Question

Given a chord AB of length 5 cm, of a circle with centre O. OL is perpendicular to chord AB
and OL = 4 cm. OM is perpendicular to chord CD such that OM = 4 cm. Then CM is equal to

Answer 1

Step-by-step explanation:

Radius of a circle is 5 cm.

Therefore OB = 5 cm.

Perpendicular drawn to the chord from the centre bisects the chord.

Therefore AM = MB

Triangle OMB is right angled triangle at M

∠OMB=90

∘

By Pythagoras theorem;

OB

2

=OM

2

+MB

2

MB=

OB

2

−OM

2

MB=

5

2

−4

2

MB=3cm

So, the length of chord AB=2.(MB)=6cm.