The tangents drawn at points A and B of a circle with centre O, meet at P. If ∠AOB = 120° and AP = 6 cm, then what is the area of triangle (in cm$$^2)$$ APB?

Given : ∠AOB = 120° and AP = 6 cm and $$\angle$$ OAP = 90°

To find : ar($$\triangle$$ APB) = ?

Solution : $$\angle$$ AOP = $$\frac{1}{2} \angle$$ AOB = $$\frac{120}{2}=60^\circ$$

In $$\triangle$$ AOP,

=> $$tan(\angle AOP)=\frac{AP}{OA}$$

=> $$tan(60^\circ)=\frac{6}{OA}$$

=> $$\sqrt3=\frac{6}{OA}$$

=> $$OA=\frac{6}{\sqrt3}=2\sqrt3$$ cm

Thus, area of $$\triangle$$ AOP = $$\frac{1}{2}\times(OA)\times(AP)$$

 $$\frac{1}{2}\times(2\sqrt3)\times(6)=6\sqrt3$$ $$cm^2$$ -------------(i)

Now, in $$\triangle$$ AOM

=> $$sin(\angle AOM)=\frac{AM}{OA}$$

=> $$sin(60^\circ)=\frac{AM}{2\sqrt3}$$

=> $$\frac{\sqrt3}{2}=\frac{AM}{2\sqrt3}$$

=> $$AM=3$$ cm

Similarly, $$OM = \sqrt3$$ cm

Thus, area of $$\triangle$$ AOM = $$\frac{1}{2}\times(OM)\times(AM)$$

 $$\frac{1}{2}\times(\sqrt3)\times(3)=1.5\sqrt3$$ $$cm^2$$ -------------(ii)

=> $$ar(\triangle AMP)=ar(\triangle AOP)-ar(\triangle AOM)$$

= $$6\sqrt3-1.5\sqrt3=4.5\sqrt3$$ $$cm^2$$

$$\therefore$$ $$ar(\triangle APB)=2ar(\triangle AMP)$$

= $$2\times4.5\sqrt3=9\sqrt3$$ $$cm^2$$

=> Ans - (D)