The tangents drawn at points A and B of a cir | vedclass

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(D) In $\Delta OAP$,$\angle OAP = 90^{\circ}$ (tangent is perpendicular to the radius).$\angle AOP = \frac{1}{2} \angle AOB = \frac{120^{\circ}}{2} =

Vedclass · Accepted Answer

$9 \sqrt{3}$

Vedclass · Answer

(D) In $\Delta OAP$,$\angle OAP = 90^{\circ}$ (tangent is perpendicular to the radius).$\angle AOP = \frac{1}{2} \angle AOB = \frac{120^{\circ}}{2} = 60^{\circ}$.In $\Delta OAP$,$	an(\angle AOP) = \frac{AP}{OA} \Rightarrow 	an(60^{\circ}) = \frac{6}{OA} \Rightarrow \sqrt{3} = \frac{6}{OA} \Rightarrow OA = \frac{6}{\sqrt{3}} = 2\sqrt{3} 	ext{ cm}$.$OP = \sqrt{OA^{2} + AP^{2}} = \sqrt{(2\sqrt{3})^{2} + 6^{2}} = \sqrt{12 + 36} = \sqrt{48} = 4\sqrt{3} 	ext{ cm}$.Let $M$ be the intersection of $AB$ and $OP$. Since $OP$ is the angle bisector of $\angle AOB$,$OP \perp AB$ and $AM = MB$.In $\Delta OAP$,$AM$ is the altitude to the hypotenuse $OP$. Area of $\Delta OAP = \frac{1}{2} 	imes OA 	imes AP = \frac{1}{2} 	imes 2\sqrt{3} 	imes 6 = 6\sqrt{3} 	ext{ cm}^{2}$.Also,Area of $\Delta OAP = \frac{1}{2} 	imes OP 	imes AM \Rightarrow 6\sqrt{3} = \frac{1}{2} 	imes 4\sqrt{3} 	imes AM \Rightarrow 6\sqrt{3} = 2\sqrt{3} 	imes AM \Rightarrow AM = 3 	ext{ cm}$.Since $AB = 2 	imes AM = 2 	imes 3 = 6 	ext{ cm}$ and $PM = OP - OM$. In $\Delta OAM$,$OM = \sqrt{OA^{2} - AM^{2}} = \sqrt{(2\sqrt{3})^{2} - 3^{2}} = \sqrt{12 - 9} = \sqrt{3} 	ext{ cm}$.$PM = 4\sqrt{3} - \sqrt{3} = 3\sqrt{3} 	ext{ cm}$.Area of $\Delta APB = \frac{1}{2} 	imes AB 	imes PM = \frac{1}{2} 	imes 6 	imes 3\sqrt{3} = 9\sqrt{3} 	ext{ cm}^{2}$.

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

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