In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig
Question.

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


In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig.

Solution:

Let the number of horizontal columns represent the x-coordinates whereas the vertical rows represent the y-coordinates.

$\therefore$ The points are : $\mathrm{A}(3,4), \mathrm{B}(6,7), \mathrm{C}(9,4)$ and $\mathrm{D}(6,1)$

$\therefore \quad A B=\sqrt{(6-3)^{2}+(7-4)^{2}}$

$=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

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

$=\sqrt{3^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

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

$=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

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

$=\sqrt{(3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

Since, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}$ i.e., All the four sides are equal

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

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

$B D=\sqrt{(6-6)^{2}+(1-7)^{2}}=\sqrt{(0)^{2}+(-6)^{2}}=6$

i.e., $\mathrm{BD}=\mathrm{AC}$

$\Rightarrow$ Both the diagonals are also equal.

$\therefore \quad \mathrm{ABCD}$ is a square.

Thus, Chameli is correct as $\mathrm{ABCD}$ is not a rectangle.
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