\(PQRS\) is a square. The points \(T\) and \(U\) are the midpoints of \(QR\) and \(RS\) respectively. The line \(QS\) cuts \(PT\) and \(PU\) at \(W\) and \(V\) respectively. What fraction of the area of the square \(PQRS\) is the area of the pentagon \(RTWVU\)?

(A) \(\dfrac{1}{3} \quad\) (B) \(\dfrac{2}{5} \quad\) (C) \(\dfrac{3}{7}\quad\) (D) \(\dfrac{5}{12}\quad\) (E) \(\dfrac{4}{15}\)

Let’s say the side of the square is length \(2\), without any loss of generality. The area of the square is then \(4\).

There are lots of different ways to solve this, mostly involving the areas of triangles. Here are two approaches.

### Approach 1: centroid of \(PRS\)

We want to know the area of triangle \(SVU\). Let’s draw in the line \(PR\).

In triangle \(PRS\), we have two of its medians (lines joining a vertex to a midpoint) and they meet at \(V\) which is therefore the centroid.

This means that \(V\) must be \(\dfrac{1}{3}\times 2\) above \(SR\).

Thus the area of triangle \(SVU\) is \(\dfrac{1}{2}\times 1 \times \dfrac{2}{3} = \dfrac{1}{3}\) and by symmetry triangle \(TQW\) is the same.

So the area of the pentagon is \(2 - 2 \times \dfrac{1}{3} = \dfrac{4}{3}\).

As a fraction of the whole square this is \(\dfrac{1}{3}\), and the answer is (A).

### Approach 2: ten triangles

Again, we’ll think about the area of \(SVU\). Let’s draw in the lines from \(R\) to the midpoints of \(PQ\) and \(PS\).

We’ll use \(A\) to mean the area of triangle \(SVU\). Triangle \(UVR\) has the same base and same height so also has area \(A\). By symmetry, the triangle \(SVV'\) also has area \(A\) as do five other triangles in the diagram. Triangle \(RVW\) has a different area which we’ll call \(B\).

The triangle \(RSV'\) makes up a quarter of the square and is made up of three small triangles. So \(3A=\frac{1}{4}\times4=1\).

Also, triangle \(SQR\) is half the square and is made up of four \(A\)s and a \(B\). So \(4A+B=2\).

We could solve these equations to find \(A\) and \(B\) but in fact all we need to know is the pentagon area which is \(B+2A=2-\frac{2}{3}=\frac{4}{3}\). So the pentagon makes up \(\frac{1}{3}\) of the square and the answer is (A).

Another approach would be to set up a coordinate system, with the origin at \(S\), \(SR\) as the \(x\)-axis and with \(SP\) as the \(y\)-axis. Then we could write down equations for lines \(PU\) and \(SQ\) and find the coordinates of their intersection as a means to finding the area of \(SVU\).