Question

In Fig. 3, if O is the centre of the circle ; OL$=$4 cm, AB $=$6 cm and OM$=$ 3 cm,then CD$=$?

• A. 4cm
• B. 8cm
• C. 6cm
• D. 10cm

Answer 1

Answer: (B)

8cm

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.