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.
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
$A=\frac{\theta}{360^{\circ}} \times \pi r^{2}$
$=\frac{110^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6$
$=\frac{110^{\circ}}{360^{\circ}} \times 113.1$
$=34.5 \mathrm{~cm}^{2}$
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