Mary draws an equiangular polygon with m sides, and George draws an equiangular polygon with g sides, where m is less than g. If the exterior angle of Mary's polygon is congruent to the interior angle of George's polygon, find g.

help, anyone???

noobieatmath Jan 14, 2019

edited by
Guest
Jan 14, 2019

#1**+2 **

It says, that the exterior angle of a polygon with m sides is equal to a regular polygon with g sides. So,

\(\frac{360}{m} = \frac{180(g + 2)}{g}\)

\(360g = 180gm + 360m\)

\(2g = gm + 2m\)

Factoring out g:

\(2g - gm = 2m\)

\(g(2-m) = 2m\)

So therefore, \(g = \frac{2m}{2-m}\).

-24

TwentyFour Jan 15, 2019

#2**+1 **

Exterior angle of Mary's polygon = 360 / m

Interior angle of George's polygon = (g - 2)*180 / g

Since these are congruent....then

360 / m = (g - 2) * 180 / g cross - multiply

360g = 180m * (g - 2)

360g = 180 gm - 360m

360m = 180gm - 360g

2m = gm - 2g

2m = g ( m - 2 )

g = 2m / ( m - 2)

CPhill Jan 15, 2019