Chapter 11.4, Problem 39HP

To determine

To find: The way to find the area of the regular octagon if only the distance is given from the centre of the figure of the vertex.

Answer to Problem 39HP

The area of the octagon is $\frac{x^2}{2\sqrt{2}}$ .

Explanation of Solution

Calculation:

Consider that there is the regular octagon. The distance from the centre to the vertex is given as $x$ units.

The first step is to find the area of the triangle formed by one side and the centre of the octagon. Since, one of the side of the triangle is $x$ , the triangle is isosceles and have equal side with each equal to $x$ units.

As the shape of the octagon and the vertex of the triangle is $\frac{360°}{8}=45°$ .

Then, the area of the octagon is,

  $\begin{array}{c}A=\frac{1}{2}\left(bc\right)\sin A\\ =\frac{1}{2}\left(x\right)\left(x\right)\sin 45°\\ =\frac{x^2}{2\sqrt{2}}\end{array}$

Thus, the area of the octagon is $\frac{x^2}{2\sqrt{2}}$ .