Question

4 friends are seated at points A, B, C, and D in a classroom, as shown in the following figure. Amy walked into the class, and after observing for a few minutes, she thought, “ABCD must be a square”. State whether Amy’s statement is correct or not.

• A. No
• B. None of the above
• C. Yes
• D. Can't be determined

Answer 1

The correct option is C Yes


Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).

We know that, the distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is represented as:

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

So, the distance of all the points given on the graph, are:

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

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

$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$


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

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

$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$


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

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

$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$



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

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

$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$


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

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


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

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

It can be observed that all sides of the quadrilateral ABCD are of same length and also the diagonals have same length.


So, ABCD is a square and hence, Amy is correct.