Question

Prove that of any two chords of a circle, the one which is greater is nearer the centre.

Answer 1

Let, AB be the greater chord and Cd be the smaller chord.

O is the centre of the circle

P is the mid-point of AB and Q is the mid-point of CD

Join OP and OQ.

Now, come to the calculations_

Given: O is the centre of the circle

          AB>CD, OP⊥AB and OQ⊥CD.

To prove: OP<OQ

Construct: Join OA and OC.

Proof:

In rt.∠d ΔAPO,

AP² + OP² = AO²

In rt.∠d ΔCOQ,

CQ²+OQ² =OC²

But, AO = OC (radii of the same circles)

∴ AO²= OC²

∴AP²+OP² = CQ²+OQ²

But, AP² > CQ² (∵AB>CD)

∴ OP² <OQ²

⇒ OP < OQ (Is proved).

Answer 2

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Please refer to attachment ☺️