Question

If the number of sides of two regular polygons having the same perimeter be n and 2n respectively, their areas are in the ratio

• A. $\frac { 2 \cos \left( \frac { \pi } { n } \right) } { \cos \left( \frac { \pi } { 2 n } \right) }$
• B. $\frac { 2 \cos \left( \frac { \pi } { n } \right) } { 1 + \cos \left( \frac { \pi } { n } \right) }$
• C. $\frac { \cos \left( \frac { \pi } { n } \right) } { \sin \left( \frac { \pi } { n } \right) }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 2 \cos \left( \frac { \pi } { n } \right) } { 1 + \cos \left( \frac { \pi } { n } \right) }$

Answer 2

Let s be the perimeter of both the polygons. Then the length of each side of the first polygon is $\frac { s } { n }$ and that of second polygon is $\frac { s } { 2 n }$ .

If $A _ { 2 }$denote their areas, then $A _ { 1 } = \frac { n } { 4 } \left[ \frac { s } { n } \right] ^ { 2 } \cot \frac { \pi } { n }$ and

$A _ { 2 } = \frac { 1 } { 4 } \cdot ( 2 n ) \left( \frac { s } { 2 n } \right) ^ { 2 } \cdot \cot \left( \frac { \pi } { 2 n } \right)$

$\frac { A _ { 1 } } { A _ { 2 } } = \frac { 2 \cot \left( \frac { \pi } { n } \right) } { \cot \left( \frac { \pi } { 2 n } \right) } = \frac { 2 \cos \left( \frac { \pi } { n } \right) \sin \left( \frac { \pi } { 2 n } \right) } { \sin \left( \frac { \pi } { n } \right) \cos \left( \frac { \pi } { 2 n } \right) }$ $= \frac { 2 \cos \left( \frac { \pi } { n } \right) \sin \left( \frac { \pi } { 2 n } \right) } { 2 \sin \left( \frac { \pi } { 2 n } \right) \cos \left( \frac { \pi } { 2 n } \right) \cos \left( \frac { \pi } { 2 n } \right) }$ ⇒ $\frac { A _ { 1 } } { A _ { 2 } } = \frac { 2 \cos \left( \frac { \pi } { n } \right) } { 1 + \cos \left( \frac { \pi } { n } \right) }$ .