Find the area of a regular octagon inscribed in a circle of radius 10cm?

Question

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Answer 1

My answer is 200√2 cm².

Divide the octagon in 8, passing through the vertices.
Each part is an isosceles triangle with equal legs of 10 cm and the middle angle is 45º.

Using one leg as base. The height of the triangle is 10sin45º.
Thus one triangle has area = ½(10)(10*√2/2) = 25√2.

Then 8 triangles = 1 octagon has area 200√2 cm².

d:

Answer 2

Area Of Regular Octagon

Answer 3

Find Area Of Octagon

Answer 4

Not much help above huh? well not all of us are geniuses. . .

Here is how:
you know that a circle has 360° divide that by 8 you get 45°
All the angles of the vertices of an octagon add 1080° (135° ea.)
You also know that an octagon has 8 identical triangles.
Two of the sides of all this triangles all measure 10cm
the area of a triangle is defined as (1/2) Base x Height but we don't have those values, so we try another way:
if ABC equal the angles of the triangle, and a,b,c the sides opposite those angles. Let's make B angle at the center of the circunference, so we have the values a=10cm, b=?, c=10, A=63.5° B=45° C=63.5°, the area can be calculated as:
1/2ab(sinC) or 1/2bc(sinA) or 1/2ca(sinB), the only formula we can use is the last one, so we use that one.
8(1/2)(10*10) sin 45°
(4)(sin 45°)(10^2)
By trigonometry we know (4sin45°) = 4(sin π/4)=2√2
So:
2(√2)*10^2
200(√2)
or:
282.84271247461900976033774484194 cm^2

Answer 5

A regular octagon is formed by 8 triangles with a common point (the center of the circle). Each one of these triangles is conformed by 2 radius (10 cm each) and 1 side of the octagon.

You also know that each triangle takes 1/8 of a total cincunference, so the angle formed by the 2 radius of each triangle is 45º

The easiest way to get the result is multiplying 8 times the area of the triangle. How to get it?. First using sinus theorem to get the unknown side X (you have 1 angle=45º, the others =67.5º; two sides R = 10cm)

X / sin 45º = 10cm / sin 67.5º --> X = 10*sin 45º/sin 67.5º

Then you need the altitude (for side X) [ This time i'll use a rectagular triangle formed by the altitude required, a radius and half of a side X]

Knowing sinus definition, in this triangle we could say

sin 67,5º =H/10 --> H = 10* sin 67.5º

Area of triangle =1/2(X*H) = 0,5*10sin67.5 *10 sin45/sin 67.5
A = 50 * sin 45º

Area of Octagon = 8 * 50 * sin 45º = 400 sin 45º
A= 282.84 cm2

And that is as clear as i can explain without any drawing.

Answer 6

Octagon is made up from 8 isosceles triangles having angle at centre of octagon of 45° and base angles of 67.5°
Area of 1 triangle
= (1/2) x base x height
= (1/2).(10.cos 67.5°) x (10.sin 67.5°)
= 17.7 cm²
Area of octagon
= 8 x 17.7 cm²
= 141.6 cm²

Answer 7

Imagine triangles joining the center and the vertices of the regular polygon and draw a perpendicular on the side from the center of the circle, this perpendicular will be the apothem of the regular polygon. Since the given figure is a regular
polygon.
Now. it is easy to find the angles formed by the so imagined triangles at the center of the circle.
Accordingly, angle between a radius and the apothem will be 22.5 degree. and given, r, the radius of the circumscribing circle = 10
Then, apothem of the given inscribed octagon is
= r cos (theta)
ie., R = 10 cos (22.5) = 9.2388 cm

Now, area of the octagon can easily be found by formula given by, A = n tan (theta) R^2
Where n is the number of sides of the regular polygon.
(*****You cannot find this formula in any of the books, since it is my invention. and this resembles the equation for the area of a circle )
Accordingly, we have ; A = 8 * tan (22.5) *9.2388 ^2
A= 3.31371*9.2388^2 = 282.8431 sq cm
( Note that 3.31371 is a universal constant w.r.t the regular Octagon (rounded off to fifth decimal)

Answer 8

area = 1/2 ( 10 / sqr(2) )( 10 / sqr(2) ) (8 )
area = 200 sq. cm

Answer 9

282.84sqcm