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In the given figure, chord $AB$ subtends an angle of $120^{\circ}$ at the centre of
nthe circle with radius $7$ cm. Find (i) perimeter of major sector $OACB$, and
n(ii) area of the shaded segment, if area of $\triangle OAB = 21.2$ cm$^2$.
nthe circle with radius $7$ cm. Find (i) perimeter of major sector $OACB$, and
n(ii) area of the shaded segment, if area of $\triangle OAB = 21.2$ cm$^2$.
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(i) Perimeter of major sector = length of major arc $ACB + 2 \times$ radius
$= \frac{(360-120)}{360} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7$ (1 Mark)
$= \frac{130}{3}$ cm or $43.3$ cm (1/2 Mark)
So, perimeter of major sector is $\frac{130}{3}$ cm or $43.3$ cm
(ii) Area of shaded segment = Area of minor sector $-$ Area of $\triangle OAB$
$= \frac{120}{360} \times \frac{22}{7} \times 7 \times 7 - 21.2$ (1 Mark)
$= 30.1$ cm$^2$ (1/2 Mark)
So, area of shaded segment is $30.1$ cm$^2$.
$= \frac{(360-120)}{360} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7$ (1 Mark)
$= \frac{130}{3}$ cm or $43.3$ cm (1/2 Mark)
So, perimeter of major sector is $\frac{130}{3}$ cm or $43.3$ cm
(ii) Area of shaded segment = Area of minor sector $-$ Area of $\triangle OAB$
$= \frac{120}{360} \times \frac{22}{7} \times 7 \times 7 - 21.2$ (1 Mark)
$= 30.1$ cm$^2$ (1/2 Mark)
So, area of shaded segment is $30.1$ cm$^2$.