In rectangle $ABCD$, $ P$ is the mid point of $AB$. $S$ and $T$ are the points of trisection
of $DC$. If area of the rectangle is $70$ square units, with reference to the figure find area
of shaded region.
Let $DS=x$ , and $AD=y$.
So $3xy=70$.
Now I've no idea how to calculate the area of shaded triangle. We don't know any of its side or altitudes.
I'm thinking that it is probably similar to some other triangle , But I can't find it.
Please help.
Best Answer
Let $d(Q,DC)$ the distance between point $Q$ and the line defined by $DC$, and $d(R,DC)$ the distance between point $R$ and the line defined by $DC$.
Note that: $\triangle DQS \sim \triangle BQP$ and $\triangle BRP \sim \triangle DRT$, hence: $$\frac{d(Q,DC)}{d(Q,AB)}= \frac{1x}{\frac{3}{2}x}= \frac{2}{3} \quad(1)$$ and $$\frac{d(R,DC)}{d(R,AB)}= \frac{2x}{\frac{3}{2}x}= \frac{4}{3}. \quad(2)$$ From $(1)$ and $(2)$ we get: $$d(Q,AB)=\frac{3}{5}y \quad(3)$$ and $$d(R,AB)=\frac{3}{7}y, \quad(4)$$ which are the heights of triangles $\triangle BQP$ and $\triangle BRP$.
We can get the area of $\triangle PQR$ from: $$A_{\triangle PQR}= A_{\triangle PQB}- A_{\triangle PRB}$$ Therefore $$A_{\triangle PQR}= \frac{9}{70}xy=3.$$