Q. 214.0( 5 Votes )

*AB* and *CD* are two parallel chords of a circle with centre *O* such that *AB*=6 cm and *CD*= 12 cm. The chords are on the same side of the centre and the distance between them in 3 cm. The radius of the circle is

A. 6 cm

B. 5 cm

C. 7 cm

D. 3 cm

Answer :

Given that,

AB || CD (Chords on same side of centre)

AO = CO (Radii)

OL and OM perpendicular bisector of CD and AB respectively

CL = LD = 6 cm

AM = MB = 3 cm

LM = 3 cm (Given)

In COL,

CO^{2} = OL^{2} + 6^{2} (i)

In AOM,

AO^{2} = AM^{2} + OM^{2}

= ౩^{2} + (OL + LM)^{2}

= 9 + OL^{2} + 9 + 6O L

OL^{2} = AO^{2} - 18 - 6O L (ii)

Using (ii) in (i),

OL = 3 cm

Putting OL in (i),

AO^{2} =

AO = 35

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PREVIOUSIf A, B, C are three points on a circle with centre O such that ∠AOB=90° and ∠BOC=120°, then ∠ABC=NEXTIn a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

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