I have a similar triangle question that has got me stumped:
I have already worked out question a)
(Let $x$ = area of $∆CDE$）
$$ \frac{x}{15} =(8/5)^2$$
$$ \frac{x}{15} =64/25$$
$$ x=64/25*15$$
$$x=38.4$$
But I can’t seem to work out part b). Can anyone help? Thanks in advance.
You already noticed that the triangles are similar.
In similar triangles, if their lengths differ by a factor of $r$, their areas differ by a factor of $r^2$.
Let's use this!
The ratio of sides of $\triangle DEC$ to $\triangle ABC$ is $\displaystyle \frac{8}{5}$.
Therefore, the area of $\triangle DEC$ is $\displaystyle 15\left(\frac{8}{5}\right)^2=\boxed{38.4}$
Similarly, notice that by the Midpoint Theorem, $PQ=4$, so the ratio of sides of $\triangle PQC$ to $\triangle ABC$ is $\displaystyle \frac{4}{5}$.
Therefore, the area of $\triangle PQC$ is $\displaystyle 15\left(\frac{4}{5}\right)^2=\boxed{9.6}$
I will explain how I figured out that $PQ=4$ a bit more.
Notice that by the Midpoint Theorem, we have $CQ=QE$.
Therefore, $\displaystyle CQ=\frac{1}{2}CE$.
Because the triangles $CQP$ and $CED$ are similar, we have the ratio of the sides of $CQP$ to $CED$ is $\displaystyle \frac{1}{2}$, so directly applying that to $DE$ and $PQ$ gives $\displaystyle PQ = \frac{1}{2}DE = 4$.
Forgot to answer your last question.
The area of PQED is the difference of the areas of $\triangle DEC$ and $\triangle PQC$, as you can see. This is $38.4 - 9.6 = \boxed{28.8}$.