Question Number 211188 by Ari last updated on 30/Aug/24
Answered by mr W last updated on 30/Aug/24
$${AB}={x} \\ $$$${BC}={y} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} \\ $$$${x}+{y}=\mathrm{16} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{16}^{\mathrm{2}} \\ $$$$\mathrm{12}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{16}^{\mathrm{2}} \\ $$$$\Rightarrow{red}\:{area}=\frac{{xy}}{\mathrm{2}}=\frac{\mathrm{16}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{28} \\ $$