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Question Number 158927 by cherokeesay last updated on 10/Nov/21

Answered by amin96 last updated on 10/Nov/21

$$areatriangle=\frac{126-16\sqrt{21}}{5}$$

Commented by cherokeesay last updated on 10/Nov/21

$$thankyousir.$$

Answered by mr W last updated on 10/Nov/21

$$\cos \alpha =\frac{3}{3+2}=\frac{3}{5}\Rightarrow \tan \alpha =\frac{4}{3}$$$$\cos \beta =\frac{2}{3+2}=\frac{2}{5}\Rightarrow \tan \beta =\frac{\sqrt{21}}{2}$$$$\frac{h}{\tan \alpha }+\frac{h}{\tan \beta }=3+2$$$$h(\frac{3}{4}+\frac{2}{\sqrt{21}})=5\Rightarrow h=\frac{20\sqrt{21}}{8+3\sqrt{21}}$$$$A_{blue}=\frac{5h}{2}=\frac{5\times 20\sqrt{21}}{2(8+3\sqrt{21})}=\frac{126-16\sqrt{21}}{5}\approx 10.54$$

Commented by cherokeesay last updated on 10/Nov/21

$$verynice!$$$$thankyousir!$$

Commented by Tawa11 last updated on 11/Nov/21

$$\mathrm{Great}\mathrm{sir}.$$

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