직교 대각선 넓이비

[출제의도] 피타고라스의 정리를 간단한 도형에 활용하기 $\displaystyle \mathrm{\overline{AB}} ^{2} = {\overline{BH}} ^{2} + {\overline{AH}} ^{2}$에서 $\displaystyle \mathrm{25} = 12 + {\overline{AH}} ^{2}$ ∴$\displaystyle {\mathrm{\overline{AH}}} ^{2} = 13$ $\displaystyle \mathrm{\overline{AD}} ^{2} = {\overline{AH}} ^{2} + {\overline{DH}} ^{2}$에서 $\displaystyle \mathrm{40} = 13 + {\overline{DH}} ^{2}$이므로 $\displaystyle \mathrm{\overline{DH}} = 3 \sqrt{3}$ $\displaystyle \mathrm{\left( {\triangle BCH \text{의} \text{넓이}} \right)} : \left( {\triangle AHD \text{의} \text{넓이}} \right) \mathrm{=} {\overline{BH}} : {\overline{DH}}$ $\displaystyle = 2 \sqrt{3} : 3 \sqrt{3} = 2 : 3$