. We can find the length of each side of the square $EGIK$: $L = \sqrt{(1)^2+(2)^2}$ $L = \sqrt{1+4}$ $L = \sqrt{5}$ We can find the ratio: $\frac{P_{EGIK}}{P_{ABCD}} = \frac{4\times \sqrt{5}}{4\times 3} = \frac{\sqrt{5}}{3}$ (b) We can find the ratio: $\frac{A_{EGIK}}{A_{ABCD}} = \frac{(\sqrt{5})^2}{(3)^2} = \frac{5}{9}$