Properties of Polygons

Let the number of sides of polygons A and B be n and 2n, respectively.

$$\ \dfrac{\ \ \dfrac{\ \left(n-2\right)180}{n}}{\ \dfrac{\ \left(2n-2\right)180}{2n}}=\dfrac{3}{4}$$

$$\ \dfrac{\ \ \ n-2}{\ \ n-1}=\dfrac{3}{4}$$

$$\ \ \ \ 4n-8=3n-3$$

$$n=5$$

The number of sides of polygon B is 2*5, i.e. 10.

The sum of the interior angles of a polygon of 'n' sides is given by $$\left(2n-4\right)\times\ 90$$, and the sum of the exterior angles of a polygon is 360 degrees.

So, the difference between them will be 120 * n

=> $$\left(2n-4\right)90-360=120n$$

=> 60n = 720 => n = 12.

We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

Let the total number of angles of the polygon be n.
Hence, the sum of the interior angles is 180*(n-2)

We know that the number of angles which are 90 degrees is 31.
Hence, the number of angles which are 270 is n-31

Hence, the sum of the interior angles is 90*31 + 270*(n-31) which equals 180(n-2)

Therefore, 90*31 + 270*n - 270*31 = 180*n - 360

Or, 90*n = 180*31 - 360

So,n = 2*31 - 4 = 58

We know that 31 angles are 90 degrees. Hence, the number of angles which are 270 degrees is 58 - 31 = 27