SOLUTION: The coordinates of the vertices of a quadrilateral ABCD are A(-2,3), B(3,3), C(3,-2), and D(-2,-2). Find the area of quadrilateral ABCD?

Question 848387: The coordinates of the vertices of a quadrilateral ABCD are A(-2,3), B(3,3), C(3,-2), and D(-2,-2). Find the area of quadrilateral ABCD?
Answer by pmesler(52) (Show Source): You can put this solution on YOUR website!
To answer this question, the first thing you want to do is establish what type of quadrilateral ABCD is. Quadrilateral is a very broad term. ABCD could be a parallelogram, a rectangle, a square, or a rhombus. Each of these has their own formula to find the area. Given the coordinates in the problem, you need to use the distance formula to find out the different lengths of the quadrilateral.
If you graph the coordinates on a piece of graph paper, you can tell right away that it is a square. It's obviously a square if you look at the x and y coordinates. In quadrants I and IV, both x coordinates are 3 units away from zero. In quadrants I and II, both y coordinates are 3 units from zero. I think you get the picture. It's a square. Now let's find the distance between two of these coordinates. We only need to find one side, because a square's sides are all the same length. Let's measure the distance AB. First, the formula for distance given some coordinates is:
sqrroot((x2-x1)^2 + (y2-y1)^2).

Plugging in the values for AB into the formula we get

sqrroot((3-(-2))^2 + (3-3)^2).
= sqrroot ((3+2)^2 + 0^2)
= sqrroot (25) = 5. So the distance of all the sides is 5 units. Now, we simply apply the formula for the area of a square to find the area

A = s^2.
Therefore, for this square A = 5^2 = 25 square units.

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