Question

6. In a circle of radius 5 cm,
AB and AC are two chords
such that AB = AC = 6 cm,
as shown in the figure.
Find the length of the chord
BC.

Answer in full steps
answer fast​

Answer 1

MATHS

In a circle of radius 5 cm , AB and AC are two chords such that AB = AC = 6 cm . Find the length of the chord BC .

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,

AB2=AP2+BP2

BP2=36−AP2 ....(1)

In △ OBP, we have

OB2=OP2+BP2

52=(5−AP)2+BP2

BP2=25−(5−AP)2 .....(2)

From 1 & 2, we get,

36−AP2=25−(5−AP)2

36=10AP

AP=3.6cm

Substitute in equation 1,

BP2=36−(3.6)2=23.04

BP=4.8cm

Answer 2

Answer:

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