In a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Given:

In a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.

To do:

We have to find which of them is correct.

Solution:

Let the points be $A (3, 4), B (6, 7), C(9, 4)$ and $D (6, 1)$

We know that,

The distance between two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Therefore,

The distance between the points $A(3, 4)$ and $B(6, 7)$

$AB=\sqrt{(6-3)^{2}+(7-4)^{2}}$

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt2$

The distance between the points $B(6, 7)$ and $C(9, 4)$

$BC=\sqrt{(9-6)^{2}+(4-7)^{2}}$

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt2$

The distance between the points $C(9, 4)$ and $D(6, 1)$

$CD=\sqrt{(6-9)^{2}+(1-4)^{2}}$

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt2$

The distance between the points $A(3, 4)$ and $D(6, 1)$

$AD=\sqrt{(6-3)^{2}+(1-4)^{2}}$

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt2$

The distance between the points $A(3, 4)$ and $C(9, 4)$

$AC=\sqrt{(9-3)^{2}+(4-4)^{2}}$

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

$=\sqrt{36}$

$=6$

The distance between the points $B(6, 7)$ and $D(6, 1)$

$BD=\sqrt{(6-6)^{2}+(1-7)^{2}}$

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

$=\sqrt{36}$

$=6$

Here, $AB = BC = CD = DA$ and $AC = BD$

This implies,

$ABCD$ is a square.

Hence, Champa is correct.

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Updated on: 10-Oct-2022

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