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:
$\frac{A_{EHIL}}{A_{ABCD}} = \frac{(2\sqrt{2})(\sqrt{2})}{(3)^2} = \frac{4}{9}$
