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Question Number 215885 by a.lgnaoui last updated on 20/Jan/25

$$\mathrm{Area}\mathrm{of}\mathrm{ABC}?$$

Commented by a.lgnaoui last updated on 20/Jan/25

Answered by A5T last updated on 21/Jan/25

Commented by A5T last updated on 21/Jan/25

$${\mathrm{OE}}^2={\mathrm{DO}}^2+{\mathrm{DE}}^2-2\times \mathrm{DO}\times \mathrm{DEcos}\mathrm{\angle }\mathrm{ODE}$$$$\mathrm{DO}=\mathrm{x}\Rightarrow 9^2={\mathrm{x}}^2+{10}^2-10\mathrm{x}\Rightarrow {\mathrm{x}}^2-10\mathrm{x}+19=0$$$$\Rightarrow \mathrm{x}=\frac{10\underset{-}{+}2\sqrt{6}}{2}=5\underset{-}{+}\sqrt{6}\Rightarrow \mathrm{x}=5-\sqrt{6}$$$$\Rightarrow \mathrm{DB}=9+5-\sqrt{6}=14-\sqrt{6}$$$$\Rightarrow \mathrm{BE}=\sqrt{162-18\sqrt{6}}$$$$\Rightarrow \mathrm{AB}=\sqrt{{18}^2-162+18\sqrt{6}}=\sqrt{162+18\sqrt{6}}$$$$9^2=162+18\sqrt{6}+9^2-2\times 9\sqrt{162+18\sqrt{6}}\mathrm{cosABD}$$$$\Rightarrow \mathrm{cosABD}=\frac{\sqrt{162+18\sqrt{6}}}{18}$$$$\Rightarrow \mathrm{AD}=\sqrt{162+18\sqrt{6}+{(14-\sqrt{6})}^2-\frac{(162+18\sqrt{6})(14-\sqrt{6})}{9}}$$$$\mathrm{AD}=2\sqrt{31-5\sqrt{6}}$$$$\mathrm{AD}\times \mathrm{DF}=\mathrm{BD}(2\mathrm{R}-\mathrm{BD})\Rightarrow \mathrm{DF}=\frac{5(5+\sqrt{6})}{\sqrt{31-5\sqrt{6}}}$$$$\Rightarrow \mathrm{EF}=\sqrt{{\mathrm{DE}}^2-{\mathrm{DF}}^2}=\frac{5\sqrt{2433(661-155\sqrt{6})}}{811}$$$$\frac{\mathrm{EF}}{\mathrm{AB}}=\frac{\mathrm{CF}}{\mathrm{CB}=\mathrm{BE}+\mathrm{CE}}[\mathrm{CF}=\mathrm{y};\mathrm{CE}=\mathrm{z}]$$$$\Rightarrow \frac{\frac{5\sqrt{3(31-10\sqrt{6})}}{\sqrt{31-5\sqrt{6}}}}{\sqrt{162+18\sqrt{6}}}=\frac{\mathrm{y}}{\sqrt{162-18\sqrt{6}}+\mathrm{z}}...\left(\mathrm{i}\right)$$$${\mathrm{CF}}^2+{\mathrm{EF}}^2={\mathrm{CE}}^2$$$$\Rightarrow {\mathrm{y}}^2+\frac{25\left[2433(661-155\sqrt{6})\right]}{{811}^2}={\mathrm{z}}^2...\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right)$$$$\Rightarrow \frac{25\left[3(31-10\sqrt{6})\right]}{(162+18\sqrt{6})(31-5\sqrt{6})}{(\sqrt{162-18\sqrt{6}}+\mathrm{z})}^2+\frac{25\left[2433(661-155\sqrt{6})\right]}{{811}^2}={\mathrm{z}}^2$$$$\Rightarrow \mathrm{z}=\mathrm{CE}=\frac{6\sqrt{6633-2687\sqrt{6}}}{5}$$$$\Rightarrow \left[\mathrm{ABC}\right]=\frac{1}{2}\mathrm{AB}\times \mathrm{BC}=\frac{1}{2}\times \mathrm{AB}\times [\mathrm{BE}+\mathrm{CE}]$$$$\Rightarrow \left[\mathrm{ABC}\right]=279\sqrt{3}-243\sqrt{2}\approx 139.5883$$

Commented by a.lgnaoui last updated on 21/Jan/25

$$\mathrm{thank}\mathrm{you}$$

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