Question

In a circle of radius $21$ cm, an arc subtends an angle of ${60}^{\circ }$ at the centre. Find:

$\left(i\right)$ the length of the arc

$\left(ii\right)$ area of the sector formed by the arc

$\left(iii\right)$ area of the segment formed by the corresponding chord

Answer 1

(i) Length of arc APB$=\frac{\theta }{360}\times 2\pi r$

$=\frac{{60}^o}{{360}^o}\times 2\times \frac{22}{7}\times 21$

$=\frac{1}{6}\times 2\times \frac{22}{7}\times 21$

$=22cm$

(ii) Area of sector OAPB$=\frac{\theta }{{360}^o}\times \pi r^2$

$=\frac{60}{360}\times \frac{22}{7}\times 21\times 21$

$=\frac{1}{6}\times 22\times 3\times 21=231c{m}^2$ ……………..$\left(1\right)$

(iii) Area of segment APB$=$Area of sector OAPB$-$ Area of $\Delta$ OAB

$=231-\frac{1}{2}\times 6\times 4$

Draw OM$\perp$AM

$\therefore \angle OMB=\angle OMA={90}^o$

In $\Delta OMA$ and $\Delta OMB$,

$\angle OMA=\angle OMB={90}^o$

$OA=OB$(radius), OM$=$OM (common)

$\therefore \Delta OMA\cong \Delta OMB$ (by RHS congruency)

$\Rightarrow \angle AOM=\angle BOM$ (by CPCT)

$\therefore \angle AOM=\angle BOM=\frac{1}{2}\angle BOA$

$\Rightarrow \angle AOM=\angle BOM=\frac{1}{2}\times 60={30}^o$

Since $\Delta OMB\equiv \Delta OMA$

$\therefore BM=AM$ by CPCT

$\Rightarrow BM=AM=\frac{1}{2}AB$ ………….$\left(2\right)$

In right angled $\Delta OMA$,

$\sin \theta =AM/AO\Rightarrow \frac{1}{2}=AM/21\Rightarrow AM=\frac{21}{2}$

In right angled $\Delta OMA$

$\cos \theta =\frac{OM}{AO}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{OM}{21}\Rightarrow OM=\frac{\sqrt{3}}{2}\times 21$

From $\left(2\right)$ $AM=\frac{1}{2}AB$

$\Rightarrow 2AM=AB$

or $AB=2AM$

$=2\times \frac{21}{2}=21$cm

Area $\left(\Delta AOB\right)=\frac{1}{2}bh=\frac{1}{2}\times AB\times OM$

$=\frac{1}{2}\times 21\times \frac{\sqrt{3}}{2}\times 21$

$=\frac{441\sqrt{3}}{4}c{m}^2$

Area of segment APB$=$ Area of sector OAPB$-$ Area of $\Delta$OAB

$=(231-\frac{441\sqrt{3}}{4})c{m}^2$.