Question

Find the number of sides in the polygon, if each interior angle of a regular polygon is greater than each exterior angle.

Column - I		Column - II
I	Interior angle is $120^{\circ}$	P	6 sides
II	Interior angle is $150^{\circ}$	Q	12 sides
		R	24 sides
		S	48 sides

• A. $I - Q , II - R$
• B. $I - P , II - Q$
• C. $I - Q , II - P$
• D. $I - Q , II - S$

Answer 1

Answer: (A)

Sum of exterior angles of polygon $=360^{\circ}$
Let exterior angle $=x^{\circ}$
Then interior angle will be $120^{\circ}+x^{\circ}$
Sum of exterior and its corresponding interior angle is $180^{\circ}$
$\Rightarrow x+(120+x)=180^{\circ} $
$2 x=60^{\circ} \Rightarrow x=30^{\circ}$
So, number of sides $=\frac{360^{\circ}}{30^{\circ}}=12$ sides
When interior angle $=150^{\circ}$
Then $x+(150+x)=180^{\circ}$
$2 x=30^{\circ} \Rightarrow x=15^{\circ}$
So, number of sides $=\frac{360}{15}=24$ sides