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Question Number 193975 by mr W last updated on 24/Jun/23

Commented by mr W last updated on 25/Jun/23

$$findtheareaofthehatchedsquare.$$

Answered by mr W last updated on 25/Jun/23

Commented by mr W last updated on 25/Jun/23

$$\cos \alpha =\frac{7^2+9^2-{\left(\sqrt{2}s\right)}^2}{2\times 7\times 9}$$$$\cos (\beta +\gamma )=\frac{2^2+6^2-{\left(\sqrt{2}s\right)}^2}{2\times 2\times 6}=-\cos \alpha$$$$\Rightarrow \frac{2^2+6^2-{\left(\sqrt{2}s\right)}^2}{2\times 2\times 6}=-\frac{7^2+9^2-{\left(\sqrt{2}s\right)}^2}{2\times 7\times 9}$$$$\Rightarrow s^2=\frac{136}{5}=areaofsquare$$

Answered by Subhi last updated on 25/Jun/23

$$pute\widehat{fb}=\alpha ,f\widehat{be}=\beta ,eachsideofthesquare=l$$$$fd=\sqrt{l^2+l^2}=\sqrt{2}l$$$$\mathrm{\Delta }fbd:{BD}^2=2l^2+6^2-12\sqrt{2}lcos(45+\alpha )$$$$2+{13}^2+36Q^2=11(18+36+2l^2-12\sqrt{2}l.cos(45+\alpha )$$$$\left({stewart}^{\prime }stheorem\right)$$$$36Q^2-22l^2+132\sqrt{2}l.cos(45+\alpha )=256\to \left(i\right)$$$$\mathrm{\Delta }AEB:h^2=Q^2+{13}^2-26Qcos\left(\beta \right)$$$${11}^2+{13}^2=2Q^2+2h^2\left(Apolloniustheorem\right)$$$${11}^2+{13}^2=4Q^2+2\times {13}^2-52Qcos\left(\beta \right)\left(1\right)$$$$\mathrm{\Delta }efb:cos\left(\beta \right)=\frac{Q^2+6^2-l^2}{12Q}\left(2\right)$$$$compute1,2$$$$108=\frac{13}{3}{l}^2-\frac{1}{3}{Q}^2\left(ii\right)$$$$compute\left(i\right),\left(ii\right)$$$$468l^2-11664-22l^2+132\sqrt{2}l.cos(45+\alpha )=256$$$$446l^2+132\sqrt{2}l.cos(45+\alpha )=11920\to \left(iii\right)$$$$\mathrm{\Delta }ADF:cos(135-\alpha )=\frac{7^2+2l^2-9^2}{14\sqrt{2}l}=-cos(45+\alpha )\left(3\right)$$$$compute\left(iii\right),\left(3\right)$$$$446l^2-132\sqrt{2}.\left(\frac{2l^2-32}{14\sqrt{2}}\right)=11920$$$$l^2=\frac{11920-\frac{132\sqrt{2}.32}{14\sqrt{2}}}{446-\frac{264}{14}}=\frac{136}{5}=areaofthesquare$$

Commented by Subhi last updated on 25/Jun/23

Commented by mr W last updated on 26/Jun/23

$$thanks!$$

Commented by Subhi last updated on 26/Jun/23

$$Youarewelcome$$

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