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.


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 $$A_{\triangle PQR}= \frac{9}{70}xy=3.$$


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We first observe that $\frac{|PB|}{|DT|} = \frac{3}{4}$. So, $\frac{S(PRB)}{S(DRT)} = \frac{9}{16}$, where $S$ denotes the area. We can assume actually $S(PRB) = 9$ and $S(DRT) = 16$ (we will do the renormalization later). I thus write $9$ for the area of $PRB$ on the figure. This makes $S(SRT) = 8$ as $|DS| = |ST|$. Also $\frac{|PR|}{|RT|} = \frac{3}{4}$, and this gives $S(PSR) = 6$ as we knew $S(SRT) = 8$. So, $S(PST) = 6+8 = 14$, which means $S(DSP) = 14$ as well. Now, we note that $\frac{|PQ|}{|QS|} = \frac{3}{2}$, which means $S(DQS) = \frac{28}{5}$. Since $S(DSR) = 8$, we get $S(QSR) = 8-\frac{28}{5}$, and finally $S(PQR) = 6-(8-\frac{28}{5}) = 3.6$. The area of the entire rectangle is on the other hand (from $PST$), $14\times 3\times 2 = 84$. Finally, we solve $\frac{3.6}{84} = \frac{x}{70}$, which gives $x=S(PQR) = 3$.

So, the general idea is to dissect like crazy, and try to identify the normalized area of each piece one by one. You can see my failed attempt as I also unnecessarily calculated the areas of $BRT$ and $BTC$.

  • $\begingroup$ In the first sentence, your $S(SRT)$ in the denominator should be $S(DRT)$. $\endgroup$ – TonyK Apr 4 '14 at 19:35
  • $\begingroup$ Also, your arbitrary choice of units is a bit confusing. (It confused me, anyway!) $\endgroup$ – TonyK Apr 4 '14 at 19:41
  • $\begingroup$ I blew your diagram up some; hope you don't mind... $\endgroup$ – colormegone Apr 4 '14 at 21:03
  • $\begingroup$ @TonyK Thanks, I have fixed it. I also tried to better explain the later-to-come renormalization. $\endgroup$ – Lord Soth Apr 4 '14 at 22:18
  • $\begingroup$ @RecklessReckoner Looks much better, thanks. $\endgroup$ – Lord Soth Apr 4 '14 at 22:18

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