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

Answer:

(a) $Area = 14.7824cm^2$

(b) $Area = 118.2592cm^2$

(c) $Area = 4nh$

Step-by-step 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:

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

$\sin(67.5) = \frac{h}{8/2}$

$\sin(67.5) = \frac{h}{4}$

Solve for h

$h =4*\sin(67.5)$

$h =4*0.9239$

$h =3.6956$

The area of each triangle is:

$Area = \frac{1}{2} *b * h$

$Area = \frac{1}{2} *3.6956 * 8$

$Area = 14.7824cm^2$

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:

$Area = \frac{1}{2} *b * h$

Replace b with n

$Area = \frac{1}{2} *n * h$

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

$Area = 8 * \frac{1}{2} *n * h$

$Area = 4nh$