CAT 2019 Slot 2 — QA Question 11

Sum of the interior angles of a n-sided Polygon = (n - 2) × 180

Measure of each interior angle = $\frac{(n-2)}{n}$× 180°

So, Interior angle of the polygon with side a = $\frac{(a-2)}{a}$ × 180°

Interior angle of the polygon with side b = $\frac{(2a-2)}{2a}$ × 180°

It is given that interior angle of side b is 3/2 times to that of the polygon with side A

⇒ $\frac{(2a-2)}{2a}$ × 180° =  $\frac{3}{2}\left[\frac{(a-2)}{a}\times 180°\right]$

⇒ (2a - 2) × 180 = (3a - 6) × 180

⇒ 2a - 2 = 3a - 6

⇒ a = 4. So, b = 8

Now, (a + b) sides = 8 + 4 = 12 sides

Interior angle =$\frac{12-2}{12}$× 180° = 150°

Hence, 150.