반원과 삼각형 넓이

[출제의도] 삼각함수를 활용하여 문제 해결하기 $\displaystyle \mathrm{\overline{OA} = \overline{OP}}$이므로 $\mathrm{\angle OAP = \angle OPA = \theta}$ $\displaystyle \mathrm{\angle APB = \frac{\pi}{2}}$이므로 $\displaystyle \mathrm{\angle CPD = \frac{\pi}{2} - \theta}$ $4 \sin \theta = 3 \cos \theta$, $\sin ^{2} \theta + \cos ^{2} \theta = 1$이므로 $\displaystyle \sin ^{2} \theta + \frac{16}{9} \sin ^{2} \theta = 1$, $\displaystyle \sin ^{2} \theta = \frac{9}{25}$ $\displaystyle 0 < \theta < \frac{\pi}{2}$이므로 $\displaystyle \sin \theta = \frac{3}{5}$, $\displaystyle \cos \theta = \frac{4}{5}$ $\displaystyle \mathrm{\overline{AB} = 10}$, $\displaystyle \mathrm{\angle APB = \frac{\pi}{2}}$이므로 $\displaystyle \mathrm{\overline{PA} = 10 \cos \theta = 8}$, $\displaystyle \mathrm{\overline{PA} = \overline{PC} = \overline{PD} = 8}$ (삼각형 $\mathrm{ADC}$의 넓이) $=$(삼각형 $\mathrm{PAD}$의 넓이)$+$(삼각형 $\mathrm{PDC}$의 넓이)$-$(삼각형 $\mathrm{PAC}$의 넓이) $\displaystyle = \frac{1}{2} \times 8 \times 8 \times \sin \theta + \frac{1}{2} \times 8 \times 8 \times \sin \left( \frac{\pi}{2} - \theta \right)$$\displaystyle - \frac{1}{2} \times 8 \times 8$ $= 32 \sin \theta + 32 \cos \theta - 32$ $\displaystyle = 32 \times \frac{3}{5} + 32 \times \frac{4}{5} - 32$ $\displaystyle = \frac{64}{5}$