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
$${BE}={BA}+{ED}\:{is}\:{clear}. \\ $$$${error}\:{in}\:{calculating}\:{x}\:{is}\:{fixed}. \\ $$
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
$${BE}={BF}+{FD}={BA}+{ED}=\mathrm{6}+\mathrm{6}−{x} \\ $$$$\sqrt{\mathrm{6}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\mathrm{12}−{x} \\ $$$$\mathrm{6}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} −\mathrm{24}{x} \\ $$$$\Rightarrow{x}=\frac{\mathrm{12}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{24}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${A}_{{shadeed}} =\frac{\mathrm{6}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{2}}=\frac{\mathrm{27}}{\mathrm{2}} \\ $$
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
$$\mathrm{BF}^{\mathrm{2}} =\mathrm{BC}^{\mathrm{2}} +\mathrm{CF}^{\mathrm{2}} \\ $$$$\left(\mathrm{6}+\mathrm{x}\right)^{\mathrm{2}} =\:\mathrm{6}^{\mathrm{2}} +\left(\mathrm{6}−\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{6}+\mathrm{x}\right)^{\mathrm{2}} −\left(\mathrm{6}−\mathrm{x}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$$\left(\mathrm{12}\right)\left(\mathrm{2x}\right)=\mathrm{36}\:\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{CF}\:=\:\mathrm{6}−\frac{\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{shaded}\:\mathrm{area}\:=\:\frac{\mathrm{6}×\frac{\mathrm{9}}{\mathrm{2}}}{\mathrm{2}}=\:\frac{\mathrm{27}}{\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\: \\ $$
Commented by bobhans last updated on 29/May/20
$$\mathrm{good}\: \\ $$
Commented by i jagooll last updated on 29/May/20
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{it}\:\mathrm{correct} \\ $$