In the given figure, PQRS is a square of side 8 cm. ∠PQO = 60°. What is the area (in cm2) of the triangle POQ?

In the given figure, PQRS is a square of side 8 cm. ∠PQO = 60°. What is the area (in cm²) of the triangle POQ?

Answer: Option D

Solution

Answer: Option D

Solution:

$$\begin{array}{rl}& \text{In }\sin \text{rule}\\ & \frac{\sin {75}^{\circ }}{PQ}=\frac{\sin {60}^{\circ }}{PO}\\ & \frac{\sin \left({30}^{\circ }+{45}^{\circ }\right)}{8}=\frac{\sin {60}^{\circ }}{PO}\\ & \frac{\frac{1}{2}\times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}}{8}=\frac{\sqrt{3}}{2PO}\\ & \frac{\sqrt{3}+1}{2\sqrt{2}\times 8}=\frac{\sqrt{3}}{2PO}\\ & PO=\frac{8\sqrt{6}}{\left(\sqrt{3}+1\right)}\\ & \text{After rationaltion}\\ & PO=\frac{8\sqrt{6}}{\left(\sqrt{3}+1\right)}=4\sqrt{6}\left(\sqrt{3}-1\right)\\ & \text{Then in }\Delta POD,\\ & \text{Area}=\frac{1}{2}PQ\times PO\sin 45^{\circ }\\ & =\frac{1}{2}\times 8\times 4\sqrt{6}\left(\sqrt{3}-1\right)\times \frac{1}{\sqrt{2}}\\ & =2\sqrt{2}\times 4\sqrt{6}\left(\sqrt{3}-1\right)\\ & =16\left(3-\sqrt{3}\right)\end{array}$$