Question

In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length. Then, AB = CD

• A. True
• B. False

Answer 1

The correct option is A

True

Given: Two concentric circles with centre O AB and CD are two cords of outer circle which touch the inner circle at P and Q respectively

Construction : $\because$ OP and OQ are the radii of the inner circle and AB and CD are tangents

$\therefore OP\perp AB$ and $OQ\perp CD$

and P and Q are the midpoints of AB and CD Now right $\Delta OAP$ and OCQ,

Side OP= OQ (radii of the inner circle)

Hyp. OA = OC (radii of the outer circle)

$\therefore \Delta OAP\cong \Delta OCQ$ (R.H.S. axiom)

$\therefore$ AP= CQ (c.p.c.t.)

But AP = $\frac{1}{2}$ AB and CQ = $\frac{1}{2}$ CD

$\therefore$ AB=CD Hence proved.