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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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 am; Thanked: 1 times; Followed by:5 members

In the coordinate plane, one of the vertices of a square is the point $(-3, -4).$ If the diagonals of that square

by M7MBA » Sat May 30, 2020 4:11 pm

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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 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]

Source: Veritas Prep

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Source: Beat The GMAT — Problem Solving | Sat May 30, 2020 4:11 pm

orthodoxparadox: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sat May 30, 2020 10:48 pm

Re: In the coordinate plane, one of the vertices of a square is the point $(-3, -4).$ If the diagonals of that square

by orthodoxparadox » Sat May 30, 2020 11:16 pm

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