Question

# $ABCD$ is a rectangle formed by the points $A\left(-1,-1\right)$ , $B\left(-1,4\right)$ , $C\left(5,4\right)$ and$D\left(5,-1\right)$ . $P,Q,R$ 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.

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Solution

## 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 $P=\left(\frac{-1+\left(-1\right)}{2}\right),\left(\frac{-1+4}{2}\right)=\left(\frac{-2}{2}\right),\left(\frac{3}{2}\right)=\left(-1,\frac{3}{2}\right)$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 $R=\left(\frac{5+5}{2}\right),\left(\frac{4+\left(-1\right)}{2}\right)=\left(\frac{10}{2}\right),\left(\frac{3}{2}\right)=\left(5,\frac{3}{2}\right)$Coordinate of $S=\left(\frac{5+\left(-1\right)}{2}\right),\left(\frac{-1+\left(-1\right)}{2}\right)=\left(\frac{4}{2}\right),\left(\frac{-2}{2}\right)=\left(2,-1\right)$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 $PQ$ $\begin{array}{rcl}& =& \sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\\ & =& \sqrt{{\left(2-\left(-1\right)\right)}^{2}+{\left(4-\frac{3}{2}\right)}^{2}}\\ & =& \sqrt{{\left(2+1\right)}^{2}+{\left(\frac{8-3}{2}\right)}^{2}}\\ & =& \sqrt{{\left(3\right)}^{2}+{\left(\frac{5}{2}\right)}^{2}}\\ & =& \sqrt{9+\frac{25}{4}}\\ & =& \sqrt{\frac{36+25}{4}}\\ & =& \sqrt{\frac{61}{4}}\\ & =& \frac{\sqrt{61}}{2}\end{array}$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 $SP$ $\begin{array}{rcl}& =& \sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\\ & =& \sqrt{{\left(2-\left(-1\right)\right)}^{2}+{\left(-1-\frac{3}{2}\right)}^{2}}\\ & =& \sqrt{{\left(2+1\right)}^{2}+{\left(\frac{-2-3}{2}\right)}^{2}}\\ & =& \sqrt{{\left(3\right)}^{2}+{\left(\frac{-5}{2}\right)}^{2}}\\ & =& \sqrt{9+\frac{25}{4}}\\ & =& \sqrt{\frac{36+25}{4}}\\ & =& \sqrt{\frac{61}{4}}\\ & =& \frac{\sqrt{61}}{2}\end{array}$Step $3:$ Finding the length of diagonals $PR$ and $SQ$:Length of $PR$$\begin{array}{rcl}& =& \sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\\ & =& \sqrt{{\left(5-\left(-1\right)\right)}^{2}+{\left(\frac{3}{2}-\frac{3}{2}\right)}^{2}}\\ & =& \sqrt{{\left(5+1\right)}^{2}+{\left(\frac{3-3}{2}\right)}^{2}}\\ & =& \sqrt{{\left(6\right)}^{2}+{\left(\frac{0}{2}\right)}^{2}}\\ & =& \sqrt{36+0}\\ & =& \sqrt{36}\\ & =& 6\end{array}$Length of $SQ$$\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-\left(-1\right)\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,$PR=6,SQ=5\phantom{\rule{0ex}{0ex}}⇒PR\ne SQ$Hence, $PQRS$ is a rhombus.

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