잘라낸 도형 둘레

$\displaystyle {\overline{\mathrm{DE}}}$$\displaystyle = \frac{\sqrt{2}}{2}$이므로 $\displaystyle {\overline{\mathrm{AE}}}$$\displaystyle = {\overline{\mathrm{AD}}} - {\overline{\mathrm{ED}}}$$\displaystyle = 4 \sqrt{2} - \frac{\sqrt{2}}{2}$$\displaystyle = \frac{7 \sqrt{2}}{2}$ $\cdots \cdots$ ㉠ 직각삼각형 $\mathrm{ECD}$에서 피타고라스 정리에 의하여 $\displaystyle {\overline{\mathrm{EC}}} ^{2}$$\displaystyle = {\overline{\mathrm{ED}}} ^{2} + {\overline{\mathrm{DC}}} ^{2} = \left( \frac{\sqrt{2}}{2} \right) ^{2} + \left( 4 \sqrt{2} \right) ^{2}$$\displaystyle = \frac{65}{2}$ 직각삼각형 $\mathrm{FCE}$에서 피타고라스 정리에 의하여 $\displaystyle {\overline{\mathrm{EC}}} ^{2} = {\overline{\mathrm{EF}}} ^{2} + {\overline{\mathrm{FC}}} ^{2}$ $\displaystyle {\overline{\mathrm{EF}}} : {\overline{\mathrm{FC}}} = 4 : 7$에서 $\displaystyle {\overline{\mathrm{EF}}} = \frac{4}{7} {\overline{\mathrm{FC}}}$이므로 $\displaystyle {\overline{\mathrm{EC}}} ^{2}$$\displaystyle = \left( \frac{4}{7} {\overline{\mathrm{FC}}} \right) ^{2} + {\overline{\mathrm{FC}}} ^{2}$$\displaystyle = \frac{65}{49} \times {\overline{\mathrm{FC}}} ^{2}$ $\displaystyle \frac{65}{49} \times {\overline{\mathrm{FC}}} ^{2} = \frac{65}{2}$에서 $\displaystyle {\overline{\mathrm{FC}}} ^{2} = \frac{49}{2}$ $\displaystyle {\overline{\mathrm{FC}}} = \frac{7 \sqrt{2}}{2}$이고 $\displaystyle {\overline{\mathrm{EF}}} = \frac{4}{7} {\overline{\mathrm{FC}}} = 2 \sqrt{2}$ $\cdots \cdots$ ㉡ 정사각형 $\mathrm{ABCD}$에서 $\displaystyle {\overline{\mathrm{AB}}} = {\overline{\mathrm{BC}}} = 4 \sqrt{2}$ $\cdots \cdots$ ㉢ ㉠, ㉡, ㉢에서 모양의 도형의 둘레의 길이는 $\displaystyle {\overline{\mathrm{AB}}} + {\overline{\mathrm{BC}}} + {\overline{\mathrm{CF}}} + {\overline{\mathrm{FE}}} + {\overline{\mathrm{EA}}}$ $\displaystyle = 4 \sqrt{2} + 4 \sqrt{2} + \frac{7 \sqrt{2}}{2} + 2 \sqrt{2} + \frac{7 \sqrt{2}}{2}$ $\displaystyle = 17 \sqrt{2}$ $\displaystyle a = 17 \sqrt{2}$ 따라서 $a ^{2} = 578$