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We have triangle $\triangle ABC$ where $AB = AC$ and $AD$ is an altitude. Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE.$ If $BC = 12$ and the area of $\triangle ABC$ is $180,$ what is the area of $ABDE$?

Aug 4, 2019

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We have triangle $$\triangle ABC$$ where $$AB = AC$$ and $$AD$$ is an altitude.
Meanwhile, $$E$$ is a point on $$AC$$ such that $$AB \parallel DE$$.
If $$BC = 12$$ and the area of  $$\triangle ABC$$ is $$180$$, what is the area of $$ABDE$$?

$$\text{Let AB = AC  } \\ \text{Let DB = CD =\dfrac{BC}{2} = 6 }$$

$$\begin{array}{|rcll|} \hline \text{area of } \triangle ABC =180 &=& \dfrac{BC*AD}{2} \\ 180 &=& \dfrac{12*AD}{2} \\ 180 &=& 6AD \\ \mathbf{ AD } &=& \mathbf{30} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{CD}{ED} &=& \dfrac{BC}{AB} \\\\ \dfrac{6}{ED} &=& \dfrac{12}{AB} \\\\ \dfrac{ED}{6} &=& \dfrac{AB}{12} \\\\ \mathbf{ED }&=& \mathbf{\dfrac{AB}{2}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{area of } ABDE &=& \dfrac{AB+ED}{2}\times H \quad| \quad H=\dfrac{DB*AD}{AB}=\dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+ED}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+\dfrac{AB}{2}}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \dfrac{3}{4}AB\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& 3* 3*15 \\\\ \mathbf{\text{area of } ABDE} &=& \mathbf{135} \\ \hline \end{array}$$ Aug 5, 2019