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Question Number 188267 by normans last updated on 27/Feb/23

Answered by mr W last updated on 02/Mar/23

Commented by mr W last updated on 02/Mar/23

$$xy=128$$$$\frac{16a-x}{2}=3a\Rightarrow x=10a$$$$\frac{x}{y}=\frac{c}{a}$$$$yc=ax=10a^2$$$${(12c-\frac{y}{2})}^2+{\left(8a\right)}^2={\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{2}\right)}^2$$$$144c^2-12\times 10a^2+39a^2=0$$$$\Rightarrow c=\frac{3a}{4}$$$$\Rightarrow \frac{x}{y}=\frac{3}{4}$$$$\Rightarrow xy=\frac{3y^2}{4}=128\Rightarrow y=\frac{16\sqrt{6}}{3}\Rightarrow x=4\sqrt{6}$$$$\Rightarrow a=\frac{4\sqrt{6}}{10}=\frac{2\sqrt{6}}{5}\Rightarrow c=\frac{3\sqrt{6}}{10}$$$$A_{triangle}=\frac{\left(25a\right)\left(25c\right)}{2}=\frac{{25}^{2}ac}{2}$$$$=\frac{{25}^2}{2}\times \frac{2\sqrt{6}}{5}\times \frac{3\sqrt{6}}{10}$$$$=25\times 9=225$$

Answered by a.lgnaoui last updated on 01/Mar/23

$$posons$$$$posonsDM=yHI=x$$$$Area\left(DEMN\right)=xy=128x=\frac{128}{y}$$$$Areaofteiangle\left(ADM\right)=\frac{MD\times AL}{2}$$$$=Area\left(ALD\right)+\left(ALM\right)$$$$△LCD$$$$\cos \alpha =\frac{25a}{25b}=\frac{a}{b}\Rightarrow \sin \alpha =\frac{\sqrt{b^2-a^2}}{b}$$$$DL=12\sqrt{b^2-a^2}CL=12b\cos \alpha =12a$$$$\Rightarrow AL=25a-12aAL=13a$$$$△ADM$$$$\measuredangle AOM=2\measuredangle ADM=2\phi$$$$$$$${AD}^2={AL}^2+{DL}^2={\left(13a\right)}^2+{12}^2(b^2-a^2)$$$$AD=\sqrt{{12}^2{b}^2+25a^2}$$$$\cos \phi =\frac{DL}{AD}=\frac{12\sqrt{b^2-a^2}}{\sqrt{{DL}^2+{AL}^2}}=$$$$\cos \phi =\frac{12\sqrt{b^2-a^2}}{\sqrt{{12}^2{b}^2+25a^2}}$$$${AM}^2={DM}^2+{AD}^2-2AD\times DM\cos \phi \left(1\right)$$$$△AMD(KM=\frac{AM}{2};\measuredangle MDK=\frac{\phi }{2})$$$$\sin \frac{\phi }{2}=\frac{MK}{MD}=\frac{AM}{2MD}AM=2MD\sin \frac{\phi }{2}\left(2\right)$$$$\left(1\right)\Rightarrow 4{MD}^2\sin {}^2\frac{\phi }{2}=$$$${DM}^2+{AD}^2-2AD\times DM\cos \phi$$$${DM}^2(1-4\sin {}^2\frac{\phi }{2})-2AD\times DM\cos \phi +{AD}^2=0$$$${\mathit{y}}^2-\frac{24\sqrt{b^2-a^2}}{(24\sqrt{b^2-a^2}-3\sqrt{{12}^2{b}^2+25a^2})}\mathit{y}+$$$$+\frac{({12}^2{b}^2+25a^2)\sqrt{{12}^2{b}^2+25a^2}}{24\sqrt{b^2-a^2}-3\sqrt{{12}^2{b}^2+25a^2}}$$$${(\mathit{y}-\frac{4\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}}{8\sqrt{b^2-a^2}-\sqrt{{12}^2{b}^2+25a^2}})}^2+$$$$\frac{\sqrt{{({12}^2{\mathit{b}}^2+25{\mathit{a}}^2)}^3}}{24\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-3\sqrt{{12}^2{\mathit{b}}^2+25{\mathit{a}}^2}}-$$$$\frac{16({\mathit{b}}^2-{\mathit{a}}^2)}{{(24\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-\sqrt{{12}^2{\mathit{b}}^2+25{\mathit{a}}^2})}^2}$$$$$$$$\mathit{y}=\frac{4\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}}{8\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-\sqrt{{12}^2{\mathit{b}}^2+25{\mathit{a}}^2}}\pm$$$$\frac{\sqrt{16({\mathit{b}}^2-{\mathit{a}}^2)-\sqrt{{(12{\mathit{b}}^2+25{\mathit{a}}^2)}^3}[(24\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-3\sqrt{12{\mathit{b}}^2+25{\mathit{a}}^2)}}}{(24\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-\sqrt{12{\mathit{b}}^2+25{\mathit{a}}^2)}}$$$$$$$$[Area=\frac{13a}{2}\mathit{y}]\left(3\right)$$$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{e}\mathit{g}\mathit{e}\mathit{o}\mathit{m}\mathit{e}\mathit{t}\mathit{r}\mathit{i}\mathit{q}\mathit{u}\mathit{e}:$$$$△DEP\tan \alpha =\frac{EP}{DE}=\frac{BI}{DI}=\frac{AB}{AC}$$$$\frac{EP}{x}=\frac{\sqrt{b^2-a^2}}{a}\left(1\right)$$$$\frac{HP}{CH}=\frac{AB}{AC}\Rightarrow \frac{\left(12\sqrt{b^2-a^2}\right)+FP}{x+12a}\left(2\right)$$$$$$$$FP=\frac{x\sqrt{b^2-a^2}}{a}$$$$\frac{FP+12\sqrt{b^2-a^2}}{x+12a}=\frac{\sqrt{b^2-a^2}}{a}$$$$\frac{\frac{x\sqrt{b^2-a^2}}{a}+12\sqrt{b^2-a^2}}{x+12a}=\frac{\sqrt{b^2-a^2}}{a}$$$$\frac{x\sqrt{b^2-a^2}+12a\sqrt{b^2-a^2}}{x+12a}=\frac{\sqrt{b^2-a^2}}{a}$$$$ax\sqrt{b^2-a^2}+12a^2\sqrt{b^2-a^2}=$$$$x\sqrt{b^2-a^2}+12a\sqrt{b^2-a^2}$$$$\mathit{x}(\mathit{a}\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-\sqrt{{\mathit{b}}^2-{\mathit{a}}^2})=$$$$12\mathit{a}(\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}-\mathit{a}\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}$$$$\mathit{x}\left(\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}\right)(\mathit{a}-1)=$$$$12\mathit{a}\sqrt{{\mathit{b}}^2-{\mathit{a}}^2}(1-\mathit{a})$$$$\mathit{x}=12\mathit{a}xy=128$$$$\mathit{y}=\frac{32}{3\mathit{a}}=$$$$\left(3\right)Area\left(triangle\right)=\frac{13a}{2}y$$$$\mathit{A}\mathit{r}\mathit{e}\mathit{a}=69,33$$$$$$

Commented by a.lgnaoui last updated on 01/Mar/23

Commented by mr W last updated on 02/Mar/23

$$clearlywrong!thetrianglelooks$$$$almostdoublesolargeasthe$$$$rectangle,butyouranswersaysthe$$$$triangleisonlyhalfsolargeasthe$$$$rectangle!$$$$youseemtohavenosenseofgeometric$$$$size.$$

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