Review question

# Why is this quadrilateral a rhombus but not a square? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6641

## Solution

Draw a diagram showing points $A(-3,1)$, $B(4,2)$, $C(9,-3)$, $D(2,-4)$.

1. For the figure $ABCD$ state and prove a fact (or facts) sufficient to ensure that it is a parallelogram.

Drawing a diagram is a good idea.

#### Approach 1

A parallelogram is a quadrilateral where opposite sides have equal lengths, so all we have to show is that $AB=CD$ and $AD=BC$.

The length of the sides can be calculated with the use of Pythagoras’ theorem by constructing right triangles between the points. We then find

\begin{align*} AB &= \sqrt{(-3-4)^2+(1-2)^2} &=\sqrt{50},\\ CD &= \sqrt{(9-2)^2+(-3+4)^2} &=\sqrt{50},\\ AD &= \sqrt{(-3-2)^2+(1+4)^2} &=\sqrt{50},\\ BC &= \sqrt{(4-9)^2+(2+3)^2} &=\sqrt{50}. \end{align*}

This certainly satisfies $AB=CD$ and $AD=BC$. The figure therefore is a parallelogram.

#### Approach 2

A quadrilateral is a parallelogram if and only if its diagonals bisect each other. Midpoint($AC$) = ($3,-1$) = Midpoint ($BD$), so $ABCD$ must be a parallelogram.

1. State and prove an additional fact sufficient to ensure that $ABCD$ is a rhombus.

A rhombus is a quadrilateral with four equal sides. We’ve already calculated all four side lengths, and they’re equal, so $ABCD$ must be a rhombus.

1. Prove that $ABCD$ is not a square.

#### Approach 1

If we can prove that any of the angles inside the figure is not a right angle, then this would show that $ABCD$ isn’t a square.

The angle at $C$ is a right angle if and only if $AC^2 = AD^2 + CD^2$. In fact, we find

\begin{align*} AC^2 &= (-3-9)^2+(1+3)^2 &= 160,\\ AD^2 + CD^2 &= 50 + 50 &= 100. \end{align*}

The figure is therefore not a square.

#### Approach 2

A square is a rhombus where diagonals have equal lengths. So all we have to consider is whether $AC=BD$. A short calculation reveals

\begin{align*} AC &= \sqrt{(-3-9)^2+(1+3)^2} &= \sqrt{160},\\ BD &= \sqrt{(4-2)^2+(2+4)^2} &= \sqrt{40}. \end{align*}

Once again, we see that $ABCD$ is not a square.

#### Approach 3

Gradient($AB$) = $1/7$, Gradient($BC$) = $-1$, so their product is not $-1$. Thus the angle at $B$ is not a right angle, and $ABCD$ is not a square.