Question

In the figure, o is the centre of the circle OA = 75 cm
OP=45 cm Seq OP perpendicular Chord AB Find length of seg AB​

Answer 1

Answer:

Given O is the centre of the circle.

O P \perp A BOP⊥AB

where AB is a chord of the circle.

Perpendicular from centre bisects the chord.

⇒ AL = LB

Similarly, O Q \perp A COQ⊥AC

where AC is a chord of the circle.

Perpendicular from centre bisects the chord.

⇒ AM = MC

But AB = AC

$\Rightarrow \frac{A B}{2}=\frac{A C}{2}

⇒ LB = MC

Now, OP = OQ (Radii of same circle)

Equal chords are equidistant from centre.

⇒ OL = OM

Then, OP – OL = OQ – OM

⇒ LP = MQ

In triangles LPB and MQC,

LB = MC (side)

LP = MQ (side)

∠PLB = ∠QMC = 90° (angle)

Therefore, ΔLPB ≅ ΔMQC (by SAS congruence rule)

Corresponding parts of congruence triangles are congruent.

⇒ PB = QC

Hence proved