Question

In the given figure, AB is a diameter of the circle with centre O and P is a point from where two tangents PT and PT′ are drawn on the circle. A semicircle is drawn taking BC as diameter, A as centre and AP = PC. If AB = 12 cm, then the length of PD is

• A.
  8 cm
• B.
  12 cm
• C.
  $2\sqrt{2}cm$
• D.
  $6\sqrt{3}cm$

Answer 1

The correct option is D

$6\sqrt{3}cm$
Given that AB and BC are diameters of the circles with centre O and A, respectively.

$\therefore AC=AB=AD=12cm...\left(i\right)$



Now, $AC=AP+PC=AP+AP(\because AP=PC)=2AP\Rightarrow 2AP=12(\because AC=12)\Rightarrow AP=6cm...\left(ii\right)In\Delta APD,\text{by pythagoras theorem,}A{D}^2=A{P}^2+P{D}^2\Rightarrow {12}^2=6^2+P{D}^2\left[From\right(i\left)and\right(ii\left)\right]\Rightarrow P{D}^2=144-36\Rightarrow P{D}^2=108\Rightarrow PD=\sqrt{108}\Rightarrow PD=6\sqrt{3}cm$

Hence, the correct answer is option d.