# Thread: Need help with Rhombus Question

1. ## Need help with Rhombus Question

Explain the steps you would follow to prove that the figure formed by connecting the adjacent midpoints of any rectangle is a rhombus.

(Don't have to prove it, but I need it explained how I would do so)

Can't seem to wrap my head around this one and need help.

I appreciate your time and response.thanks
Joanne

2. ## rhombus properties

A rhombus has one major property about the lengths of its sides.

Here's a good starting point: Draw a rectangle with width of 2a and length of 2b. Add in the midpoints of each side and write down how long each segment is in your picture. Then, use the pythagorean theorem to decide how long each side of the rhombus is. Don't just find one of them and try and conclude anything, find at least two of them before you stop to consider what's going on.

3. Hello, Joanne!

Prove that the figure formed by connecting the adjacent midpoints of any rectangle is a rhombus.

First, make a sketch . . .
Code:
                  P
A * - - - - - * - - - - - * B
|        *     *        |
|     *           *     |
|  *                 *  |
S *                       * Q
|  *                 *  |
|     *           *     |
|        *     *        |
D * - - - - - * - - - - - * C
R

Let $ABCD$ be the rectangle and $P,Q,R,S$ be the midpoints.

There are right triangles in each corner.
We can prove that they are all congruent.
Therefore: . $PQ = QR = RS = SP$ . and $PQRS$ is a rhombus.

For example, consider $\Delta PAS$ and $\Delta PBQ$

$AD = BC$ . . . Opposite sides of a rectangle are equal.

$S$ is the midpoint of $AD$; $Q$ is the midpont of $BC.$
. . $AS = \tfrac{1}{2}AD,\;BQ = \tfrac{1}{2}BC \quad\Rightarrow\quad AS = BQ$

$\angle A \:=\:\angle B \:=\:90^o$

$P$ is the midpont of $AB \quad\Rightarrow\quad AP = PB$

Hence: . $\Delta PAS \cong \Delta PBQ\;\;\text{(s.a.s.)}$

Therefore: . $PS = PQ$

Continue in the same manner and prove all the hypotenuses are equal.