Question

Jana
AB and CD ARE two Chords
two choods of a a Circle
such that AB abem, CD = 12 cm, and AB
11 60 18 the distance between AB and
Co is 3cm. Find the radius of the circle
​

Answer 1

Answer:

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm CD = 12 cm. Let the radius of the circle be r cm. Drwa OP ⊥ AB and OQ ⊥ CD. Since,  AB ∥ CD and OP ⊥ AB, OQ ⊥ CD. Therefore points  O, Q  and P are collinear. Clearly, PQ = 3 cm.

Let OQ = x cm. Then, OP = (x + 3) cm

In right triangles OAP and OCQ, we have

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

⇒ r2 = (x + 3)2 + 32   and  r2 = x2 + 62

[∵  AP = ½ AB = 3 cm and CQ = ½ CD = 6  cm

⇒  (x + 3)2 + 32 = x2 + 62      (on equating the value of r2)

⇒  x2 + 6x + 9 + 9  = x2 + 36

⇒  6x = 18  ⇒  x = 3 cm

Putting the values of x in r2 = x2 + 62, we get

r2 = 32 + 62 = 45

⇒  r = √45 cm = 6.7 cm

Hence, the radius of the circle is 6.7 cm.

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Answer 2

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Answer:

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm CD = 12 cm. Let the radius of the circle be r cm. Drwa OP ⊥ AB and OQ ⊥ CD. Since,  AB ∥ CD and OP ⊥ AB, OQ ⊥ CD. Therefore points  O, Q  and P are collinear. Clearly, PQ = 3 cm.

Let OQ = x cm. Then, OP = (x + 3) cm

In right triangles OAP and OCQ, we have

OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2

⇒ r2 = (x + 3)2 + 32   and  r2 = x2 + 62

[∵  AP = ½ AB = 3 cm and CQ = ½ CD = 6  cm

⇒  (x + 3)2 + 32 = x2 + 62      (on equating the value of r2)

⇒  x2 + 6x + 9 + 9  = x2 + 36

⇒  6x = 18  ⇒  x = 3 cm

Putting the values of x in r2 = x2 + 62, we get

r2 = 32 + 62 = 45

⇒  r = √45 cm = 6.7 cm

Hence, the radius of the circle is 6.7 cm.