Question

The area of sector is $1540sq.cm$. The sector subtends ${50}^o$ at centre, then find the radius of circle.

Answer 1

Angle formed at the centre $\theta ={50}^o$, let $r$ be the radius of circle.
According to question
Area of sector $=1540$
$\Rightarrow \frac{\theta }{360}\times \pi r^2=1540$
$\Rightarrow \frac{50}{360}\times \frac{22}{7}{r}^2=1540$
$\Rightarrow r^2=\frac{1540\times 36\times 7}{5\times 22}$
$\Rightarrow r^2=14\times 36\times \times 7$
$\Rightarrow r=\sqrt{14\times 36\times \times 7}$
$\Rightarrow r=\sqrt{2\times 7\times 6\times 6\times 7}$
$\Rightarrow r=7\times 6\sqrt{2}=42\sqrt{2}$
Thus, the radius of circle $=42\sqrt{2}cm$