Area-of-ABC-

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

Area of 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

{OE}^{2} = {DO}^{2} + {DE}^{2} - 2 \times DO \times DEcos ∠ ODE

DO = x \Rightarrow 9^{2} = x^{2} + 10^{2} - 10 x \Rightarrow x^{2} - 10 x + 19 = 0

\Rightarrow x = \frac{10 \underline{+} 2 \sqrt{6}}{2} = 5 \underline{+} \sqrt{6} \Rightarrow x = 5 - \sqrt{6}

\Rightarrow DB = 9 + 5 - \sqrt{6} = 14 - \sqrt{6}

\Rightarrow BE = \sqrt{162 - 18 \sqrt{6}}

\Rightarrow 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}} cosABD

\Rightarrow cosABD = \frac{\sqrt{162 + 18 \sqrt{6}}}{18}

\Rightarrow AD = \sqrt{162 + 18 \sqrt{6} + {(14 - \sqrt{6})}^{2} - \frac{(162 + 18 \sqrt{6}) (14 - \sqrt{6})}{9}}

AD = 2 \sqrt{31 - 5 \sqrt{6}}

AD \times DF = BD (2 R - BD) \Rightarrow DF = \frac{5 (5 + \sqrt{6})}{\sqrt{31 - 5 \sqrt{6}}}

\Rightarrow EF = \sqrt{{DE}^{2} - {DF}^{2}} = \frac{5 \sqrt{2433 (661 - 155 \sqrt{6})}}{811}

\frac{EF}{AB} = \frac{CF}{CB = BE + CE} [CF = y; CE = z]

\Rightarrow \frac{\frac{5 \sqrt{3 (31 - 10 \sqrt{6})}}{\sqrt{31 - 5 \sqrt{6}}}}{\sqrt{162 + 18 \sqrt{6}}} = \frac{y}{\sqrt{162 - 18 \sqrt{6}} + z} \dots (i)

{CF}^{2} + {EF}^{2} = {CE}^{2}

\Rightarrow y^{2} + \frac{25 [2433 (661 - 155 \sqrt{6})]}{811^{2}} = z^{2} \dots (ii)

(i) & (ii)

\Rightarrow \frac{25 [3 (31 - 10 \sqrt{6})]}{(162 + 18 \sqrt{6}) (31 - 5 \sqrt{6})} {(\sqrt{162 - 18 \sqrt{6}} + z)}^{2} + \frac{25 [2433 (661 - 155 \sqrt{6})]}{811^{2}} = z^{2}

\Rightarrow z = CE = \frac{6 \sqrt{6633 - 2687 \sqrt{6}}}{5}

\Rightarrow [ABC] = \frac{1}{2} AB \times BC = \frac{1}{2} \times AB \times [BE + CE]

\Rightarrow [ABC] = 279 \sqrt{3} - 243 \sqrt{2} \approx 139.5883

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

thank you