Please solve question number 4 please
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Let the number of sides of the regular polygon with the fewer number of sides be x, and y be the number of sides of the polygon with the greater number of sides.
xy=12xy=12 solving for y: y=2xy=2x
The measure of an interior angle of a regular polygon (I) is:
I=(n−2)180nI=(n−2)180n
(x−2)180x(y−2)180y=34(x−2)180x(y−2)180y=34
Simplifying:
y(x−2)x(y−2)=34y(x−2)x(y−2)=34
Substituting for y:
2x(x−2)x(2x−2)=342x(x−2)x(2x−2)=34
Cross multiplying and distributing:
8x2−16x=6x2−6x8x2−16x=6x2−6x
Combine like terms:
2x2−10x=02x2−10x=0
Factor:
2x(x−5)=02x(x−5)=0
Therefore, x = 0 or x = 5.
Eliminate x = 0, thus, x = 5 and y = 10, or the number of sides of the two regular polygons are 5 and 10.
xy=12xy=12 solving for y: y=2xy=2x
The measure of an interior angle of a regular polygon (I) is:
I=(n−2)180nI=(n−2)180n
(x−2)180x(y−2)180y=34(x−2)180x(y−2)180y=34
Simplifying:
y(x−2)x(y−2)=34y(x−2)x(y−2)=34
Substituting for y:
2x(x−2)x(2x−2)=342x(x−2)x(2x−2)=34
Cross multiplying and distributing:
8x2−16x=6x2−6x8x2−16x=6x2−6x
Combine like terms:
2x2−10x=02x2−10x=0
Factor:
2x(x−5)=02x(x−5)=0
Therefore, x = 0 or x = 5.
Eliminate x = 0, thus, x = 5 and y = 10, or the number of sides of the two regular polygons are 5 and 10.
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