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Consider a quadrilateral ABCD whose vertices are A (x_1, y_1), B (x_2, y_2), C (x_3, y_3) and D (x_4, y_4) taken in order.
Draw a diagonal BD and split the quadrilateral into two triangles ABD and BCD.
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Therefore, we have:
Area of the quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD.
We know that the area of a triangle having its vertices (x_1, y_1), (x_2, y_2), (x_3, y_3)is
given by:
Area of the triangle = \frac{1}{2}\left[ \left(x_1y_2 + x_2y_3 + x_3y_1\right) - \left(x_2y_1 + x_3y_2 + x_1y_3\right)\right] square units.
Therefore, Area of the quadrilateral ABCD = \frac{1}{2}\left[ \left(x_1y_2 + x_2y_4 + x_4y_1\right) - \left(x_2y_1 + x_4y_2 + x_1y_4\right)\right] + \frac{1}{2}\left[ \left(x_2y_3 + x_3y_4 + x_4y_2\right) - \left(x_3y_2 + x_4y_3 + x_2y_4\right)\right]
Factor out the common terms and rearrange the terms.
Thus, we have:
Area of the quadrilateral ABCD = \frac{1}{2}\left[\left(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1\right) - \left(x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4\right)\right]
The formula can be easily remembered using the following pictorial representation.
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Another form of the formula:
Area of the quadrilateral ABCD = \frac{1}{2}\left[\left(x_1 - x_3\right)\left(y_2 - y_4\right) - \left(x_2 - x_4\right)\left(y_1 - y_3\right)\right] square units.
Example:
Find the area of the quadrilateral joining the vertices (- 3, 4), (- 5, - 6), (4, - 1) and (1, 2).
Solution:
By the formula, we have:
Area of the quadrilateral = \frac{1}{2}\left[\left(x_1 - x_3\right)\left(y_2 - y_4\right) - \left(x_2 - x_4\right)\left(y_1 - y_3\right)\right] square units
Substitute the required values in the above formula.
Area of the quadrilateral = \frac{1}{2}\left[\left(-3 - 4\right)\left(-6 - 2\right) - \left(-5 - 1\right)\left(4 - (-1)\right)\right]
= \frac{1}{2}\left[(-7)(- 8) - (-6)(4)\right]
= \frac{1}{2}\left[56 + 24\right]
= \frac{80}{2}
= 40
Therefore, area of the quadrilateral joining the vertices (- 3, 4), (- 5, - 6), 4, - 1) and (1, 2) is 40 square units
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