Question Number 208554 by efronzo1 last updated on 18/Jun/24
Answered by liuxinnan last updated on 18/Jun/24
Commented by efronzo1 last updated on 18/Jun/24
Answered by A5T last updated on 18/Jun/24
Commented by A5T last updated on 18/Jun/24
![DF=AE=h;EF=10 FC=x⇒BE=6−x h^2 =8^2 −x^2 =6^2 −(6−x)^2 ⇒x=((16)/3) ⇒h=(√(64−(((16)/3))^2 ))=((8(√5))/3) [ABCD]=(((10+16)×((8(√5))/3))/2)=((104(√5))/3)](https://www.tinkutara.com/question/Q208588.png)
$${DF}={AE}={h};{EF}=\mathrm{10} \\ $$$${FC}={x}\Rightarrow{BE}=\mathrm{6}−{x} \\ $$$${h}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} −\left(\mathrm{6}−{x}\right)^{\mathrm{2}} \Rightarrow{x}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\Rightarrow{h}=\sqrt{\mathrm{64}−\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\left[{ABCD}\right]=\frac{\left(\mathrm{10}+\mathrm{16}\right)×\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{104}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$