# 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 ${\tfrac {1}{2}}$ 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 $area={\sqrt {\left({\frac {p}{2}}-a\right)\left({\frac {p}{2}}-b\right)\left({\frac {p}{2}}-c\right)\left({\frac {p}{2}}\right)}}$ 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 $area={\sqrt {\left({\frac {16}{2}}-3\right)\left({\frac {16}{2}}-6\right)\left({\frac {16}{2}}-7\right)\left({\frac {16}{2}}\right)}}$ and we would get the answer $area=8.94units^{2}$ ### 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.

### 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: $area={\sqrt {\left({\frac {p}{2}}-a\right)\left({\frac {p}{2}}-b\right)\left({\frac {p}{2}}-c\right)\left({\frac {p}{2}}-d\right)}}$ , 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:

Using these side lengths: 4, 6, 6, & 7, we would use the formula to find the area --> p=23 $area={\sqrt {\left({\frac {23}{2}}-4\right)\left({\frac {23}{2}}-6\right)\left({\frac {23}{2}}-6\right)\left({\frac {23}{2}}-7\right)}}$ and get the answer to be $area=31.95units^{2}$ ### 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: ### 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: $area={\tfrac {b}{2}}+n-1$ 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: 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 $area={\tfrac {15}{2}}+14-1$ . So the area=20.5 square units.