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Question
Show that the quadrilateral with vertices (3, 2), (0, 5), (- 3, 2) and (0, -1) is a square.
Solution
Let A(3, 2), B(0, 5), C(-3, 2) and D(0, -1) are the verticles of quadrilateral.
Now AB = `sqrt((3 - 0)^2 + (2 - 5)^2)`
= `sqrt(18)"units"`.
⇒ AB2 = 18
BC = `sqrt((0 + 3)^2 + (5 - 2)^2)`
= `sqrt(18)"units"`.
⇒ BC2 = 18
CD = `sqrt((-3 - 0)^2 + (2 + 1)^2)`
= `sqrt(18)"units"`.
⇒ CD2 = 18
Also AD = `sqrt((3 - 0)^2 + (2 + 1)^2)`
= `sqrt(18)"units"`.
⇒ AD2 = 18.
Here AB = BC = CD = DA
= `sqrt(18)"units"`.
Also
AC2 = (3 + 3)2 + (2 - 2)2
AC2 = 36
or
BD2 = (0 - 0)2 + (5 + 1)2
= 36
⇒ Diagonal AC = BD
Hence, ABCD is a square.
Hence proved.
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