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Question Number 161578 by Ari last updated on 19/Dec/21

Commented by blackmamba last updated on 20/Dec/21

$$\cos 75°=\frac{\frac{1}{2}a}{10}=\frac{a}{20}$$$$\frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}.\frac{1}{2}=\frac{a}{20}$$$$\Rightarrow \sqrt{6}-\sqrt{2}=\frac{a}{5}$$$$\Rightarrow a=5\sqrt{2}(\sqrt{3}-1)cm$$$$\therefore S\left[BDHF\right]=a^2\sqrt{2}{cm}^2$$$$=50\sqrt{2}(4-2\sqrt{3}){cm}^2$$$$=100\sqrt{2}(2-\sqrt{3}){cm}^2$$

Answered by Rasheed.Sindhi last updated on 20/Dec/21

$$a^2={10}^2+{10}^2-2.10.10.\cos 30$$$$a=\sqrt{200-200\cdot \frac{\sqrt{3}}{2}}=10\sqrt{2-\sqrt{3}}$$$$d=\sqrt{a^2+a^2}=a\sqrt{2}=10\sqrt{2-\sqrt{3}}\cdot \sqrt{2}$$$$S_{HDBF}=ad=\left(10\sqrt{2-\sqrt{3}}\right)(10\sqrt{2-\sqrt{3}}\cdot \sqrt{2})$$$$=100(2-\sqrt{3})\sqrt{2}{\mathrm{cm}}^2$$

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