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
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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
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