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Question Number 95999 by i jagooll last updated on 29/May/20

Answered by mr W last updated on 29/May/20

Commented by mr W last updated on 29/May/20

$$BE=BA+EDisclear.$$$$errorincalculatingxisfixed.$$

Commented by john santu last updated on 29/May/20

$$\mathrm{check}..\mathrm{sir}.\mathrm{it}\mathrm{wrong}$$

Commented by mr W last updated on 29/May/20

$$BE=BF+FD=BA+ED=6+6-x$$$$\sqrt{6^2+x^2}=12-x$$$$6^2={12}^2-24x$$$$\Rightarrow x=\frac{{12}^2-6^2}{24}=\frac{9}{2}$$$$A_{shadeed}=\frac{6}{2}\times \frac{9}{2}=\frac{27}{2}$$

Commented by bobhans last updated on 29/May/20

$$\mathrm{what}\mathrm{formula}\mathrm{BE}=\mathrm{BA}+\mathrm{ED}?$$

Commented by john santu last updated on 29/May/20

$$\mathrm{oo}\mathrm{yes}.\mathrm{great}$$

Answered by john santu last updated on 29/May/20

Commented by john santu last updated on 29/May/20

$${\mathrm{BF}}^2={\mathrm{BC}}^2+{\mathrm{CF}}^2$$$${(6+\mathrm{x})}^2=6^2+{(6-\mathrm{x})}^2$$$${(6+\mathrm{x})}^2-{(6-\mathrm{x})}^2=36$$$$\left(12\right)\left(2\mathrm{x}\right)=36\Rightarrow \mathrm{x}=\frac{3}{2}$$$$\mathrm{CF}=6-\frac{3}{2}=\frac{9}{2}$$$$\mathrm{shaded}\mathrm{area}=\frac{6\times \frac{9}{2}}{2}=\frac{27}{2}$$$$\mathrm{is}\mathrm{the}\mathrm{answer}$$

Commented by bobhans last updated on 29/May/20

$$\mathrm{good}$$

Commented by i jagooll last updated on 29/May/20

$$\mathrm{yes}\mathrm{sir}\mathrm{it}\mathrm{correct}$$

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