Q6 of 63 Page 145

In figure, O and O’ are the centre of the given circle. AB || OO’. Prove that AB = 2 OO’.

Given O and O’ are the centres of the given circles.


Also AB || OO’


We have to prove that AB = 2OO’.


Construction:


Draw perpendicular CP from point C on OO’.



Proof:


We know that perpendicular from the centre of a circle to a chord bisects the chord.


⇒ BE = EC and CD = DA


Since AB || OO’ and O’E || PC || OD, [since all are perpendiculars on line AB]


⇒ EC = O’P and CD = PO


⇒ OO’ = OP + PO’


= CD + EC


= 1/2 BC + 1/2 AC


= 1/2 (BC + AC)


= 1/2 AB


∴ 2OO’ = AB


Hence proved.


More from this chapter

All 63 →
4

In figure, two equal chords AB and CD intersect each other at E. Prove that arc DA = arc CB.

 5

In figure, AB and CD are equal chords of a circle. O is the centre of the circle of OM ⊥ AB and ON ⊥ CD. Then prove that ∠OMN = ∠ONM.

 7

AB and CD are two chords of a circle such that AB = 10 cm, CD = 24 cm and AB || CD. The distance between AB and CD is 17 cm. Find the radius of the circle.

 8

In a circle of radius 10 cm, the lengths of two parallel chords are 12 cm and 16 cm respectively. Find the distance between AB and CD. If chords (a) are on the same side of the centre (b) are on the opposite sides of the centre.