The angle in one regular polygon is to that in another as \(3:2\) and the number of sides in the first is twice that in the second. Determine the number of sides of the two polygons.
Solution:
Let the number of sides of the second polygon be \(n\).
Then the number of sides of the first polygon is \(2n\).
Interior angle of a regular polygon:
\[ \frac{(n-2)\times180^\circ}{n} \]
First polygon angle:
\[ \frac{(2n-2)\times180^\circ}{2n} \]
\[ =\frac{(n-1)\times180^\circ}{n} \]
Second polygon angle:
\[ \frac{(n-2)\times180^\circ}{n} \]
Given ratio:
\[ \frac{(n-1)\times180^\circ}{n} : \frac{(n-2)\times180^\circ}{n} = 3:2 \]
\[ \frac{n-1}{n-2}=\frac{3}{2} \]
\[ 2(n-1)=3(n-2) \]
\[ 2n-2=3n-6 \]
\[ n=4 \]
Therefore,
Second polygon has \(4\) sides.
First polygon has \(2n=8\) sides.
Hence, the polygons are:
\[ \text{Quadrilateral and Octagon} \]