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Question Number 64829 by Tawa1 last updated on 22/Jul/19

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tony Lin last updated on 22/Jul/19

$$let\mathrm{\angle }BDC=\theta$$$$BC=\sqrt{{17}^2+{11}^2-2\times 17\times 11cos\theta }$$$$=\sqrt{410-374cos\theta }$$$$AB=\sqrt{{17}^2+{31}^2-2\times 17\times 31cos(\pi -\theta )}$$$$=\sqrt{{17}^2+{31}^2+2\times 17\times 31cos\theta }$$$$=\sqrt{1250+1054cos\theta }$$$$areaof△ABC$$$$=BC\times AB\times \frac{1}{2}$$$$=\frac{1}{2}\times (BD\times DC\times sin\theta +BD\times AD\times sin(\pi -\theta )$$$$=\frac{1}{2}\times 17\times (11+31)sin\theta$$$$=357sin\theta$$$$\Rightarrow \sqrt{(410-374cos\theta )(1250+1054cos\theta )}$$$$=714sin\theta$$$$\Rightarrow \sqrt{512500-35360cos\theta -394196{cos}^2\theta }$$$$=714sin\theta$$$$letcos\theta =t$$$$\Rightarrow \sqrt{512500-35360t-394196t^2}$$$$=714\sqrt{1-t^2}$$$$\Rightarrow 512500-35360t-394196t^2$$$$=509796-509796t^2$$$$\Rightarrow 115600t^2-35360t+2704=0$$$$\Rightarrow t=\frac{13}{85}$$$$areaof△BDC$$$$=\frac{1}{2}\times BD\times DC\times sin\theta$$$$=\frac{1}{2}\times 17\times 11\times \sqrt{1-{\left(\frac{13}{85}\right)}^2}$$$$=92.4$$

Commented by mathmax by abdo last updated on 23/Jul/19

$$whereareyoufromsirtony...$$

Commented by Tony Lin last updated on 23/Jul/19

$$Taiwan$$

Answered by MJS last updated on 22/Jul/19

$$\mathrm{put}\frac{A+C}{2}\mathrm{in}\mathrm{the}\mathrm{center}$$$$\Rightarrow A=\left(\begin{array}{c}-21\\ 0\end{array}\right)\mathrm{C}=\left(\begin{array}{c}21\\ 0\end{array}\right)D=\left(\begin{array}{c}10\\ 0\end{array}\right)$$$$B\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{circumcircle}{x}^2+y^2={21}^2$$$$B=\left(\begin{array}{c}x\\ \sqrt{441-x^2}\end{array}\right)$$$$\mid BD\mid =17$$$$\sqrt{{(x-10)}^2+{(\sqrt{441-x^2}-0)}^2}=17$$$$\sqrt{541-20x}=17$$$$x=\frac{63}{5}$$$$B=\left(\begin{array}{c}\frac{63}{5}\\ \frac{84}{5}\end{array}\right)$$$$\mid BC\mid =\sqrt{{(\frac{63}{5}-21)}^2+{(\frac{84}{5}-0)}^2}=\frac{42\sqrt{5}}{5}$$$$\Rightarrow \mathrm{the}\mathrm{shaded}\mathrm{triangle}\mathrm{has}\mathrm{sides}$$$$11,17,\frac{42\sqrt{5}}{5}$$$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{sides}a,b,c\mathrm{is}$$$$\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$$$\Rightarrow \mathrm{shaded}\mathrm{area}=\frac{462}{5}=92.4$$$$$$

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by ajfour last updated on 22/Jul/19

Commented by ajfour last updated on 22/Jul/19

$$\cos \theta =\frac{{21}^2-{10}^2-{17}^2}{2\times 10\times 17}=\frac{13}{85}$$$$\sin \theta =\frac{\sqrt{7225-169}}{85}=\frac{84}{85}$$$$A_{rea}=\frac{1}{2}\times 11\times 17\sin \theta$$$$A_{rea}=\frac{11\times 17\times 84}{2\times 85}=\frac{11\times 42}{5}=\frac{462}{5}$$$$A_{rea}=92.4squnits◼$$

Commented by ajfour last updated on 22/Jul/19

$$soyoudontseemtolikethis$$$$solution,Tawa.$$

Commented by MJS last updated on 22/Jul/19

$${\mathrm{she}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{yet}\mathrm{through}\mathrm{with}\mathrm{it}$$

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{Hahaha}.\mathrm{No}\mathrm{sir},\mathrm{i}\mathrm{am}\mathrm{going}\mathrm{through}\mathrm{it}.$$

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{And}\mathrm{i}\mathrm{am}\mathrm{grateful}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

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