Question

A chord PQ of a length 12 cm subtends an angle of ${120}^{\circ }$ at the center of a circle. Find the area of the minor segment cut off by chord PQ.

Answer 1

In $△AOB,\angle AOB=120°$

$OM\perp AB,\therefore \angle AOM=\angle MOB=60°$ & $MA=MB=6$

Now, $\sin 60°=\frac{6}{r}\Rightarrow \frac{\sqrt{3}}{2}=\frac{6}{r}\Rightarrow r=\frac{12}{\sqrt{3}}=4\sqrt{3}$

Angle of minor segment=Area of sector-Area of $△AOB$

Area of sector$=\frac{\theta }{360°}\times \pi r^2=\frac{120°}{360°}\times \pi \times 4\sqrt{3}\times 4\sqrt{3}=16\pi c{m}^2$

Area $△AOB=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{4\sqrt{3}+4\sqrt{3}+12}{2}=4\sqrt{3}+6$

Area of $△AOB=\sqrt{(4\sqrt{3}+6)(4\sqrt{3}+6-4\sqrt{3})(4\sqrt{3}+6-4\sqrt{3})(4\sqrt{3}+6-12)}$

$=\sqrt{6\times 6(4\sqrt{3}+6)(4\sqrt{3}-6)}=6\sqrt{{\left(4\sqrt{3}\right)}^2-{\left(6\right)}^2}$$=6\sqrt{48-36}=6\sqrt{12}=12\sqrt{3}c{m}^2$

$\therefore$Area of minor segment$=16\pi -12\sqrt{3}=4(4\pi -3\sqrt{3})c{m}^2$