1. ## octagon

Hello everyone, I do not know if I am solving this problem well,

perhaps not so simple as I see it:

Two octagons equal have a common side. Find the angle between the sides of both polygons consecutive common.Find the area of the figure if the side is 2m.

I draw on the figure I made:

And according to the following theory:

In a regular octagon, with sides and angles that are equal, the sides are joined at an angle of 135 degrees or $\displaystyle \displaystyle\frac{3\pi}{4} rad$. Each outer corner of the regular octagon is 45 º or $\displaystyle \displaystyle\frac{\pi}{4} rad.$

And so I think then that the sum 45 + 45 = 90 degrees to the exterior angle formed by the two side keys names to common.

Furthermore, the area of an octagon side $\displaystyle t$ is:

$\displaystyle A = 2t^2(1 + \sqrt[ ]{2})$

I am well on these considerations?

A greeting and thank you very much

2. Hello, Dogod11!

You're right on both counts!

That exterior angle is 90°.

And your area formula is correct.

Good work!

3. Originally Posted by Dogod11
Hello everyone, I do not know if I am solving this problem well,

perhaps not so simple as I see it:

Two octagons equal have a common side. Find the angle between the sides of both polygons consecutive common.Find the area of the figure if the side is 2m.

I draw on the figure I made:

And according to the following theory:

In a regular octagon, with sides and angles that are equal, the sides are joined at an angle of 135 degrees or $\displaystyle \displaystyle\frac{3\pi}{4} rad$. Each outer corner of the regular octagon is 45 º or $\displaystyle \displaystyle\frac{\pi}{4} rad.$

And so I think then that the sum 45 + 45 = 90 degrees to the exterior angle formed by the two side keys names to common.

Furthermore, the area of an octagon side $\displaystyle t$ is:

$\displaystyle A = 2t^2(1 + \sqrt[ ]{2})$

I am well on these considerations?

A greeting and thank you very much

Is correct!!The area if the octagon have to be right!

4. Thanks!!

5. ## Nice

I found no fault in your calculations, good job.