Area of a Quadrilateral

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Consider the following four points:

\[\begin{array}{l}A\left( { - 3,\;1} \right), & B\left( { - 1,\;4} \right)\\C\left( {3,\;2} \right), & D\left( {1,\; - 2} \right)\end{array}\]

These four points are the vertices of a quadrilateral:

Vertices of a quadrilateral

How do we calculate the area of this quadrilateral using the coordinates of the four vertices? The answer is simple. We divide the quadrilateral in two triangles (using either of the diagonals), calculate the (positive value of) the areas of each triangle, and add these values to obtain the total area. In the following figure, quadrilateral ABCD has been divided into \({\rm{\Delta ABD}}\) and  \({\rm{\Delta ADC}}\).

Two triangles make a quadrilateral

Now, we separately calculate the areas of the two triangles.

Area of Triangle ABC

\[\begin{align}&{\rm{Area}}\left( {\Delta ABC} \right) = \frac{1}{2}\;\left| {\,\begin{gathered}{ - 3}&{ - 1}&3\\1&4&2\\1&1&1\end{gathered}\,} \right|\\ &= \frac{1}{2}\;\left| { - 3 \times \left( {4 - 2} \right) + \left( { - 1} \right) \times \left( {2 - 1} \right) + 3 \times \left( {1 - 4} \right)} \right|\\ & = \frac{1}{2}\;\left| { - 6 - 1 - 9} \right| = \frac{1}{2} \times 16 = 8\;{\rm{sq}}{\rm{.}}\;{\rm{units}}\end{align}\]

Area of Triangle ACD

\[\begin{align}&{\rm{Area}}\;\left( {\Delta ADC} \right) = \frac{1}{2}\;\left| {\,\begin{array}{*{20}{c}}{ - 3}&1&3\\1&{ - 2}&2\\1&1&1\end{array}\,} \right|\\&= \frac{1}{2}\left| { - 3 \times \left( { - 2 - 2} \right) + 1 \times \left( {2 - 1} \right) + 3 \times \left( {1 - \left( { - 2} \right)} \right)} \right|\\& = \frac{1}{2}\;\left| {12 + 1 + 9} \right| = \frac{1}{2} \times 22 = 11\,\,{\rm{sq}}{\rm{.}}\;{\rm{units}}\end{align}\]

Area of Quadrilateral ABCD

\[\begin{array}{l}{\rm{Area}}\left( {ABCD} \right)\\ = {\rm{Area}}\;\left( {\Delta ABC} \right) + {\rm{Area}}\;\left( {\Delta ADC} \right)\\ = 8 + 11 = 19\;{\rm{sq}}{\rm{.}}\;{\rm{units}}\end{array}\]

Example 1: A farmer has a pentagonal field ABCDE, which he divides into three triangular plots, as shown below:

Three triangles make a pentagon

Arrange the plots in the order of increasing areas.

Solution: We calculate the area of each plot separately:

Area of Plot I

\[\begin{align}&{\rm{Area}}\left( {\Delta ABC} \right) = \frac{1}{2}\;\left| {\,\begin{array}{*{20}{c}}{ - 3}&{ - 2}&2\\1&5&3\\1&1&1\end{array}\,} \right|\\&= \frac{1}{2}\;\left| { - 3\left( {5 - 3} \right) + \left( { - 2} \right)\left( {3 - 1} \right) + 2\left( {1 - 5} \right)} \right|\\ &= \frac{1}{2}\;\left| { - 6 - 4 - 8} \right| = \frac{1}{2} \times 18 = 9\;{\rm{sq}}{\rm{.}}\;{\rm{units}}\end{align}\]

Area of Plot II

\[\begin{align}&{\rm{Area}}\left( {\Delta ACD} \right) = \frac{1}{2}\;\left| {\,\begin{array}{*{20}{c}}{ - 3}&2&3\\1&3&{ - 1}\\1&1&1\end{array}\,} \right|\\& = \frac{1}{2}\;\left| { - 3\left( {3 - \left( { - 1} \right)} \right) + 2\left( { - 1 - 1} \right) + 3\left( {1 - 3} \right)} \right|\\&= \frac{1}{2}\;\left| { - 12 - 4 - 6} \right| = \frac{1}{2} \times 22 = 11\;{\rm{sq}}{\rm{.}}\;{\rm{units}}\end{align}\]

Area of Plot III

\[\begin{align}&{\rm{Area}}\left( {\Delta ADE} \right) = \frac{1}{2}\;\left| {\,\begin{array}{*{20}{c}}{ - 3}&3&{ - 1}\\1&{ - 1}&{ - 3}\\1&1&1\end{array}\,} \right|\\& = \frac{1}{2}\;\left| { - 3\left( { - 1 - \left( { - 3} \right)} \right) + 3\left( { - 3 - 1} \right) + \left( { - 1} \right)\left( {1 - \left( { - 1} \right)} \right)} \right|\\&= \frac{1}{2}\;\left| { - 6 - 12 - 2} \right| = \frac{1}{2} \times 20 = 10\;{\rm{sq}}{\rm{.}}\;{\rm{units}}\end{align}\]

Thus, the plots ordered by increasing areas will be: I, III and II.