Question

What is the perimeter of a regular octagon with a radius of length 20?

Answer 1

Hint: Here in this question, we have to find the perimeter of a regular octagon of given radius of length $$r=20$$ . First, we have to find the length $$l$$ of each side of octagon using a distance formula $$l=\sqrt{{\left(y_2-y_1\right)}^2+{\left(x_2-x_1\right)}^2}$$ and later find a perimeter using a formula $$P=8\times l$$ . If we are finding the perimeter of a regular octagon, then we know that all eight sides are equal lengths, so we can simplify the formula using multiplication operation to get the required solution.

Complete step-by-step answer:
In geometry, perimeter can be defined as the path or the boundary that surrounds a shape. It can also be defined as the length of the outline of a shape.
If a octagon is regular, then all the sides are equal in length, and eight angles are of equal measures
Consider a regular octagon having radius of length $$r=20$$ , which is same for vertices of pentagons.
In figure, the red circle circumscribes the outer radius and the green circle the inner one.

Let consider $$r=20$$ be the outer radius - that is the radius of the red circle.
Then, the vertices of the octagon centred at origin i.e., $$\left(0,0\right)$$ are at $$\left(\pm r,0\right)$$ , $$\left(0,\pm r\right)$$ and $$\left(\pm {\displaystyle \frac{r}{\sqrt{2}}},\pm {\displaystyle \frac{r}{\sqrt{2}}}\right)$$ .
So the length of one side of regular octagon is distance between $$\left(r,0\right)$$ and $$\left({\displaystyle \frac{r}{\sqrt{2}}},{\displaystyle \frac{r}{\sqrt{2}}}\right)$$
Let consider a distance formula $$l=\sqrt{{\left(y_2-y_1\right)}^2+{\left(x_2-x_1\right)}^2}$$ , on substituting we have
 $$\Rightarrow l=\sqrt{{\left({\displaystyle \frac{r}{\sqrt{2}}}-0\right)}^2+{\left({\displaystyle \frac{r}{\sqrt{2}}}-r\right)}^2}$$
 $$\Rightarrow l=\sqrt{{\left({\displaystyle \frac{r}{\sqrt{2}}}\right)}^2+{\left({\displaystyle \frac{r}{\sqrt{2}}}-r\right)}^2}$$
 $$\Rightarrow l=\sqrt{r^2{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2+r^2{\left({\displaystyle \frac{1}{\sqrt{2}}}-1\right)}^2}$$
Take $$r^2$$ to outside, then
 $$\Rightarrow l=r\sqrt{{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{2}}}-1\right)}^2}$$
Simplify using a algebraic identity $${\left(a-b\right)}^2=a^2+b^2-2ab$$ , we have
 $$\Rightarrow l=r\sqrt{{\displaystyle \frac{1}{{\left(\sqrt{2}\right)}^2}}+{\displaystyle \frac{1}{{\left(\sqrt{2}\right)}^2}}+1-{\displaystyle \frac{2}{\sqrt{2}}}}$$
On simplification, we get
 $$\Rightarrow l=r\sqrt{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}+1-{\displaystyle \frac{2}{\sqrt{2}}}}$$
 $$\Rightarrow l=r\sqrt{1+1-{\displaystyle \frac{2}{\sqrt{2}}}}$$
 $$\Rightarrow l=r\sqrt{2-\sqrt{2}}$$ ------(1)
Which is the length of the each side of the regular octagon when $$r$$ be the outer radius.
Then perimeter of regular octagon is
 $$\Rightarrow P=8\times l$$
On substituting equation (1), we have
 $$\Rightarrow P=8\times r\sqrt{2-\sqrt{2}}$$
given $$r=20$$ , then
 $$\Rightarrow P=8\times 20\sqrt{2-\sqrt{2}}$$
 $$\Rightarrow P=160\sqrt{2-\sqrt{2}}$$
On using calculator, we get the exact value
 $$\Rightarrow P=160\sqrt{2-\sqrt{2}}\simeq 122.46$$
Now, consider $$r_1=20$$ be the inner radius - that is the radius of the green circle
The inner radius will be $$r_1=r\cos \theta$$
From the figure we have $$\theta ={\displaystyle \frac{\pi }{8}}$$ , then
 $$\Rightarrow r_1=r\cos \left({\displaystyle \frac{\pi }{8}}\right)$$
By using a calculator, the value of $$\cos \left({\displaystyle \frac{\pi }{8}}\right)={\displaystyle \frac{\sqrt{2+\sqrt{2}}}{2}}$$ , then
 $$\Rightarrow r_1=r\left({\displaystyle \frac{\sqrt{2+\sqrt{2}}}{2}}\right)$$
On cross multiplication, we have
 $$\Rightarrow r={\displaystyle \frac{2r_1}{\sqrt{2+\sqrt{2}}}}$$ ------(2)
On substituting equation (2) in (1), we get the length of each side i.e.,
 $$\Rightarrow l={\displaystyle \frac{2r_1}{\sqrt{2+\sqrt{2}}}}\sqrt{2-\sqrt{2}}$$
 $$\Rightarrow l=2r_1{\displaystyle \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}}$$
To rationalize the denominator, we have to multiply and divide the RHS by $$\sqrt{2+\sqrt{2}}$$ , then
 $$\Rightarrow l=2r_1{\displaystyle \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}}\times {\displaystyle \frac{\sqrt{2+\sqrt{2}}}{\sqrt{2+\sqrt{2}}}}$$
 $$\Rightarrow l=2r_1{\displaystyle \frac{\left(\sqrt{2-\sqrt{2}}\right)\left(\sqrt{2+\sqrt{2}}\right)}{\left(\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2+\sqrt{2}}\right)}}$$
 $$\Rightarrow l=2r_1{\displaystyle \frac{\sqrt{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}}{{\left(\sqrt{2+\sqrt{2}}\right)}^2}}$$
On simplification, we get
 $$\Rightarrow l=2r_1{\displaystyle \frac{\sqrt{4+2\sqrt{2}-2\sqrt{2}-2}}{2+\sqrt{2}}}$$
 $$\Rightarrow l=2r_1{\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}$$ -----------(3)
Which is the length of each side of the regular octagon when $$r$$ be the inner radius.
Then perimeter of regular octagon is
 $$\Rightarrow P=8\times l$$
On substituting equation (1), we have
 $$\Rightarrow P=8\times 2r_1{\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}$$
 $$\Rightarrow P=16r_1{\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}$$
given $$r_1=20$$ , then
 $$\Rightarrow P=16\left(20\right){\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}$$
 $$\Rightarrow P=320{\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}$$
On using calculator, we get the exact value
 $$\Rightarrow P=320{\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}\simeq 132.55$$
Hence, the perimeter of regular octagon of radius 20 is
If the outer radius is 20, then the perimeter is: $$P=160\sqrt{2-\sqrt{2}}\simeq 122.46$$
If the inner radius is 20, then the perimeter is: $$P=320{\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}}}\simeq 132.55$$
So, the correct answer is “$\simeq 132.55$”.

Note: While determining the perimeter we use the formula. The unit for the perimeter will be the same as the unit of the length of a side or polygon. Whereas the unit for the area will be the square of the unit of the length of a polygon. We should not forget to write the unit with a final answer and we should also know about regular and irregular polygons.