Page 3 - Lesson note 4 Equal Chords-equidistant- circle Ch.- 10 Circle

Given :A circle with centre O , OP⊥ AB and OQ ⊥ CD
OP=OQ
To Prove: AB=CD
Const.
Join OA and OC
Proof:
In Δ’s OPA and OQC,
OP=OQ (given)
OA = OC (Radii of same circle)
∠OPA = ∠OQC = 90⁰ (given)
∴ ΔOPA ≅ ΔOQC (By RHS congruence criterion)
Hence AP=CQ (CPCT)
Now 2AP= 2CQ
AB=CD (The perpendicular from the centre of a circle to a chord bisects the chord).

Hence Proved.

Example 1.
Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If
the distance between AB and CD is 3 cm, find the radius of the circle.
Solution:
Given: AB = 5 cm, CD = 11 cm, PQ = 3 cm
Draw perpendiculars OP on CD and OQ on AB
Let OP = x cm and OC = OA = r cm

We know, perpendicular from centre to chord bisects it.
Since OP⊥CD, we have
CP = PD = 11/2 cm
And OQ⊥AB
AQ = BQ = 5/2 cm
In ΔOCP:
2
2
OC = OP + CP 2 By Pythagoras theorem
2
2
2
r = x + (11/2) …..(1)
In ΔOQA:
2
2
OA =OQ +AQ 2 By Pythagoras theorem,