Q17 of 31 Page 65

AB and AC are two equal chords of a circle having the radius of 5 cm. The centre of the circle

is situated at the outside of the triangle ABC. If AB = AC = 6 cm. then let us calculate the length of the chord BC.


Given, AC = AB, BAF = CAF, AB and AC are two equal Chords of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.


In ABF and CAF


⇒ AC = AB


⇒ BAF = CAF


⇒ AF = AF


⇒ ABF CAF


⇒ BFA = CFA


⇒ FB = FC


AE is a perpendicular bisector of chord BC.


In ΔABF, by Pythagoras theorem,


⇒ AB2 = AF2 + BF2


⇒ BF2 = 62 - AF2 .............(1)


In OBF


⇒ OB2 = OF2 + BF2


⇒ 52 = (5 - AF)2 + BF2


⇒ BF2 = 25 - (5 - AP)2 ...........(2)


Equating (1) and (2), we get


⇒ 62 - AF2 = 25 - (5 - AF)2


⇒ 11 - AF2 = -25 - AF2 + 10AF


⇒ 36 = 10AF


⇒ AF = 3.6 cm


Putting AF in (1), we get


⇒ BF2 = 62 - (3.6)2 = 23.04


⇒ BF = 4.8 cm


⇒ BC = 2BF = 2 × 4.8 = 9.6 cm


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Let us fill in the blanks:

i. If the ratio of two chords PQ and RS of a circle with its centre O is 1 : 1, then ∠POQ: ∠ROS=_______________


ii. The perpendicular bisector of any chord of a circle is __________ of that circle.

 17

Two equal circles of radius 10 cm. intersect each other and the length of their common chord is 12 cm. Let us determine the distance between the two centers of two circle

 17

The lengths of two chords AB and CD of a circle with its centre O are equal. If ∠AOB = 60° and CD = 6 cm. then let us calculate the length of the radius of the circle.

 17

P is any point in a circle with its centre O. If the length of the radius is 5 cm. and OP = 3 cm., then let us determine the least of the chord passing through the point P.