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inna [77]
2 years ago
5

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.

Mathematics
1 answer:
Katyanochek1 [597]2 years ago
3 0

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

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