Question

The area of the circle and the area of a regular polygon of n sides and its perimeter equal to that of the circle are in the ratio of

• A. $\tan \left( \frac { \pi } { n } \right) : \frac { \pi } { n }$
• B. $\cos \left( \frac { \pi } { n } \right) : \frac { \pi } { n }$
• C. $\sin \left( \frac { \pi } { n } \right) : \frac { \pi } { n }$
• D. $\cot \left( \frac { \pi } { n } \right) : \frac { \pi } { n }$

Answer 1

Answer: (A)

$\tan \left( \frac { \pi } { n } \right) : \frac { \pi } { n }$

Answer 2

Let r be the radius of the circle and $A _ { 1 }$ be its area $\therefore A _ { 1 } = \pi r ^ { 2 }$

Since the perimeter of the circle is the same as the perimeter of a regular polygon of n sides $\therefore 2 \pi r = n a$, where 'a' is the length of one side of the regular polygon, $\therefore a = \frac { 2 \pi r } { n }$

Let $A _ { 2 }$ be the area of the polygon, then

$A _ { 2 } = \frac { 1 } { 4 } n a ^ { 2 } \cdot \cot \frac { \pi } { n }$ = $\frac { 1 } { 4 } n \cdot \frac { 4 \pi ^ { 2 } r ^ { 2 } } { n ^ { 2 } } \cot \frac { \pi } { n } = \pi r ^ { 2 } \cdot \frac { \pi } { n } \cdot \cot \frac { \pi } { n }$

$\therefore$ $A _ { 1 } : A _ { 2 } = \pi r ^ { 2 } : \pi r ^ { 2 } \cdot \frac { \pi } { n } \cdot \cot \frac { \pi } { n }$ = $1 : \frac { \pi } { n } \cot \frac { \pi } { n } = \tan \frac { \pi } { n } : \frac { \pi } { n }$ .