$ABCD$ 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.

AcademicMathematicsNCERTClass 10

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 the mid-point formula, we get,

The coordinates of $P$ are $(\frac{-2}{2}, \frac{3}{2})$

$=(-1, \frac{3}{2})$

Similarly,

The coordinates of $\mathrm{Q}$ are $(\frac{-1+5}{2}, \frac{4+4}{2})$

$=(\frac{4}{2}, \frac{8}{2})$

$=(2,4)$

The coordinates of $\mathrm{R}$ are $(\frac{5+5}{2}, \frac{4-1}{2})$

$=(\frac{10}{2}, \frac{3}{2})$

$=(5, \frac{3}{2})$

The coordinates of $\mathrm{S}$ are $(\frac{5-1}{2}, \frac{-1-1}{2})$

$=(\frac{4}{2}, \frac{-2}{2})$

$=(2,-1)$

Using distance formula, we get, $\mathrm{PQ}=\sqrt{(2+1)^{2}+(4-\frac{3} {2})^{2}}$

$=\sqrt{(3)^{2}+(\frac{5}{2})^{2}}$

$=\sqrt{9+\frac{25}{4}}$

$=\sqrt{\frac{36+25}{4}}$

$=\sqrt{\frac{61}{4}}$

$=\frac{\sqrt{61}}{2}$

$\mathrm{QR}=\sqrt{(5-2)^{2}+(\frac{3}{2}-4)^{2}}$ $=\sqrt{(3)^{2}+(\frac{-5}{2})^{2}}$

$=\sqrt{9+\frac{25}{4}}$

$=\sqrt{\frac{36+25}{4}}$

$=\sqrt{\frac{61}{4}}$

$=\frac{\sqrt{61}}{2}$

$O$ is the mid-point of $PR$. Coordinates of $\mathrm{O}$ are $(\frac{-1+5}{2}, \frac{\frac{3}{2}+\frac{3}{2}}{2})$

$=(\frac{4}{2}, \frac{3}{2})$

$=(2, \frac{3}{2})$

Similarly,

$\mathrm{O}$ is the mid-point of $\mathrm{QS}$,

Coordinates of $\mathrm{O}$ are $(\frac{2+2}{2}, \frac{4+(-1)}{2})$

$=(\frac{4}{2}, \frac{3}{2})$

$=(2, \frac{3}{2})$

We see that the coordinates of $\mathrm{O}$ in both the cases is same and adjacent sides are also equal.

This implies it may be a square or a rhombus.

$\mathrm{PR}=\sqrt{(5+1)^{2}+(\frac{3}{2}-\frac{3}{2})^{2}}$

$=\sqrt{(6)^{2}+(0)^{2}}$

$=\sqrt{36+0}$

$=\sqrt{36}$

$=6$

$\mathrm{QS}=\sqrt{(2-2)^{2}+(-1-4)^{2}}$

$=\sqrt{(0)^{2}+(-5)^{2}}$

$=\sqrt{0+25}$

$=\sqrt{25}$

$=5$

Here, diagonals are not equal. 

Therefore, PQRS is a rhombus.

raja
Updated on 10-Oct-2022 13:22:10

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