Question

There are two regular polygons with number of sides equal to (n 1) and (n + 2). Their exterior angles differ by 6. The value of n is:

Answer 1

The value of n is 13

• Number of sides of the two regular polygons are (n-1) and (n+2).
• Now we know that the sum of exterior angles of a regular polygon is 360 degrees.
• So in this two cases individual exterior angles for the two polygons are $\frac{360}{n-1}$ and $\frac{360}{n+2}$ .
• Now according to the problem  $\frac{360}{n-1} -\frac{360}{n+2} = 6$.
• Dividing both sides by 360 we get $\frac{1}{n-1} -\frac{1}{n+2} = \frac{1}{60}$  
• Now manipulating a little bit we arrive at the quadratic equation $n^{2}+n-182 = 0$
• Factoring it we get (n - 13) (n + 14) = 0.
• This yields two solutions as 13 and -14. But n can't be negative as it is number of sides of polygon.
• So the value of n is 13

Answer 2

Answer:

13

Step-by-step explanation:

Number of sides of the two regular polygons are (n-1) and (n+2).

Now we know that the sum of exterior angles of a regular polygon is 360 degrees.

So in this two cases individual exterior angles for the two polygons are  and  .

Now according to the problem  .

Dividing both sides by 360 we get   

Now manipulating a little bit we arrive at the quadratic equation

Factoring it we get (n - 13) (n + 14) = 0.

This yields two solutions as 13 and -14. But n can't be negative as it is number of sides of polygon.

So the value of n is 13