Question

12. In a circle of radius 25 cm, AB and AC are two chords, such that AB = AC = 30 cm . Find the
length of the chord,
(a) 40 cm (b) 48 cm (c) 60 cm (d) 50 cm​

Answer 1

Answer:

ANSWER

AB and AC are two equal chord of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

OA is the bisector of ∠BAC.

Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.

P divides BC in the ratio 6:6=1:1.

P is mid-point of BC.

OP ⊥ BC.

In △ ABP, by pythagoras theorem,

AB

2

=AP

2

+BP

2

BP

2

=36−AP

2

....(1)

In △ OBP, we have

OB

2

=OP

2

+BP

2

5

2

=(5−AP)

2

+BP

2

BP

2

=25−(5−AP)

2

.....(2)

From 1 & 2, we get,

36−AP

2

=25−(5−AP)

2

36=10AP

AP=3.6cm

Substitute in equation 1,

BP

2

=36−(3.6)

2

=23.04

BP=4.8cm

BC=2×4.8=9.6cm