Question

$ABCD$ is a rectangle formed by the points $A(-1,-1)$ , $B(-1,4)$ , $C(5,4)$ and$D(5,-1)$ . $P,Q,R$ and $S$ are the midpoints of $AB,BC,CD$ and $DA$ respectively. Is the quadrilateral $PQRS$ a square? a rectangle? or a rhombus? Justify your answer.

Answer 1

Step $1:$Finding the midpoints $P,Q,R$ and $S$:

$P,Q,R$ and $S$ are the midpoints of $AB,BC,CD$ and $DA$ respectively.

Coordinate of $P=\left(\frac{-1+(-1)}{2}\right),\left(\frac{-1+4}{2}\right)=\left(\frac{-2}{2}\right),\left(\frac{3}{2}\right)=\left(-1,\frac{3}{2}\right)$

Coordinate of $Q=\left(\frac{-1+5}{2}\right),\left(\frac{4+4}{2}\right)=\left(\frac{4}{2}\right),\left(\frac{8}{2}\right)=\left(2,4\right)$

Coordinate of $R=\left(\frac{5+5}{2}\right),\left(\frac{4+(-1)}{2}\right)=\left(\frac{10}{2}\right),\left(\frac{3}{2}\right)=\left(5,\frac{3}{2}\right)$

Coordinate of $S=\left(\frac{5+(-1)}{2}\right),\left(\frac{-1+(-1)}{2}\right)=\left(\frac{4}{2}\right),\left(\frac{-2}{2}\right)=\left(2,-1\right)$

Step $2:$ Finding the length of $PQ,QR,RS$ and $SP$

Using distance formula $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Length of $PQ$

$\begin{array}{rcl}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-(-1)\right)}^2+{\left(4-\frac{3}{2}\right)}^2}\\ & =& \sqrt{{\left(2+1\right)}^2+{\left(\frac{8-3}{2}\right)}^2}\\ & =& \sqrt{{\left(3\right)}^2+{\left(\frac{5}{2}\right)}^2}\\ & =& \sqrt{9+\frac{25}{4}}\\ & =& \sqrt{\frac{36+25}{4}}\\ & =& \sqrt{\frac{61}{4}}\\ & =& \frac{\sqrt{61}}{2}\end{array}$

Length of $QR$

$\begin{array}{rcl}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-5\right)}^2+{\left(4-\frac{3}{2}\right)}^2}\\ & =& \sqrt{{\left(-3\right)}^2+{\left(\frac{8-3}{2}\right)}^2}\\ & =& \sqrt{9+{\left(\frac{5}{2}\right)}^2}\\ & =& \sqrt{9+\frac{25}{4}}\\ & =& \sqrt{\frac{36+25}{4}}\\ & =& \sqrt{\frac{61}{4}}\\ & =& \frac{\sqrt{61}}{2}\end{array}$

Length of $RS$

$\begin{array}{rcl}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-5\right)}^2+{\left(-1-\frac{3}{2}\right)}^2}\\ & =& \sqrt{{\left(-3\right)}^2+{\left(\frac{-2-3}{2}\right)}^2}\\ & =& \sqrt{9+{\left(\frac{-5}{2}\right)}^2}\\ & =& \sqrt{9+\frac{25}{4}}\\ & =& \sqrt{\frac{36+25}{4}}\\ & =& \sqrt{\frac{61}{4}}\\ & =& \frac{\sqrt{61}}{2}\end{array}$

Length of $SP$

$\begin{array}{rcl}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-(-1)\right)}^2+{\left(-1-\frac{3}{2}\right)}^2}\\ & =& \sqrt{{\left(2+1\right)}^2+{\left(\frac{-2-3}{2}\right)}^2}\\ & =& \sqrt{{\left(3\right)}^2+{\left(\frac{-5}{2}\right)}^2}\\ & =& \sqrt{9+\frac{25}{4}}\\ & =& \sqrt{\frac{36+25}{4}}\\ & =& \sqrt{\frac{61}{4}}\\ & =& \frac{\sqrt{61}}{2}\end{array}$

Step $3:$ Finding the length of diagonals $PR$ and $SQ$:

Length of $PR$

$\begin{array}{rcl}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(5-(-1)\right)}^2+{\left(\frac{3}{2}-\frac{3}{2}\right)}^2}\\ & =& \sqrt{{\left(5+1\right)}^2+{\left(\frac{3-3}{2}\right)}^2}\\ & =& \sqrt{{\left(6\right)}^2+{\left(\frac{0}{2}\right)}^2}\\ & =& \sqrt{36+0}\\ & =& \sqrt{36}\\ & =& 6\end{array}$

Length of $SQ$

$\begin{array}{rcl}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-2\right)}^2+{\left(4-(-1)\right)}^2}\\ & =& \sqrt{{\left(0\right)}^2+{\left(4+1\right)}^2}\\ & =& \sqrt{0+{\left(5\right)}^2}\\ & =& \sqrt{25}\\ & =& 5\end{array}$

Step 4: Finding $PQRS$ is a square or rectangle or rhombus:

As , the lengths $PQ=QR=RS=SP=\frac{\sqrt{61}}{2}$,

But, the lengths of diagonals are not equal as,

$$\begin{gathered} PR=6,SQ=5 \\ \Rightarrow PR\ne SQ \end{gathered}$$

Hence, $PQRS$ is a rhombus.