In the coordinate plane, one of the vertices of a square is the point \((-3, -4).\) If the diagonals of that square intersect at point \((3, 2),\) what is the area of that square?
A. 36
B. 72
C. 108
D. 144
E. 180
[spoiler]OA=D[/spoiler]
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In the coordinate plane, one of the vertices of a square is the point \((-3, -4).\) If the diagonals of that square
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orthodoxparadox
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Let the length of the side of the square be s.
Note that diagonals of a square intersect at the center of the square.
Therefore, by Pythagoras theorem (s/2) $$\left(\frac{s}{2}\right)^2\ +\ \left(\frac{s}{2}\right)^2$$ = square of distance between vertex and center.
So $$\left(\frac{s}{2}\right)^2\ +\ \left(\frac{s}{2}\right)^2\ =\ \left(3\ -\ \left(-3\right)\right)^2\ +\ \left(2\ -\ \left(-4\right)\right)^2\ =\ 72$$
$$\left(\frac{s}{2}\right)^2\ =\ 36$$
$$\left(\frac{s}{2}\right)\ =\ 6$$
$$s\ =\ 12$$
Area = $$s\ \cdot\ s\ =\ 144$$ D
Note that diagonals of a square intersect at the center of the square.
Therefore, by Pythagoras theorem (s/2) $$\left(\frac{s}{2}\right)^2\ +\ \left(\frac{s}{2}\right)^2$$ = square of distance between vertex and center.
So $$\left(\frac{s}{2}\right)^2\ +\ \left(\frac{s}{2}\right)^2\ =\ \left(3\ -\ \left(-3\right)\right)^2\ +\ \left(2\ -\ \left(-4\right)\right)^2\ =\ 72$$
$$\left(\frac{s}{2}\right)^2\ =\ 36$$
$$\left(\frac{s}{2}\right)\ =\ 6$$
$$s\ =\ 12$$
Area = $$s\ \cdot\ s\ =\ 144$$ D
