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# In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). 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.

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get

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

Therefore, All the sides of ABCD are equal here. … (1)

Now, we will check the length of its diagonals.

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

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

So, Diagonals of ABCD are also equal. … (2)

From (1) and (2), we can definitely say that ABCD is a square.

Therefore, Champa is correct.