Ideas in Geometry/Area

3.2.1 Areas

There are simple equations to find the area of common shapes such as the triangle or parallelogram.

Heron's Formula

We learn at a young age that the area of a triangle can be expressed by the equation $\frac{1}{2}$ base $\cdot$ height. This uses two sides of a triangle on either side of an angle but sometimes it can be difficult to find the height of a triangle that is perpendicular to the base, creating a $90^{\circ }$ angle, which is what you need for the usual formula for the area of a triangle. But there is also a way to find the area of a triangle using the lengths of all three sides. This can be expressed in Heron's Formula where the area can be found using the equation $area=\sqrt{(\frac{p}{2}-a)(\frac{p}{2}-b)(\frac{p}{2}-c)\left(\frac{p}{2}\right)}$ where a,b &c represent the sides of the triangle and p=a+b+c, the perimeter of the triangle.

Here is an example to show what we mean:

To find the area of this triangle, we would use the equation with the sides lengths: 3,6 & 7 and find the area --> p=16 $area=\sqrt{(\frac{16}{2}-3)(\frac{16}{2}-6)(\frac{16}{2}-7)\left(\frac{16}{2}\right)}$ and we would get the answer $area=8.94unit{s}^2$

A Quadrilateral Circumscribed in a Circle

If we know that there is a formula that works for triangles given the three lengths of the sides, we can find a formula in a similar way for the area of a quadrilateral. As long as the quadrilateral can be circumscribed in a circle, which means each vertex touches the inside of the circle and the opposite angles must sum to $180^{\circ }$, the area of the quadrilateral can be solved. See picture below of a quadrilateral circumscribed in a circle.

File:Cyclic quad.png

Brahmagupta's Formula

In this formula, Brahmagupta's Formula, if given a quadrilateral that can be circumscribed in a circle (also known as cyclic), the area of the quadrilateral can be expressed by the equation: $area=\sqrt{(\frac{p}{2}-a)(\frac{p}{2}-b)(\frac{p}{2}-c)(\frac{p}{2}-d)}$, where a,b,c & d represent the sides of the quadrilateral and p=a+b+c+d, the perimeter of the quadrilateral.

Here is an example to show what we mean:

File:Cyclic quad2.tif

Using these side lengths: 4, 6, 6, & 7, we would use the formula to find the area --> p=23 $area=\sqrt{(\frac{23}{2}-4)(\frac{23}{2}-6)(\frac{23}{2}-6)(\frac{23}{2}-7)}$ and get the answer to be $area=31.95unit{s}^2$

Lattice Points

There is also a way of finding the area of shapes besides triangles and quadrilaterals that involves lattice points. Lattice points are points that are spaced 1 unit apart, horizontally and vertically in a plane. Here are some lattice points:

                                                      File:Lattice points-1.png

Pick's Theorem

If the vertices of a polygon are located on the lattice points, we can find the area of the polygon by using Pick's Theorem: $area=\frac{b}{2}+n-1$ where b=the number of lattice points on the border of the polygon and n=the number of lattice points on the inside of the polygon. Here is an example:

                                           File:Math img011.jpg

Since the vertices of the polygon lie on the lattice points, we can use Pick's Theorem. So in this example: b=15, n=14. And $area=\frac{15}{2}+14-1$. So the area=20.5 square units.