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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.