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Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.

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Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.

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Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.

We know that AB bisects BM as the perpendicular from the centre bisects chord.

Since AB = 5

BM = AB/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

OM = 6−x.

Consider ΔMOB,

OB2 = OM2 + MB2

\(OB^2 = 36 + x^2 - 12x + \frac{25}{4}\) .......(1)

Consider ΔNOD,

OD= ON+ ND2

\(OD^2 = x^2 + \frac{121}{4}\) ......(2)

We know, OB = OD (radii)

From equation 1 and equation 2 we get

\(36 + x^2 - 12x + \frac{25}{4} = x^2 + \frac{121}{4}\)

12x = \(36 + \frac{25}{4} - \frac{121}{4}\)

= \(\frac{144+25-121}{4}\)

12x = \(\frac{48}{4}\) = 12

x = 1

Now, from equation (2) we have,

OD2= 1+ \(\frac{121}{4}\)

OD = (5/2) × √5 cm

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