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Answer :
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.