Answer:
(a) [tex]Area = 14.7824cm^2[/tex]
(b) [tex]Area = 118.2592cm^2[/tex]
(c) [tex]Area = 4nh[/tex]
Step-by-step explanation:
Given
[tex]\theta = 67.5[/tex]
[tex]b= 8cm[/tex] --- base
[tex]n = 8[/tex] --- triangles
Solving (a) The area of each triangle
First, calculate the height (h) of each triangle using:
[tex]\sin(67.5) = \frac{h}{b/2}[/tex] --- i.e. we consider half of the triangle
[tex]\sin(67.5) = \frac{h}{8/2}[/tex]
[tex]\sin(67.5) = \frac{h}{4}[/tex]
Solve for h
[tex]h =4*\sin(67.5)[/tex]
[tex]h =4*0.9239[/tex]
[tex]h =3.6956[/tex]
The area of each triangle is:
[tex]Area = \frac{1}{2} *b * h[/tex]
[tex]Area = \frac{1}{2} *3.6956 * 8[/tex]
[tex]Area = 14.7824cm^2[/tex]
Solving (b): Area of the octagon
This is calculated as:
Area = 8 * area of 1 triangle
[tex]Area = 8 * 14.7824cm^2[/tex]
[tex]Area = 118.2592cm^2[/tex]
Solving (c): Area of octagon of side length n
In (a), we have:
[tex]Area = \frac{1}{2} *b * h[/tex]
Replace b with n
[tex]Area = \frac{1}{2} *n * h[/tex]
Multiply by 8 (the sides) to get the area of the octagon
[tex]Area = 8 * \frac{1}{2} *n * h[/tex]
[tex]Area = 4nh[/tex]
