I am stumped on the following question

PQRS is a parallelogram and ST=TR. What is the ratio of area of triangle QST to the area of parallelogram (Ans 1:4)

enter image description here

I need the height of the triangle, how would I get that? Any suggestions would be appreciated.

  • 1
    $\begingroup$ The height of the triangle is the same as the height of the parallelogram. $\endgroup$ – Martin Sleziak Jul 25 '12 at 9:19
  • $\begingroup$ How do I go about finding the height of the parallelogram. None of the angles are 90. $\endgroup$ – Rajeshwar Jul 25 '12 at 9:21
  • $\begingroup$ You don't need to know the exact value of the height, since you are only interested in the ration of the two areas. See the answer that has already been posted. $\endgroup$ – Martin Sleziak Jul 25 '12 at 9:24


where $H$ is the orthogonal projection of $Q$ on the line containing $S$ and $R$.

  • $\begingroup$ I still dont get it. Could you please explain how you got QH. ? $\endgroup$ – Rajeshwar Jul 25 '12 at 9:25
  • $\begingroup$ The second comment from @MartinSleziak to your question applies here. $\endgroup$ – enzotib Jul 25 '12 at 9:28

The diagonal divides the parallelogram into two parts of equal area. In fact, $\triangle SRQ$ and $\triangle QPS$ are congruent (they have matching sides).

Look at $\triangle STQ$ and $\triangle TRQ$. Think of them as having bases $ST$ and $TR$. Then they have the same height. Since $ST=TR$, they also have equal bases. so they have the same area.

Thus our shaded $\triangle STQ$ has area half the area of $SRQ$, which has half the area of the parallelogram. Therefore the area of $\triangle STQ$ is $\frac{1}{2}\cdot \frac{1}{2}$, that is, $\frac{1}{4}$ of the area of the parallelogram.

Remark: The height of the parallelogram cannot be computed from the given information. But we do not need it to find the ratio of the areas.

  • $\begingroup$ Why are we assuming that QS is a diagonal? we're not given this info. $\endgroup$ – fullMoon Jan 27 '16 at 0:58
  • $\begingroup$ From the supplied picture that seesms to be the intension. $\endgroup$ – André Nicolas Jan 27 '16 at 1:38

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