Review question

# What fraction of the square is the pentagon? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5492

## Solution

$PQRS$ 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 $SVU$.