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siddhartharao77:
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Let the common ratio be x.
Let 3x be the number of sides of the 1st polygon.
Let 2x be the number of sides of the 2nd polygon.
We know that the measure of an interior angle of an n-sided polygon is 180 - 360/n.
So, the interior angles of the 1st polygon = 180 - 360/(3x).
So, the interior angles of the 2nd polygon = 180 - 360/(2x).
Given that the interior angles are in the ratio 10:9.
(180 - 360)/((3x))/(180 - 360)/(2x) = 10/9
10(180 - 360/(2x)) = 9(180 - 360/(3x))
1800 - 1800/x = 1620 - 1080/x
1800x - 1800 = 1620x - 1080
1800x = 1620x + 720
180x = 720
x = 4.
Therefore,
the number of sides of the 1st polygon = 3 * 4 = 12.
the number of sides of the 2nd polygon = 2 * 4 = 8
Hope this helps!
Let 3x be the number of sides of the 1st polygon.
Let 2x be the number of sides of the 2nd polygon.
We know that the measure of an interior angle of an n-sided polygon is 180 - 360/n.
So, the interior angles of the 1st polygon = 180 - 360/(3x).
So, the interior angles of the 2nd polygon = 180 - 360/(2x).
Given that the interior angles are in the ratio 10:9.
(180 - 360)/((3x))/(180 - 360)/(2x) = 10/9
10(180 - 360/(2x)) = 9(180 - 360/(3x))
1800 - 1800/x = 1620 - 1080/x
1800x - 1800 = 1620x - 1080
1800x = 1620x + 720
180x = 720
x = 4.
Therefore,
the number of sides of the 1st polygon = 3 * 4 = 12.
the number of sides of the 2nd polygon = 2 * 4 = 8
Hope this helps!
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