Question

In triangle $ABC$, with $\overline{AB}=\overline{AC}=3.6$, a point $D$ is taken on $AB$ at a distance $1.2$ from $A$. Point $D$ is joined to point $E$ in the prolongation of $AC$ so that triangle $AED$ is equal in area to triangle $ABC$. Then $\overline{AE}$ equals.

• A. $4.8$
• B. $5.4$
• C. $7.2$
• D. $10.8$
• E. $12.6$

Answer 1

Answer: (D)

$10.8$

Let the altitude from $B$ to $AC$ be $h$. Then since. $\overline{AD}=\frac{1}{3}\overline{AB}$
the altitude from $D$ to $AE$ is $\frac{1}{3}h$. Let $AE=x$. Then
$\frac{1}{2}\cdot \frac{1}{3}hx=\frac{1}{3}h\left(3.6\right)=\therefore x=10.8$.