#### Chapter 5 Geometry

Section 5.4 Polygons, Area and Circumference

# 5.4.5 Area

The area of a surface equals the number of unit squares required to cover this surface completely.
 Let us first consider rectangles. If the sides of the rectangle are of lengths $a$ and $b$, then the rectangle contains $b$ rows with $a$ unit squares, i.e. $b·a$ unit squares.

##### Area of a Rectangle 5.4.10
The area $F$ of a rectangle with sides of lengths $a$ and $b$ is

$F=b·a=a·b .$

 With this, the area of a right triangle can be calculated easily. Let $ABC$ be a right triangle rotated by an angle of $180{}^{\circ }$. If the original and the rotated triangle are merged along the hypotenuse, one obtains a rectangle.
The area of the right triangle is then half the area of the rectangle, i.e. $F=\frac{1}{2}·a·b$.
And how is the area calculated if the triangle is not right-angled?
Every triangle can be divided into two right triangles by drawing a line from one vertex to the opposite side such that this line is perpendicular to the side. This line is called the altitude ${h}_{i}$ of a triangle on a specific side $i$, where $i$ is the index of the side $a$, $b$, or $c$.
Depending on whether the new line is interior or exterior to the triangle, the area of the triangle equals the sum or the difference of the areas of the two resulting right triangles:

Thus, on the left, we have (if ${F}_{\Delta }$ is the area of the triangle $\Delta$)

${F}_{ABC}={F}_{DBC}+{F}_{ADC}=\frac{1}{2}·{h}_{c}·{c}_{2}+\frac{1}{2}·{h}_{c}·{c}_{1}=\frac{1}{2}·{h}_{c}·\left({c}_{2}+{c}_{1}\right)=\frac{1}{2}·{h}_{c}·c .$

On the right, we have

${F}_{UVW}={F}_{XVW}-{F}_{XUW}=\frac{1}{2}·{h}_{w}·{w}_{2}-\frac{1}{2}·{h}_{w}·{w}_{1}=\frac{1}{2}·{h}_{w}·\left({w}_{2}-{w}_{1}\right)=\frac{1}{2}·{h}_{w}·w .$

Thus the area can always be calculated from the length of one side and the length of the altitude perpendicular to the corresponding side.
##### Area of a Triangle 5.4.11

The area ${F}_{ABC}$ of a triangle equals half the product of the length of a side and the length of the corresponding altitude of the triangle:

${F}_{ABC}=\frac{1}{2}·a·{h}_{a}=\frac{1}{2}·b·{h}_{b}=\frac{1}{2}·c·{h}_{c} .$

Here, the altitude of a triangle on a side denotes the line segment from the vertex opposite the side to the line containing the side itself, perpendicular to this side.

##### Example 5.4.12

 For the triangle to the right, the altitude corresponding to the side of length $8.6$ is given. The given values are rounded numerical values. Hence, the area $F$ of the triangle is approximately $F=\frac{8.6·5.5}{2}=23.65 .$

##### Exercise 5.4.13
Calculate the area of the triangle below.

Using the formula for the area of triangles, areas of polygons can also be calculated. This is due to the fact that every polygon can be divided into triangles by adding diagonals to the polygon until all subareas are triangles. However, the considerations will remain restricted here to a few simple shapes. In the following example, the polygon can be divided into a triangle and a rectangle. As a result, the calculation will be particularly easy.
##### Example 5.4.14

 Consider the polygon to the right, namely a trapezoid. In this example, the polygon can be divided into a right triangle with the legs $\left(a-c\right)$ and $b$ and the hypotenuse $d$ as well as a rectangle with sides of length $b$ and $c$. Then, the area of the polygon is:

$F={F}_{\text{triangle}}+{F}_{\text{rectangle}}=\frac{1}{2}\left(a-c\right)·b+b·c=\frac{1}{2}\mathrm{ab}-\frac{1}{2}\mathrm{bc}+\mathrm{bc}=\frac{1}{2}\left(a+c\right)·b .$

##### Exercise 5.4.15

 Calculate the area of the parallelogram to the right for $a=4$ and $h=5$.

Finally, we will calculate the area of a circle. Info Box 5.2.6 introduced the number $\pi$ describing the ratio of the circumference of a circle to its radius. The formula for the area of the circle also involves $\pi$.
##### Area of a Circle 5.4.16
The area of a circle with radius $r$ is

$F=\pi ·{r}^{2} .$

##### Example 5.4.17
Let the area of a circle with radius $r=2$ be $12.566$. This fact can be used to calculate an approximate value of the number $\pi$: We have $F=\pi ·{r}^{2}$, hence $\pi =\frac{F}{{r}^{2}}$. Inserting the given values results in the approximate value

$\pi =\frac{F}{{r}^{2}}\approx \frac{12.566}{4}=3.1415 .$