## Task B-1.4.

On the side $\overline{A B}$ of triangle $A B C$, points $P$ and $Q$ are given such that $|A P|=|P Q|=|Q B|$.

On the side $\overline{B C}$, point $R$ is given such that $|R C|=\frac{1}{3}|B C|$, and on the side $\overline{A C}$, point $S$ is given such that $|S C|=\frac{1}{3}|A C|$. If the area of quadrilateral $P Q R S$ is $16 \mathrm{~cm}^{2}$, what is the area of triangle $A B C$?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-05.jpg?height=348&width=605&top_left_y=951&top_left_x=197)

Notice that $\triangle S R C \sim \triangle A B C$ by the SAS similarity theorem because:

$\frac{|S C|}{|A C|}=\frac{|R C|}{|B C|}=\frac{1}{3} \text{ and } \angle A C B = \angle S C R$.

1 point

From the similarity mentioned, it follows that $S R \| A B$ and $\frac{|S R|}{|A B|}=\frac{1}{3}$, that is, $|S R|=\frac{1}{3}|A B|$.

Since $|P Q|=\frac{1}{3}|A B|=|S R|$ and $P Q \| S R$, we conclude that quadrilateral $P Q R S$ is a parallelogram.

Let $v_{c}$ be the height from vertex $C$ of triangle $A B C$, and $v$ be the height to side $\overline{P Q}$ of parallelogram $P Q R S$.

Then, $v=\frac{2}{3} v_{c}$ because the height $v$ is equal to the difference between the heights from vertex $C$ of triangle $A B C$ and $S R C$.

The area of parallelogram $P Q R S$ is:

$P(P Q R S)=v \cdot|P Q|=\frac{2}{3} v_{c} \cdot \frac{1}{3}|A B|$

so, $16=\frac{2}{9} \cdot|A B| \cdot v_{c}$, that is, $|A B| \cdot v_{c}=72$.

Finally, the desired area is $P(A B C)=\frac{|A B| \cdot v_{c}}{2}=36 \text{ cm}^{2}$.