Question

In the adjoining figure PA and PB are tangents drawn to a circle whose centre is at O and whose radius is 8 cm. If $\angle APO=30$, then area of $\Delta AOB$ is

• A. $12\sqrt{3}$
• B. $8\sqrt{3}$
• C. $16\sqrt{3}$
• D. $Noneofthese$

Answer 1

The correct option is C $16\sqrt{3}$

REF.Image.

area of $\Delta AOB=?$

from figure

$PA=PB$ (tangent property)

and $AO=OB=8$ (given)

From $\Delta AOP$ and $\Delta BOP$

$OP$ is common side

$PA=PB$ and $AO=OB$

$\therefore \Delta AOP\sim \Delta BOP$ (by sss)

Hence, $\Delta AOL$ and $\Delta BOL$ are similar.

$\therefore AL=BL$ & $\angle AOL=\angle BOL={60}^o$

from $\Delta AOB$ (isolate triangle)

Let $\angle OAL=\angle OBL$ be $x$

$\therefore x+x+{60}^o+\angle BOL={180}^o$

$2x={180}^o-{120}^o$

$x={30}^o$

from $\Delta DBL$

$\cos {30}^o=\frac{BL}{OB}=\frac{BL}{8}$

$\frac{\sqrt{3}}{2}=\frac{BL}{8}$

$BL=4\sqrt{3}$

$\therefore AB=2\times 4\sqrt{3}=8\sqrt{3}cm$ ...(i)

Similarly,

$\sin {30}^o=\frac{OL}{OB}=\frac{OL}{8}$

$\frac{1}{2}=\frac{OL}{8}$

$OL=4cm$ .....(ii)

from (i) and (ii)

Base of $\Delta AOB=8\sqrt{3}$

Height of $\Delta AOB=4$

Area of $\Delta AOB=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times 8\sqrt{3}\times 4$

$A=16\sqrt{3}c{m}^2$