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 mid-points of $AB$, $BC$, $CD$ and $DA$ respectively. Is the quadrilateral $PQRS$ a square? a rectangle? or a rhombus? Justify your answer.

Answer 1

Let $P,Q,R$ and $S$ are the midpoint of $AB,BC,CD$ and$DA$.

$\therefore$ Co-ordinates of $P=[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]=[\frac{-1-1}{2},\frac{-1+4}{2}]=[-1,\frac{3}{2}]$

$\therefore$ Co-ordinates of $Q=[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]=[\frac{-1+5}{2},\frac{4+4}{2}]=[2,4]$

$\therefore$ Co-ordinates of $R=[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]=[\frac{5+5}{2},\frac{4-1}{2}]=[5,\frac{3}{2}]$

$\therefore$ Co-ordinates of $S=[\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]=[\frac{5-1}{2},\frac{-1-1}{2}]=[4,-1]$

Now, length of $PQ=\sqrt{(-1-2{)}^2+(\frac{3}{2}-4{)}^2}=\sqrt{(-3{)}^2+(-\frac{5}{2}{)}^2}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$

Length of $QR=\sqrt{(2-5{)}^2+(4-\frac{3}{2}{)}^2}=\sqrt{(-3{)}^2+(\frac{5}{2}{)}^2}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$

Length of $RS=\sqrt{(5-2{)}^2+(\frac{3}{2}+1{)}^2}=\sqrt{(3{)}^2+(\frac{5}{2}{)}^2}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$

Length of $S=\sqrt{(2+1{)}^2+(-1-\frac{3}{2}{)}^2}=\sqrt{(3{)}^2+(-\frac{5}{2}{)}^2}=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}$

Length of diagonal $PR=\sqrt{(-1-5{)}^2+(\frac{3}{2}-\frac{3}{2}{)}^2}=\sqrt{\left(6^2\right)}=\sqrt{36}=6$

Length of diagonal QS=$\sqrt{(2-2{)}^2+(4+1{)}^2}=\sqrt{(5{)}^2}=\sqrt{25}=5$

Hence the all the sides of the quadrilateral $PQRS$ are equal but the diagonals are not equal then $PQRS$ is a rhombus.