Question

Chords AB and CD of a circle are parallel to each other and lie on opp sides of the centre of the circle.if AB=36cm,CD=48cm and the distance between the chords is 42cm find the radius

Answer 1

Radius = 30 cm

Step-by-step explanation:

Given:

Here, chords AB = 6 cm, CD = 12 cm and AB║CD    

Draw OP⊥ AB. Let it intersect CD at Q and AB at P

∴ AP = PB = 18 cm and CQ = DQ = 24 cm [Since perpendicular draw from the centre of the chord bisects the chord]

Let OD = OB = r

In right Δ OQD,

$OD^2=OQ^2+QD^2$

$r^2 = x^2 + 24^2$    [By Pythagoras theorem]

$r^2 = x^2 + 576$                       (1)

In right ΔOPB,  

$OB^2=OP^2+PB^2$             [given distance between the chords, QP = 42 cm ]

$r^2 = (x + 42)^2 + 18^2$       [By Pythagoras theorem]

$r^2 = x^2 + 84x+1764 + 324$

$r^2 = x^2 + 84x+2088$           (2)

From (1) and (2) we get  

$x^2 + 576 = x^2 + 84x + 2088$

⇒ 576 = 84x + 2088

⇒ 84x = 576 - 2088

⇒ 84 x = -1512

⇒ $x = -\frac{1512}{84}$

⇒ x = - 18

Put x = - 18 in (1), we get

$r^2 = (-18)^2 + 576$

$r^2 = 324 + 576$

$r^2 = 900$

$r = \sqrt{900}$

r = 30 cm

Therefore radius = 30 cm.