Question

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.

Answer 1

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,

OM² + MB² = OB²

(6 - x)² + (5/2)² = OB²

36 + x² - 12x + 25/4 = OB²                 ........(1)

In ΔNOD,

ON² + ND² = OD²

x² + (11/2)² = OD²

x² + 121/4 = OD²                          .........(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."`