The area of a regular polygon of `n` sides is (where `r` is inradius, `R` is circumradius, and `a` is side of the triangle) `(n R^2)/2sin((2pi)/n)` (b) `n r^2tan(pi/n)` `(n a^2)/4cotpi/n` (d) `n R^2tan(pi/n)`

If a regular polygon of n sides has circum radius R and inradius r then each side of polygon is:

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)sin(pi)/(n):(pi)/(n)(d)cot((pi)/(n)):(pi)/(n)

the sum of the radii of inscribed and circumscribed circle of an n sides regular polygon of side a is (A) a/2 cot (pi/(2n)) (B) acot(pi/(2n)) (C) a/4 cos, pi/(2n)) (D) a cot (pi/n)

If r is the radius of inscribed circle of a regular polygon of n-sides ,then r is equal to

Find the sum sum_(r=1)^(n)(cos((2r-1)pi))/((2n+1))

The value of sum_(r=0)^(n)C_(n-r)(n-r)sin((r pi)/(n))

A general solution of tan^(2)theta+cos2 theta=1 is (n in Z)n pi=(pi)/(4) (b) 2n pi+(pi)/(4)n pi+(pi)/(4) (d) n pi

Find the sum of sum_(r=1)^(n)cos(((2r-1)pi)/(2n+1))

If a circle touching all the n sides of a polygon of perimeter 2p has radius r, then the area of the poly-gon is (p-n)r( b) pr(c)(2p-n)r (d) (p+n)r