Areas of Polygons

7.3

Areas of Polygons

General Formulas for Area Calculation

Shape	Formula
Triangle	\(\text{Area}=\dfrac{1}{2}\times \text{base}\times\text{height}\)
Rectangle	\(\text{Area}= \text{length}\times\text{width}\)
Square	\(\text{Area}= \text{side}\times\text{side}=\text{side}^2\)
Parallelogram	\(\text{Area}=\text{base}\times\text{height}\)
Trapezium (Trapezoid)	\(\text{Area}=\dfrac{1}{2}\times \text{(sum of parallel sides)}\times\text{height}\)
Regular Polygon (e.g., regular pentagon, hexagon)	\(\text{Area}=\dfrac{1}{2}\times \text{Perimeter}\times\text{Apothem}\)

Apothem is the distance from the center to the midpoint of a side.

Coordinate Geometry Method (Shoelace Theorem)

In general, when the coordinates of each vertex of a polygon are known, we can determine the area of the polygon using:
\(\text{Area}=\dfrac{1}{2} \begin{vmatrix} x _1&& x_2 &&x_3 & &... & &x_n& &x_1\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } \\ y_{1}&& y_2 & &y_3 && ... && y_n && y_1 \end{vmatrix}\)
\(\text{Area}=\dfrac{1}{2}\begin{vmatrix}(x_1y_2+x_2y_3+...+x_ny_1)-(x_2y_1+x_3y_2+...+x_1y_n) \end{vmatrix}\)

Example

Area of Triangle

Triangle with vertices at coordinates (x1, y1), (x2, y2), (x3, y3).

Area of \(\Delta{ABC}\):

\(=\dfrac{1}{2} \begin{vmatrix} x _1&& x_2 &&x_3 & &x_1\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }\\ y_{1}&& y_2 & &y_3 && y_1 \end{vmatrix}\)

\(=\dfrac{1}{2}|(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)|\)

Area of Quadrilateral

Quadrilateral plotted with coordinates of vertices included.

Area of quadrilateral \(PQRS\):

\(=\dfrac{1}{2} \begin{vmatrix} x _1&& x_2 &&x_3 & &x_4 & &x_1\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }\\ y_{1}&& y_2 & &y_3 && y_4 && y_1 \end{vmatrix}\)

\(=\dfrac{1}{2}\begin{vmatrix}(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4) \end{vmatrix}\)

Example Problem

Question

Find the area of the triangle with the vertices given:

\(A(-7,5),\, B(2,-4), \, C(4,3)\)

Solution

Based on the question,

\(A(-7,5),\, B(2,-4), \, C(4,3)\)

Then, the area of the triangle \(\Delta{ABC}\),

\(=\dfrac{1}{2} \begin{vmatrix} -7&& 2 &&4 & &-7\\ &\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }&&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} } &&\hspace{-0.3cm}\large{{\color{red}\searrow} \hspace{-0.4cm}{\nearrow} }\\ 5&& -4 & &3 && 5 \end{vmatrix}\)

\(\begin{aligned} &=\dfrac{1}{2}\begin{vmatrix}(28+6+20)-(10-16-21)\end{vmatrix} \\\\ &=\dfrac{1}{2}|54+27| \\\\ &=40.5 \text{ unit}^2. \end{aligned}\)