Similar Triangle Challenge

Observe that the two triangles $\triangle ABC$ and $\triangle CDE$ are similar. Then, since the ratio of side lengths is $5:13,$ the ratio of their areas is $25:169.$ Therefore, if we let $\lvert\triangle CDE\rvert$ denote the area of $\triangle CDE,$ then $\begin{aligned} \lvert\triangle CDE\rvert&= \lvert\triangle ABC\rvert \times \frac{169}{25} \\ &=11 \times \frac{169}{25} \\ &=\frac{1859}{25}. \end{aligned}$

The areas of similar plane figures or similar surfaces ( $A_1,A_2)$ have the same ratio as the squares of any corresponding lines ( $x_1,x_2)$ . From the figure the two triangles are similar. So we have

$\frac{A_{ABC}}{A_{CDE}} = \frac{5^2}{13^2}$

$\frac{11}{A_{CDE}} = \frac{25}{169}$

$169(11) = (25)(A_{CDE})$

$A_{CDE} = \frac{1859}{25}$

The areas of similar plane figures have the same ratio as the squares of any two corresponding sides. We have

$\dfrac{A_{ABC}}{A_{CDE}}=\dfrac{(AC)^2}{(CD)^2}$

$\dfrac{11}{A_{CDE}}=\dfrac{5^2}{13^2}$

$A_{CDE}=\dfrac{11(169)}{25}=$ $\boxed{\dfrac{1859}{25}}$