InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The area of the circle and the area of a regular polygon of `n`sides and of perimeter equal to that of the circle are in the ratio of`tan(pi/n):pi/n`(b) `cos(pi/n):pi/n``sinpi/n :pi/n`(d)`cot(pi/n):pi/n`A. `tan((pi)/(n)): (pi)/(n)`B. `cos ((pi)/(n)) : (pi)/(n)`C. `sin.(pi)/(n): (pi)/(n)`D. `cot((pi)/(n)): (pi)/(n)` |
|
Answer» Correct Answer - A Let r be the radius of the circle and `A_(1)` be its area. Then `A_(1) = pir^(2)`. Since the perimeter of the circle is same as the perimeter of a regular polygon of n sides, we have `2pi r = na`, Where a is the length of one side of the regular polygon. Thus, `a = (2pi r)/(n)` Let `A_(2)` be the area of the polygon. Then, `A_(2) = (1)/(4) na^(2) cot ((pi)/(n)) = (pi^(2) r^(2))/(n) cot ((pi)/(n))` `rArr A_(1) : A_(2) = pi r^(2) : (pi^(2) r^(2))/(n) cot ((pi)/(n)) = tn ((pi)/(n)) : (pi)/(n)` |
|