Perimeter of a regular

If the polygon has $n$ equal sides each of length $S$, then the perimeter $P$ is:

$P = n \times S = 2nr \sin\left(\frac{180^\circ}{n}\right)$

If we let $2r = d$, where $d$ is the diameter of the circle, then:

$P = nd \sin\left(\frac{180^\circ}{n}\right)$

Example 1

Find the length of one side of an eight-sided regular polygon inscribed in a circle with radius 7 cm.

Solution:

We use the formula:

$S = 2r \sin\left(\frac{180^\circ}{n}\right)$

Substitute $r = 7$ and $n = 8$:

$S = 2 \times 7 \times \sin\left(\frac{180^\circ}{8}\right) = 14 \times \sin(22.5^\circ)$

$S \approx 14 \times 0.3827 = 5.358 \text{ cm}$

Therefore, the length of one side of the eight-sided regular polygon is approximately 5.358 cm.

To find the area of a regular polygon with $n$ sides inscribed in a circle of radius $r$, we consider that connecting each vertex to the center $O$ divides the polygon into $n$ identical isosceles triangles.

Each triangle has a central angle of $\frac{360^\circ}{n}$, and the area of one such triangle is:

$\text{Area of one triangle} = \frac{1}{2} r^2 \sin\left(\frac{360^\circ}{n}\right)$

Since there are $n$ such triangles in the polygon:

$\text{Area of polygon} = \frac{n}{2} r^2 \sin\left(\frac{360^\circ}{n}\right)$

Therefore, the formula to calculate the area of a regular polygon is:

$\boxed{\text{Area} = \frac{n}{2} r^2 \sin\left(\frac{360^\circ}{n}\right)}$

Example 3

Find the area of a twelve-sided regular polygon inscribed in a circle of radius 14 cm.

Solution:

We use the formula for the area of a regular polygon:

$\text{Area} = \frac{n}{2} r^2 \sin\left(\frac{360^\circ}{n}\right)$

Substitute $n = 12$ and $r = 14$:

$\text{Area} = \frac{12}{2} \times 14^2 \times \sin(30^\circ)$

$= 6 \times 196 \times 0.5 = 588 \text{ cm}^2$

Therefore, the area of the polygon is 588 cm².

The circumference of a circle is the distance around it. The circumference can be estimated by using a regular polygon with many sides inscribed in a circle with radius $r$.

We know the perimeter of the regular polygon is given by:

$P = nS = n \times 2r \sin\left(\frac{180^\circ}{n}\right)$

As $n$ increases, the value of $n \sin\left(\frac{180^\circ}{n}\right)$ approaches the value of $\pi$.

When $n$ is very large, the perimeter of the regular polygon approaches the circumference of the circle. Therefore, the value of $n \sin\left(\frac{180^\circ}{n}\right)$ can be replaced by $\pi$, because it approaches $\pi$ when $n$ is very large.

Thus, the circumference $C$ of the circle is given by:

$C = 2\pi r$

Area of a circle

In a similar way, we can derive the formula for calculating the area of a circle by considering the area of a regular polygon inscribed in a circle of radius $r$.

We know that the area of a regular polygon inscribed in a circle is given by:

$\text{Area of polygon} = \frac{n}{2} r^2 \sin\left(\frac{360^\circ}{n}\right)$

As $n$ increases, the number of sides of the polygon increases, and the area approaches the area of the circle. Therefore, we can write the area of the circle as:

$\text{Area of circle} = \pi r^2$

Example 4

Find the circumference of a circle of radius 21 cm. (take $\pi = 3.14$).

Solution:

The formula for the circumference of a circle is:

$C = 2\pi r$

Substituting the values for $\pi$ and $r$:

$C = 2 \times 3.14 \times 21 \, \text{cm} = 131.88 \, \text{cm}$

The ratio of areas of similar polygons

Let $ABC$ and $A'B'C'$ be two similar triangles. If we find the ratio of their sides, we get:

Generally, if the ratio of the lengths of the corresponding sides of two similar polygons is $k$, then the ratio of their areas is $k^2$.

Example 5

We are given two triangles that are similar. The length of one side of the smaller triangle is 8 cm, and the length of the corresponding side of the larger triangle is 14 cm. If the area of the smaller triangle is 24 cm², find the area of the larger triangle.

Solution:

The ratio of the corresponding sides of the two similar triangles is:

$k = \frac{14}{8} = \frac{7}{4}$

Since the ratio of the areas of two similar polygons is the square of the ratio of their corresponding sides, we can calculate the area of the larger triangle as follows:

$\text{Area of larger triangle} = 24 \times \left( \frac{7}{4} \right)^2 = 24 \times \frac{49}{16} = 73.5 \, \text{cm}^2$