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:

(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]

What type of triangle is this triangle?

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