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The Angle in One Regular Polygon is to that in Another as 3 : 2 and the Number of Sides in First is Twice that in the Second. Determine the Number of Sides of Two Polygons.

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Question

The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.

 
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Solution

Let the number of sides in the first polygon be 2x and the number of sides in the second polygon is x.
We know:
Angle of an n-sided regular polygon = \[\left( \frac{n - 2}{n} \right)\pi\] radian

∴ Angle of the first polygon =

\[\left( \frac{2x - 2}{2x} \right)\pi = \left( \frac{x - 1}{x} \right)\pi\] radian
 Angle of the second polygon = \[\left( \frac{x - 2}{x} \right)\pi\] radian
Thus, we have: \[\frac{\left( \frac{x - 1}{x} \right)\pi}{\left( \frac{x - 2}{x} \right)\pi} = \frac{3}{2}\]
\[ \Rightarrow \frac{x - 1}{x - 2} = \frac{3}{2}\]
\[ \Rightarrow 2x - 2 = 3x - 6\]
\[ \Rightarrow x = 4\]
Thus,
Number of sides in the first polygon = 2x = 8
Number of sides in the first polygon = x = 4
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Chapter 4: Measurement of Angles - Exercise 4.1 [Page 15]

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R.D. Sharma Mathematics [English] Class 11
Chapter 4 Measurement of Angles
Exercise 4.1 | Q 8 | Page 15

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