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, Dont you think $ABCD$ is a square? Chameli disagrees. Using distance formula, find which of them is correct. why?

Answer 1

According to the fig. $A(3,4$),$B(6,7)$,$C(9,4)$ and $D(6,1)$ are the position of $4$ friends .

If we join all these points then

By using distance formula

$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{(3-6{)}^2+(4-7{)}^2}=\sqrt{(-3{)}^2+(-3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$BC=\sqrt{(6-9{)}^2+(7-4{)}^2}=\sqrt{(-3{)}^2+(3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$CB=\sqrt{(9-6{)}^2+(4-1{)}^2}=\sqrt{(3{)}^2+(3^2)}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$AD=\sqrt{(3-6{)}^3+(4-1{)}^2}=\sqrt{-3{)}^2+(3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

if we join the point $AC$ and $BD$ than

Diagonal$AC=\sqrt{(3-9{)}^2+(4-4{)}^2}=\sqrt{(-6{)}^2+(0{)}^2}=\sqrt{36}=6$

Diagonal $BD=\sqrt{(6-6{)}^2+(7-1{)}^2}=\sqrt{(0{)}^2+(6{)}^2}=\sqrt{36}=6$

Hence all the sides of a quadrilateral are the same and the diagonals are also the same so $ABCD$ is a square.

Hence Champa is correct.