# $ABCD$ 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.

#### Complete Python Prime Pack for 2023

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack 2023

9 Courses     2 eBooks

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.

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