# Problem of the Week Problem C and Solution Partitioned Pentagon

## Problem

Consider pentagon $$PQRST$$. Starting at $$P$$ and moving around the pentagon, the vertices are labelled $$P$$, $$Q$$, $$R$$, $$S$$, and $$T$$, in order.

The pentagon has right angles at $$P$$, $$Q$$, and $$R$$, obtuse angles at $$S$$ and $$T$$, and an area of $$1000\mbox{ cm}^2$$.

Point $$V$$ lies inside the pentagon such that $$\angle PTV$$, $$\angle TVS$$, and $$\angle VSR$$ are right angles.

Point $$U$$ lies on $$TV$$ such that $$\triangle STU$$ has an area of $$210\mbox{ cm}^2$$. Also, it is known that $$PQ=50$$ cm, $$SR=15$$ cm, and $$TU=30$$ cm.

Determine the length of $$PT$$.

## Solution

Extend $$TV$$ to meet $$QR$$ at $$W$$. We mark this and all of the given information on the diagram.

To find the area of a triangle, multiply the length of the base by the height and divide by 2. In $$\triangle STU$$, the base $$TU$$ has length $$30$$ cm. The corresponding height of $$\triangle STU$$ is the perpendicular distance from $$TU$$ (extended) to vertex $$S$$, namely $$SV$$.

Since the area of $$\triangle STU$$ is given to be $$210$$ cm$$^2$$, \begin{aligned} 210 &=\frac{30\times SV}{2}\\ 210&=15\times SV\\ 14&=SV\end{aligned}

We know that $$TW=PQ=50$$, $$VW=SR=15$$, and $$TW = TU+UV+VW$$.

It follows that $$50=30+UV+15$$ and $$UV=5$$ cm.

Now we can relate the total area of the pentagon to the areas of the shapes inside. \begin{aligned} \text{Area }PQRST&=\text{Area }PQWT+\text{Area }RSVW+\text{Area }\triangle SUV+\text{Area }\triangle STU\\ 1000&=PQ\times PT+SV\times SR+\frac{UV\times SV}{2}+210\\ 1000&=50\times PT+14\times 15+\frac{5\times 14}{2}+210\\ 1000&=50\times PT+210+35+210\\ 1000&=50\times PT+455\\ 1000-455&=50\times PT\\ 545&=50\times PT\\ \frac{545}{50}&=PT\end{aligned} Therefore, $$PT=10.9$$ cm.