Question

If 'O' is the centre of the circle ; OL $=$ 4 cm, AB $=$ 6 cm and OM $=$ 3 cm, then CD $=$___________.

• A. 4 cm
• B. 8 cm
• C. 6 cm
• D. 10 cm

Answer 1

Answer: (B)

8 cm

Perpendicular from the center of a circle to any chord bisects the chord, so

$LB=\frac{AB}{2}=3cm$

In $△OLB$, Pythagoras' theorem gives:

$OB=\sqrt{O{L}^2+L{B}^2}=5cm$

Clearly, $OB=OC$, so in $△OCM$, Pythagoras' theorem gives:

$CM=\sqrt{O{C}^2-O{M}^2}=4cm$

Perpendicular from center of a circle to any chord bisects the chord, so

$CD=2\times CM=8cm$

So option B is the right answer.