General Polygons - Area

A polygon is a plane figure that is bounded by a finite chain of straight line segments closing in a loop to form a closed polygonal chain or circuit. These segments are called its edges or sides, and the points where two edges meet are the polygon's vertices or corners. The interior of the polygon is sometimes called its body. An $n$-gon is a polygon with $n$ sides; for example, a triangle is a 3-gon. A polygon is a 2-dimensional example of the more general polytope in any number of dimensions.

There are many ways to find the area of a polygon.

Find the area of a regular hexagon of side length $5$.

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SOLUTION

A regular hexagon is composed of $6$ congruent equilateral triangles. The area of an equilateral triangle is $\frac{\sqrt{3}}{4}x^2,$ where $x$ is the side length.

So the area of one equilateral triangle is $\frac{\sqrt{3}}{4}\big(5^2\big)=\frac{25}{4}\sqrt{3}$.

It follows that the area of the hexagon is $6\left(\frac{25}{4}\sqrt{3}\right)=\frac{75}{2}\sqrt{3}.$ $_\square$

Find the area of an irregular decagon having consecutive vertices as $(1,-3),(1,-1),(10,-3),(13,4),(2,4),(2,1),(-4,4),(-10,2),(-11,0),(-7,-3).$

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SOLUTION

We can use the formula

$A=\dfrac{1}{2} \begin{vmatrix} x_1 & x_2 & ... & x_n & x_1 \\ y_1 & y_2 & ... & y_n & y_1 \end{vmatrix},$

where $A$ is half the determinant of the matrix.

We then have

$\begin{aligned} A&=\dfrac{1}{2} \begin{vmatrix} 1 & 1 & 10 & 13 & 2 & 2 & -4 & -10 & -11 & -7 & 1 \\ -3 & -1 & -3 & 4 & 4 & 1 & 4 & 2 & 0 & -3 & -3 \end{vmatrix}\\\\ &=\dfrac{1}{2}\big[-1-3+40+52+2+8-8+0+33+21-(-3-10-39+8+8-4-40-22-0-3)\big]\\\\ &=\dfrac{1}{2}\big[144-(-105)\big]\\\\ &=\dfrac{1}{2}(249)\\\\ &=124.5.\ _\square \end{aligned}$

Note that the area of a convex polygon is defined to be positive if the points are arranged in counterclockwise order, and negative if they are in clockwise order (Beyer 1987).