Two chords AB and CD of lengths 5 cm 11cm 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 cm, find the radius of the circle.

#### Solution

Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.

BM = AB/2 = 5/2 (Perpendicular from the centre bisects the chord)

ND = CD/2 = 11/2

Let ON be *x*. Therefore, OM will be 6− *x*.

In ΔMOB,

OM^{2} + MB^{2} = OB^{2}

(6 - x)^{2} + (5/2)^{2} = OB^{2}

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

In ΔNOD,

ON^{2} + ND^{2} = OD^{2}

x^{2} + (11/2)^{2} = OD^{2}

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

We have OB = OD (Radii of the same circle)

Therefore, from equation (1) and (2),

`36+x^2-12x+25/4=x^2+121/4`

`12x=36+24/4-121/4`

`=(144+25-121)/4`

`=48/4`

**= 12**

x = 1

From equation (2),

`(1)^2+(121/4)=OD^2`

`OD^2 = 1+121/4=125/4`

`OD=5/2sqrt5`

Therefore, the radius of the circle is `5/2sqrt5" cm."`