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December 18, 2014

December 18, 2014

Posted by **Erin** on Monday, February 7, 2011 at 2:27pm.

- Maths -
**Terry**, Monday, February 7, 2011 at 3:26pmHm... 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 -
**Terry**, Monday, February 7, 2011 at 3:27pmI meant: NOT what your teacher intend you to do. Sorry for mistype. :p

- Maths -
**Reiny**, Monday, February 7, 2011 at 4:18pmYou 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

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