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 $180^{\circ }$
$\Rightarrow x+(120+x)=180^{\circ }$
$2x=60^{\circ }\Rightarrow x=30^{\circ }$
So, number of sides $=\frac{360^{\circ }}{30^{\circ }}=12$ sides
When interior angle $=150^{\circ }$
Then $x+(150+x)=180^{\circ }$
$2x=30^{\circ }\Rightarrow x=15^{\circ }$
So, number of sides $=\frac{360}{15}=24$ sides