In a circle of radius 6 cm,

Question:

In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find:

(i) the circumference of the circle

(ii) the area of the circle

(iii) the length of the arc AB,

(iv) the area of the sector OAB.

Solution:

It is given that the radius of circle $r=6 \mathrm{~cm}$, length of chord $=10 \mathrm{~cm}$ and angle at the centre of circle $\theta=110^{\circ}$.

(i) We know that the Circumference C of circle of radius r is,

$C=2 \pi \mathrm{r}$

$=2 \times \frac{22}{7} \times 6$

$=\frac{264}{7}$

$=37.71 \mathrm{~cm}$

(ii) We know that the Area A of circle of radius r is,

$\mathrm{A}=\pi r^{2}$

$=\frac{22}{7} \times 6 \times 6$

$=\frac{792}{7}$

$A=113.1 \mathrm{~cm}^{2}$

(iii) We know that the arc length l of a sector of an angle θ in a circle of radius r is

$l=\frac{110^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 6$

$=\frac{110^{\circ}}{360^{\circ}} \times 37.68$

$=11.51 \mathrm{~cm}$

(iv) We know that the area A of a sector of an angle θ in the circle of radius r is given by

$A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

$=\frac{110^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6$

$=\frac{110^{\circ}}{360^{\circ}} \times 113.1$

$=34.5 \mathrm{~cm}^{2}$