Determining the numbers sides a regular polygon with has

[Tony orlando]

Hello, this is my first post here. I came across a geometry problem that I wanted to confirm the solution for. Each interior angle A in a regular n-gon (n is the number of sides a regular polygon has) has a measure of A = 20n. How many sides does the polygon have? I recalled that the number of sides a regular polygon has can related to the sum of its interior angles with following expression:

(n-2) x 180^o = S_IN, where n is the number of sides of a regular polygon and S_IN is the sum of the internal angles.

Since each angle in the n-gon has a measure of 20n, the sum of the internal angles can be represented as n x 20n or 20n². Therefore, substituting S_IN for 20n² , the equation now becomes (n-2) x 180^o = 20n². I then multiplied 180 through n-2 and subtracted the resulting quantities from the left side to form a quadratic equation 20n²-180n+360=0. Factoring the equation, I came up with 2 solutions: n=3 or n=6. I wanted to confirm if this solution ( and method) are correct.

Thank you kindly.

 

[Dr.Peterson]

Hello, this is my first post here. I came across a geometry problem that I wanted to confirm the solution for. Each interior angle A in a regular n-gon (n is the number of sides a regular polygon has) has a measure of A = 20n. How many sides does the polygon have? I recalled that the number of sides a regular polygon has can related to the sum of its interior angles with following expression:

(n-2) x 180^o = S_IN, where n is the number of sides of a regular polygon and S_IN is the sum of the internal angles.

Since each angle in the n-gon has a measure of 20n, the sum of the internal angles can be represented as n x 20n or 20n². Therefore, substituting S_IN for 20n² , the equation now becomes (n-2) x 180^o = 20n². I then multiplied 180 through n-2 and subtracted the resulting quantities from the left side to form a quadratic equation 20n²-180n+360=0. Factoring the equation, I came up with 2 solutions: n=3 or n=6. I wanted to confirm if this solution ( and method) are correct.

Thank you kindly.

Sounds good to me.

You can, of course, check your answer easily enough. For n=3, you know each angle is 60, which is indeed 20*3; and for n=6, each angle is 120, which is 20*6.

 

[Tony orlando]

Sounds good to me.

You can, of course, check your answer easily enough. For n=3, you know each angle is 60, which is indeed 20*3; and for n=6, each angle is 120, which is 20*6.

Thanks Dr. Peterson, I figured as much. I'm trying to upgrade my high school maths and it has been many years since I've been in a class room so I wanted to make sure I was on the right track.

Cheers.

 

[Dr.Peterson]

Thanks Dr. Peterson, I figured as much. I'm trying to upgrade my high school maths and it has been many years since I've been in a class room so I wanted to make sure I was on the right track.

Cheers.

Yes, you're on the right track. Keep going in the same direction ...

It's nice seeing people who are successfully returning to math.