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Question Number 185900 by Mingma last updated on 29/Jan/23

Answered by mr W last updated on 29/Jan/23

$$sayAB=BC=a$$$$\mathrm{\angle }ABQ=15°$$$$DQ=a(1-\tan 15°)$$$$DR=\frac{DQ}{\tan 15°}$$$$orange=\frac{DQ\times DR}{2}=\frac{a^2{(1-\tan 15°)}^2}{2\tan 15°}$$$$=a^2\times \frac{{(1-2+\sqrt{3})}^2}{2(2-\sqrt{3})}=a^2\times \frac{2(2-\sqrt{3})}{2(2-\sqrt{3})}=a^2$$$$=yellow✓$$

Answered by HeferH last updated on 29/Jan/23

$$letAB=a$$$$\mathrm{\angle }BQA=75°,\mathrm{\angle }ABQ=15°\Rightarrow$$$$AQ=a(2-\sqrt{3})\Rightarrow$$$$QD=a-a(2-\sqrt{3})=a(\sqrt{3}-1)\Rightarrow$$$$DR=\frac{a(\sqrt{3}-1)}{2-\sqrt{3}}$$$$Orangearea=\frac{QD\cdot DR}{2}=\left[\frac{a^2{(\sqrt{3}-1)}^2}{2-\sqrt{3}}\right]\frac{1}{2}$$$$=\frac{a^2(4-2\sqrt{3})}{4-2\sqrt{3}}=a^2=Yellowarea$$

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