Question

If AB = 10 cm, CD = 24 cm are two chords of a circle. AB || CD and distance between AB and CD is 17 cm. Find the radius of circle.

Answer 1

Let O be center and r be radius of the circle. 

OP ⊥ AB and OQ ⊥ AB

∵ AB || CD so P, O, Q will lie on same line

PQ = 17 cm

AB = 10 cm

CD = 24 cm

∵ OP ⊥ AB and

OQ ⊥ CD

∴ AP = BP = \(\frac { 1 }{ 2 }\)AB

= \(\frac { 1 }{ 2 }\) × 10 = 5 cm

and CQ = QD = \(\frac { 1 }{ 2 }\) × CD

= \(\frac { 1 }{ 2 }\) × 24 = 12 cm

∵ PQ = 17 cm

Let OP = x cm then OQ = 17 – x cm

In right angled ∆OPA

by Pythagoras theorem

OA² = OP² + AP²

r² = x² + (5)² = x² + 25 …(i)

Similarly on right angled ∆OQC

OC² = OQ² + CQ²

r² = (17 – x)² + (12)²

= (17 – x)² + 144 …(ii)

Equating value of r’ from equations (i) and (ii)

x² + 25 = (17 – x)² + 144

= 289 – 34x + x² + 144

25 = 289 – 34x + 144

34x = 289 + 144 – 25

= 433 – 25

34x = 408

x = 12 cm

Putting value of x in equation (i)

r² = x² + 25 = (12)² + 25

= 144 +25

r² = 169

r = \(\sqrt { 169 }\) = 13 cm

Thus, radius of circle = 13 cm.