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$ABCD$ is a rectangle formed by joining 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 sides $AB, BC, CD$ and $DA$ respectively. Is the quadrilateral $PQRS$ a square? a rectangle? or a rhombus? Justify your answer.
Given:
$ABCD$ is a rectangle formed by joining 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 sides $AB, BC, CD$ and $DA$ respectively.
To do:
We have to determine whether $PQRS$ is a square or a rectangle or a rhombus.
Solution:
Join $PR$ and $QS$. Let the point of intersection of $PR$ and QS$ be $O$.
Using mid-point formula, we get,
The coordinates of $P$ are \( \left(\frac{-2}{2}, \frac{3}{2}\right) \)
\( =\left(-1, \frac{3}{2}\right) \)
Similarly,
The coordinates of \( \mathrm{Q} \) are \( \left(\frac{-1+5}{2}, \frac{4+4}{2}\right) \)
\( =\left(\frac{4}{2}, \frac{8}{2}\right) \)
\( =(2,4) \)
The coordinates of \( \mathrm{R} \) are \( \left(\frac{5+5}{2}, \frac{4-1}{2}\right) \)
\( =\left(\frac{10}{2}, \frac{3}{2}\right) \)
\( =(5, \frac{3}{2}) \)
The coordinates of \( \mathrm{S} \) are \( \left(\frac{5-1}{2}, \frac{-1-1}{2}\right) \)
\( =\left(\frac{4}{2}, \frac{-2}{2}\right) \)
\( =(2,-1) \)
Using distance formula, we get,
\( \mathrm{PQ}=\sqrt{(2+1)^{2}+\left(4-\frac{3} {2}\right)^{2}} \)
\( =\sqrt{(3)^{2}+\left(\frac{5}{2}\right)^{2}} \)
\( =\sqrt{9+\frac{25}{4}} \)
\( =\sqrt{\frac{36+25}{4}} \)
\( =\sqrt{\frac{61}{4}} \)
\( =\frac{\sqrt{61}}{2} \)
\( \mathrm{QR}=\sqrt{(5-2)^{2}+\left(\frac{3}{2}-4\right)^{2}} \)
\( =\sqrt{(3)^{2}+\left(\frac{-5}{2}\right)^{2}} \)
\( =\sqrt{9+\frac{25}{4}} \)
\( =\sqrt{\frac{36+25}{4}} \)
\( =\sqrt{\frac{61}{4}} \)
\( =\frac{\sqrt{61}}{2} \)
\( O \) is the mid-point of \( P R \).
Coordinates of \( \mathrm{O} \) are \( \left(\frac{-1+5}{2},\left(\frac{3}{2}+\frac{3}{2}\right) \frac{1}{2}\right) \)
\( =\left(\frac{4}{2}, \frac{3}{2}\right) \)
\( =\left(2, \frac{3}{2}\right) \)
Similarly,
\( \mathrm{O} \) is the mid-point of \( \mathrm{QS} \),
Coordinates of \( \mathrm{O} \) are \( \left(\frac{2+2}{2}, \frac{4+(-1)}{2}\right) \)
\( =\left(\frac{4}{2}, \frac{3}{2}\right) \)
\( =\left(2, \frac{3}{2}\right) \)
We see that the coordinates of \( \mathrm{O} \) in both the cases is same and adjacents sides are also equal.
This implies, it may be a square or a rhombus.
\( \mathrm{PR}=\sqrt{(5+1)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}} \)
\( =\sqrt{(6)^{2}+(0)^{2}} \)
\( =\sqrt{36+0} \)
\( =\sqrt{36} \)
\( =6 \)
\( \mathrm{QS}=\sqrt{(2-2)^{2}+(-1-4)^{2}} \)
\( =\sqrt{(0)^{2}+(-5)^{2}} \)
\( =\sqrt{0+25} \)
\( =\sqrt{25} \)
\( =5 \)
Here, diagonals are not equal.
Therefore, PQRS is a rhombus.
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