# Maths

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Two polygons have te number of their sides in the ratio 1:3.The sums of their interior angles are in the ratio of 4;1. Find the number of sides of each polygon.

• Maths - ,

Hm... quite a complicate question. A less complicate way to do this is trial and error. You can make a table with "n" sides and the sum of the angles in degree for each polygon (ie. triangle (3 sides) = 180, 4 sides = 360, 5 sides = 540... and so on). And base on the values, you can find the matching pairs that fit the ratio criteria. But I guess that's what your teacher intend you to do. So let's try to figure this out mathematically:

You can come up with a formula to calculate the sum of the angles from the small table you created above:

Angle = 180n - 360.

Now you know that the ratio of the angles is 4:1 and the sides is 3:1, and so you have another formula for the other polygon:

4(Angle) = (3n)180 - 360.
Angle = 135n - 90

Now using substituion and plug it back into the first equation, you get:

135n - 90 = 180n - 360
270 = 45n
n = 6

Now that you know that one polygon has 6 sides, the other one must have three time as many, thus 18 sides. Check the sum of the angles for each, and you should come up with a ratio of 1:4 on the angles for both polygons. Problem solved!

• Maths - ,

I meant: NOT what your teacher intend you to do. Sorry for mistype. :p

• Maths - ,

You surely meant that the sum of their angle sum is 1:4 , not 4:1

let the number of sides of the first polygon be n
then the numer of sides of the second is 3n
(notice n : 3n = 1 : 3 )

angle sum of first = 180(n-2)
angle sum of second = 180(3n-2)

so,
180(n-2)/(180(3n-2)) = 1/4
(n-2)/(3n-2) = 1/4
4n-8 = 3n-2
n = 6

first polygon has 6 sides,
the second has 18 sides