In the given figure, boc is a diameter of a circle with centre O if AB and CD are two chords

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Question 19 Circles Exercise 12A

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Answer:

Construct OL ⊥ AB and OM ⊥ CD

Consider △ OLB and △ OMC

We know that ∠OLB and ∠OMC are perpendicular bisector

∠OLB = ∠OMC = 90

We know that AB || CD and BC is a transversal

From the figure we know that ∠OBL and ∠OCD are alternate interior angles

∠OBL = ∠OCD

So we get OB = OC which is the radii

By AAS congruence criterion

△ OLB ≅ △ OMC

OL = CM (c. p. c. t)

We know that the chords equidistant from the centre are equal

So we get

AB = CD

Therefore, it is proved that AB = CD.

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In the given figure, BOC is a diameter of a circle with centre O.

Question:

In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. If AB = 10 cm, then CD = ?
(a) 5 cm
(b) 12.5 cm
(c) 15 cm
(d) 10 cm

Solution:

(d) 10 cm

Draw OE ⊥ AB and OF ⊥ CD.
In Δ OEB and ΔOFC, we have:
OB =  OC              (Radius of a circle)
∠BOE = ∠COF     (Vertically opposite angles)
∠OEB = ∠OFC     (90° each)
∴ ΔOEB ≅ ΔOFC (By AAS congruency rule)
∴ OE = OF
Chords equidistant from the centre are equal.
∴ CD = AB = 10 cm

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In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB||CD and AB = 10 cm then CD = ? a 5 cm b 12.5 cm c 15 cm d 10 cm