Question

In the following figure, AD is a straight line. OP $\perp$ AD and O is the centre of both the circles. If OA = 34 cm, OB = 20 cm and OP = 16 cm; find the length of AB.

• A. 20 cm
• B. 18 cm
• C. 30 cm
• D. 10 cm

Answer 1

The correct option is B

18 cm

In right $\Delta OAP$,

$O{A}^2=O{P}^2+A{P}^2$ (Pythagoras Theoram)

$\Rightarrow (34{)}^2=(16{)}^2+A{P}^2\Rightarrow 1156=256+A{P}^2$

$\Rightarrow A{P}^2=1156-256=900=(30{)}^2$

$\therefore$ AP = 30 cm.

Similarly in right $\Delta OBP$,

$\Rightarrow O{B}^2=O{P}^2+B{P}^2$

$\Rightarrow (20{)}^2=(16{)}^2+B{P}^2\Rightarrow 400=256+B{P}^2$

$\Rightarrow B{P}^2=400-256=144=(12{)}^2$

$\therefore$ BP = 12 cm

Now AB = AP - BP = 30 - 12 = 18 cm