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Question Number 193099 by Mingma last updated on 04/Jun/23

Answered by som(math1967) last updated on 04/Jun/23

Commented by som(math1967) last updated on 04/Jun/23

$$b^2=\frac{16}{\sqrt{3}},c^2=\frac{36}{\sqrt{3}}$$$$cos60=\frac{b^2+c^2-a^2}{2bc}$$$$\Rightarrow \frac{1}{2}=\frac{\frac{16}{\sqrt{3}}+\frac{36}{\sqrt{3}}-a^2}{2\times \frac{4\times 6}{\sqrt{3}}}$$$$\Rightarrow \frac{24}{\sqrt{3}}=\frac{52}{\sqrt{3}}-a^2$$$$\Rightarrow a^2=\frac{52-24}{\sqrt{3}}$$$$\therefore \frac{\sqrt{3}}{4}\times a^2=\frac{28}{\sqrt{3}}\times \frac{\sqrt{3}}{4}=7squ=redarea$$

Commented by Mingma last updated on 04/Jun/23

Perfect ��

Commented by Mingma last updated on 04/Jun/23

What's your solution for the green area?

Commented by math1234 last updated on 04/Jun/23

Commented by math1234 last updated on 04/Jun/23

Commented by math1234 last updated on 04/Jun/23

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