Chapter 13, Problem 21CT

To determine

To prove: The segments joining the midpoints of consecutive sides of a rectangle form a rhombus.

Explanation of Solution

Proof:

Consider rectangle OABC. The mid-points of side OA, AB, BC and CO are P, Q, R and S respectively.

  

In the given figure, $\text{PQ}\parallel \text{OB}$ and $\text{SR}\parallel \text{OB}$ . So, $\text{PQ}\parallel \text{SR}$ .

By the midpoint formula, the coordinates of point P is:

  $\left(\frac{0+a}{2},\frac{0+0}{2}\right)=\left(\frac{a}{2},0\right)$

By the midpoint formula, the coordinates of point Q is:

  $\left(\frac{a+a}{2},\frac{0+b}{2}\right)=\left(a,\frac{b}{2}\right)$

The length of PQ can be calculated by the distance formula as:

  $\begin{array}{c}\text{PQ}=\sqrt{{\left(a-\frac{a}{2}\right)}^2+{\left(\frac{b}{2}-0\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\end{array}$

By the midpoint formula, the coordinates of point R is:

  $\left(\frac{a+0}{2},\frac{b+b}{2}\right)=\left(\frac{a}{2},b\right)$

By the midpoint formula, the coordinates of point S is:

  $\left(\frac{0+0}{2},\frac{0+b}{2}\right)=\left(0,\frac{b}{2}\right)$

The length of SR can be calculated by the distance formula as:

  $\begin{array}{c}\text{SR}=\sqrt{{\left(\frac{a}{2}-0\right)}^2+{\left(b-\frac{b}{2}\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\end{array}$

Thus, $\text{PQ}\parallel \text{SR}$ and $\text{PQ}=\text{SR}$ . The quadrilateral PQRS is a parallelogram.

The length of PS can be calculated by the distance formula as:

  $\begin{array}{c}\text{PS}=\sqrt{{\left(0-\frac{a}{2}\right)}^2+{\left(\frac{b}{2}-0\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\end{array}$

It can be observed that the adjacent sides PQ and PS are equal. As PQRS is a parallelogram with adjacent sides of equal length. So, PQRS is a rhombus.

Hence, the segments joining the midpoints of consecutive sides of a rectangle form a rhombus.