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Question-84899




Question Number 84899 by Power last updated on 17/Mar/20
Commented by Tony Lin last updated on 17/Mar/20
A(ADE)=9  ⇒h=6  A(EFC)  =(1/2)×EC×(h×(6/7))  =(1/2)×6×((36)/7)  =((108)/7)
$${A}\left({ADE}\right)=\mathrm{9} \\ $$$$\Rightarrow{h}=\mathrm{6} \\ $$$${A}\left({EFC}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{EC}×\left({h}×\frac{\mathrm{6}}{\mathrm{7}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}×\frac{\mathrm{36}}{\mathrm{7}} \\ $$$$=\frac{\mathrm{108}}{\mathrm{7}} \\ $$

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