Height of triangle inside a parallelogram

Question

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)

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

Answer 1

$$A_T=\frac{1}{2}\overline{ST}\,\overline{QH}=\frac{1}{2}\,\frac{1}{2}\overline{SR}\,\overline{QH}=\frac{1}{4}A_P$$

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

Answer 2

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.