Question

1)An octagon can be divided into 8 congruent triangles with base angles that are 67.5°. This octagon has a side length that is 8 cm.

2)What is the area of each triangle?

3)What is the area of the octagon?

4) Use your work from above to derive a generic formula for any octagon with side length that is n units.

Answer 1

(a)
`Area = 14.7824cm^2`

(b)
`Area = 118.2592cm^2`

(c)
`Area = 4nh`

Explanation:

Given

`\theta = 67.5`

`b= 8cm` --- base

`n = 8` --- triangles

Solving (a) The area of each triangle

First, calculate the height (h) of each triangle using:

`\sin(67.5) = (h)/(b/2)` --- i.e. we consider half of the triangle

`\sin(67.5) = (h)/(8/2)`

`\sin(67.5) = (h)/(4)`

Solve for h

`h =4*\sin(67.5)`

`h =4*0.9239`

`h =3.6956`

The area of each triangle is:

`Area = (1)/(2) *b * h`

`Area = (1)/(2) *3.6956 * 8`

`Area = 14.7824cm^2`

Solving (b): Area of the octagon

This is calculated as:

Area = 8 * area of 1 triangle

`Area = 8 * 14.7824cm^2`

`Area = 118.2592cm^2`

Solving (c): Area of octagon of side length n

In (a), we have:

`Area = (1)/(2) *b * h`

Replace b with n

`Area = (1)/(2) *n * h`

Multiply by 8 (the sides) to get the area of the octagon

`Area = 8 * (1)/(2) *n * h`

`Area = 4nh`