Find the number of sides of a regular polygon in which the interior angle is four times the exterior angle

Printable View

- Sep 12th 2006, 06:32 AMhelkpangles of polygon
Find the number of sides of a regular polygon in which the interior angle is four times the exterior angle

- Sep 12th 2006, 06:39 AMThePerfectHacker
The sum of the angles a regular polygon is

180(n-2)

And there are n angles as well as sides.

Thus, each angle has,

Code:`180(n-2)`

----------

n

The sum of all exterior angles is always 360 degree.

Divided by the number of angles n thus each one has,

Code:`360`

----

n

Code:`180(n-2) 4*360`

------------ = ----

n n

- Sep 12th 2006, 07:42 AMSoroban
Hello, helkp!

It would help if you knew some formulas.

If you don't, you could wait for*years*for Divine Inspiration . . .

Quote:

Find the number of sides of a regular polygon in which the interior angle

is four times the exterior angle

A regular polygon of*n*sides has a total of*180(n - 2)*degrees in its interior.

So each interior angle has:**180(n - 2)/n**degrees.

An exterior angle is the__supplement__of an interior angle.

Hence, an exterior angle has: .*180 - 180(n - 2)/2 =***360/n**degrees.

We are told that: .[Interior angle] .= .4 × [Exterior angle]

. . . .So we have: .**180(n - 2)/n . = .4 × . (360/n)**

Can you finish it now?