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(A) Let the vertices be $A(2,2), B(5,2), C(5,5)$ and $D(2,5)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB^2 = (5-2)^2 + (2-2

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(A) Let the vertices be $A(2,2), B(5,2), C(5,5)$ and $D(2,5)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB^2 = (5-2)^2 + (2-2)^2 = 3^2 + 0^2 = 9 \implies AB = 3$.
$BC^2 = (5-5)^2 + (5-2)^2 = 0^2 + 3^2 = 9 \implies BC = 3$.
$CD^2 = (2-5)^2 + (5-5)^2 = (-3)^2 + 0^2 = 9 \implies CD = 3$.
$DA^2 = (2-2)^2 + (2-5)^2 = 0^2 + (-3)^2 = 9 \implies DA = 3$.
Since $AB = BC = CD = DA = 3$,the quadrilateral is a rhombus.
Now,check the diagonals:
$AC^2 = (5-2)^2 + (5-2)^2 = 3^2 + 3^2 = 9 + 9 = 18 \implies AC = \sqrt{18} = 3\sqrt{2}$.
$BD^2 = (2-5)^2 + (5-2)^2 = (-3)^2 + 3^2 = 9 + 9 = 18 \implies BD = \sqrt{18} = 3\sqrt{2}$.
Since all sides are equal and the diagonals are equal $(AC = BD)$,the quadrilateral $ABCD$ is a square.

Show that $(2,2), (5,2), (5,5)$ and $(2,5)$ are the vertices of a square.

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