Tony orlando
New member
- Joined
- Nov 5, 2018
- Messages
- 2
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 180o = SIN, where n is the number of sides of a regular polygon and SIN 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 SIN for 20n2 , the equation now becomes (n-2) x 180o = 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.
(n-2) x 180o = SIN, where n is the number of sides of a regular polygon and SIN 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 SIN for 20n2 , the equation now becomes (n-2) x 180o = 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.