Question

In the given figure ,O is the centre of the circle and chord AB is parallel to CD. If AB is 18 cm and CD is 24 cm and distance between chords i.e PQ is 21 cm , then the radius of the circle is

Answer 1

Given: Chord AB is parallel to CD, AB is 18 cm and CD is 24 cm, PQ is 21 cm

To find: Radius of the circle=?

Solution:

• Let the radius of the circle be r, let OQ = x and OP = 21- x.
• Join OA and OC, So, OA = OC = r.
• Since perpendicular drawn from the centre of the circle bisects the chord.
• So, CQ = QD =12cm, and AP = PB = 9cm.
• Now in triangle OQC, using pythagoras theorem, we get:

          CQ² + QO² = OC²

          12² + x² = r²    .............(i)

• In triangle OPA, we get:

          OP² + PA² = OA²

          (21-x)² + 9² = r²     ................(ii)

• From i and ii, we get:

          144 +x² = 441 + x² - 42x + 81

• Cancelling x from both sides, we get

          42x = 441 + 81 - 144

          42x = 378

          x = 378/42

          x = 9

• Put x = 9 in eqn i we get:

          144 + 81 = r²

          r² = 225

          r = 15 cm

Answer:

              The radius of the circle is 15 cm.