Question

The ratio of angles in one regular polygon to that in another is 3: 2.The number of sides in the first is twice the second. How many sides do the second polygon have?

Answer 1

Hint- Recall the formula for each interior angle of polygon with respect to the right angles present in it.

Since we have to tell about the number of sides of the second regular polygon hence let the number of sides of second polygon $\text{ = n}$
Now it is being given that the number of sides in the first polygon is twice the sides in the first polygon.
So the number of sides of the second polygon $\text{ = 2n}$
Now any n sided polygon can be divided into $\text{(n - 2)}$triangles. Now the sum of angles of a triangle is 180, therefore the sum of interior angles of a polygon having n sides is $\text{(2n - 4)}$right angles. Thus each interior angle of the polygon is $\left({\displaystyle \frac{2n-4}{n}}\right)$right angles.
Hence each interior of first polygon $\text{ = }\left({\displaystyle \frac{4n-4}{2n}}\right)$right angles
Interior angle of the second polygon $\left({\displaystyle \frac{2n-4}{n}}\right)$right angles
Now it’s given in problem that the angles are in the ratio 3: 4 so we can say that
$\left({\displaystyle \frac{4n-4}{2n}}\right):\left({\displaystyle \frac{2n-4}{n}}\right)=3:2$
Or
 ${\displaystyle \frac{\left({\displaystyle \frac{4n-4}{2n}}\right)}{\left({\displaystyle \frac{2n-4}{n}}\right)}}={\displaystyle \frac{3}{2}}$
On solving we get
$\text{2}\left(\left({\displaystyle \frac{4n-4}{2}}\right)\right)=3\left(2n-4\right)$
That is $\text{4n - 4 = 6n - 12}$
On solving we get $\text{n = 4}$
Thus the number of sides in the first polygon is 8 and the second polygon is 4.

Note- Every time we encounter such problems the key concept involved is that the sum of interior angles of a polygon having n sides is $\text{(2n - 4)}$right angles. Thus each interior angle of the polygon is ${\displaystyle \frac{2n-4}{n}}$angles.