Using `lim_(theta -> 0) (sintheta/theta)=1` prove that the area of circle of radius `R` is `piR^2` (Figure)

Explanation

In the following figure regular polygon on n sides is inscribed in the circle of radius R.

In the figure, `OA_(1)=OA_(2)=R" and "angleA_(1)OA_(2)=2pi//n` Clearly area of polygon
`Delta=nxx("Area of the "DeltaOA_(1)A_(2))`
`=nxx(1)/(2)RxxRxx"sin"(2pi)/(n)`
`:. " "Delta=(nR^(2))/(2)"sin"(2pi)/(n)`
We know that circle is regular palygon having infinite number of sides. So, from above formula
`underset(ntooo)lim(nR^(2))/(2)"sin"(2pi)/(n)=underset(ntooo)limpiR^(2)("sin"(2pi)/(n))/((2pi)/(n))`
`piR^(2)underset(ntooo)lim("sin"(2pi)/(n))/((2pi)/(n))`
`=piR^(2)xx1" "( :.underset(thetato0)lim(sintheta)/(theta)=1)`
`=piR^(2)`
Thus, area of circle having radius R is `piR^(2).`