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Question Number 186782 by Rupesh123 last updated on 10/Feb/23

Answered by HeferH last updated on 10/Feb/23

$$\frac{121A}{60}-\frac{2S}{3}+\frac{121A}{40}-S=\frac{17A}{8}$$$$\frac{121A}{24}-\frac{5S}{3}=\frac{17A}{8}$$$$\frac{121A}{24}-\frac{17\cdot 3A}{8\cdot 3}=\frac{5S}{3}$$$$\frac{70A}{24}=\frac{5S}{3}$$$$\frac{70A\cdot 3}{24\cdot 5}=S$$$$S=\frac{7}{4}A$$$$A=\frac{96}{17}$$$$S=\frac{7}{4}\cdot \frac{12\cdot 8}{17}=\frac{7\cdot 24}{17}=\frac{168}{17}$$

Commented by HeferH last updated on 10/Feb/23

Answered by nikif99 last updated on 10/Feb/23

$$A_{ABC}=36\left(Heron\right)$$$$\cos A=\frac{{AB}^2+{AC}^2-{BC}^2}{2\cdot \left(AB\right)\left(AC\right)}=-\frac{3}{5}\Rightarrow \sin A=\frac{4}{5}$$$$Similarly,\sin B=\frac{15}{17}and\sin C=\frac{77}{85}$$$$A_{AMP}=\frac{1}{2}\left(AM\right)\left(AP\right)\sin A=7.2$$$$A_{BMN}=\frac{192}{17}and{A}_{CNP}=\frac{648}{85}$$$$S=A_{ABC}-(A_{AMP}+A_{BMN}+A_{CNP})=\frac{168}{17}$$

Answered by mr W last updated on 10/Feb/23

Commented by mr W last updated on 10/Feb/23

$$\mathrm{\Delta }=areaof\mathrm{\Delta }ABC$$$$\mathrm{\Delta }=\frac{\sqrt{(9+17+10)(-9+17+10)(9-17+10)(9+17-10)}}{4}$$$$=36$$$$\frac{A}{\mathrm{\Delta }}=\frac{3\times 6}{9\times 10}=\frac{1}{5}$$$$\frac{B}{\mathrm{\Delta }}=\frac{6\times 8}{9\times 17}=\frac{16}{51}$$$$\frac{C}{\mathrm{\Delta }}=\frac{9\times 4}{17\times 10}=\frac{18}{85}$$$$\frac{S+A+B+C}{\mathrm{\Delta }}=\frac{\mathrm{\Delta }}{\mathrm{\Delta }}=1$$$$\Rightarrow \frac{S}{\mathrm{\Delta }}=1-\frac{1}{5}-\frac{16}{51}-\frac{18}{85}=\frac{14}{51}$$$$\Rightarrow S=\frac{14\times 36}{51}=\frac{168}{17}$$

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