Question

A chord of radius $12cm$ subtends an angle of ${120}^{\circ }$ at the centre. Find the area of corresponding segment of the circle. $(22/7=3.14$ and root $3=1.73)$.

Answer 1

Area of segment APB=Area of sector OAPB-area of $△OAB=$

Area of OAPB$=\frac{\theta }{360°}\times \pi r^2=\frac{120°}{360°}\times 3.14\times {12}^2=150.72㎠$

Draw $OM\perp AB,\therefore \angle OMB=\angle OMA=90°$

In $△OMA$ &$△OMB$

$\angle OMA=\angle OMB$ [Each$90°$]

$OM=OM$ (common)

$OA=OB$ (radii)

$\therefore △OMA\cong △OMB$ [By RHS congruency]

$\Rightarrow \angle AOM=\angle BOM$ [By CPCT]

$\therefore \angle AOM=\angle BOM=\frac{1}{2}\angle AOB=30°$

$AM=BM$

$BM=AM=\frac{1}{2}AB\to 1$

$\sin 60°=\frac{AM}{AO}\Rightarrow \frac{\sqrt{3}}{2}=\frac{AM}{12}\Rightarrow AM=6\sqrt{3}$

& $\cos 60°=\frac{OM}{AO}\Rightarrow \frac{1}{2}=\frac{OM}{12}\Rightarrow OM=6$

$\therefore AB=2AM=2\times 6\sqrt{3}=12\sqrt{3}$ (From $1$)

$ar△AOB=\frac{1}{2}\times AB\times OM=\frac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}=36\times 1.73$

$ar△AOB=62.28㎠$

Therefore, area segment $APB=150.72-62.28=88.44㎝$