Explanation
Let $$r$$ be the radius of the circle and a be the side of polygon.
The perimeter of circle $$= 2\pi r$$
The perimeter of a polygon with n sides $$= na$$
So $$2\pi r = na$$
So $$a = \dfrac {2\pi r}{n}$$
Area of the circle, $$A_1= \pi r^2$$
Area of a regular polygon, $$A_2=\bigg( \dfrac{1}{4} \bigg) na^2 \cot \bigg( \dfrac{\pi}{n}\bigg)$$
$$=\dfrac14\ n\ \bigg(\dfrac {4\pi ^2 r^2}{n^2}\bigg) \cot \bigg(\dfrac{\pi}{n}\bigg)$$
$$=\bigg( \dfrac{\pi ^2 r^2}{n} \bigg) \cot \bigg( \dfrac{\pi}{n} \bigg)$$
$$\Rightarrow \dfrac{A_1}{A_2}=\dfrac{\pi r^2}{\dfrac{\pi^2 r^2}{n} \cot \bigg( \dfrac{\pi}{n} \bigg)}$$
$$=\dfrac{\tan \bigg( \dfrac{\pi}{n}\bigg)}{\dfrac{\pi}{n}}$$
So the ratio is $$\tan \bigg( \dfrac{\pi}{n}\bigg) : \dfrac{\pi}{n}$$
The length of the rectangular field is $$50 m$$ and breadth of the field is $$15 m$$.
The ratio of breadth and length is,
$${\rm{breadth}}:{\rm{length}} = 15:50=\dfrac{15}{50}=\dfrac{3}{10}$$
$$ = 3:10$$
$${\textbf{Step - 1: Converting rupees to paise}}$$
$${\text{We know that, 1 Rupee = 100 paise}}$$
$$\therefore {\text{ 5 Rupees = }}\dfrac{{{\text{5 }} \times {\text{ 100}}}}{{\text{1}}}{\text{ paise}}$$
$$ \Rightarrow {\text{ 5 Rupees = 500 paise}}$$ $$\quad \text{.........eqn(i)}$$
$${\textbf{Step - 2: Substitute above result and get required ratio}}$$
$${\text{Ratio of 50 paise to 5 Rupees = }}$$$$ {\text{ }}\dfrac{{{\text{50 paise}}}}{{{\text{5 Rupees}}}}$$
$$ ={\text{ }}\dfrac{{{\text{50 paise}}}}{{{\text{500 paise}}}}$$ $$\quad \textbf{[From eqn(i)]}$$
$$= {\text{ }}\dfrac{{\text{1}}}{{{\text{10}}}}$$
$${\textbf{Thus, the ratio of 50 paise to 5 Rupees is 1:10. Option C is correct.}}$$
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