8. AB and CD are two chords of a circle such that
AB = 6 cm, CD = 12 cm and AB || CD. If the distance
between AB and CD is 3 cm, find the radius of the
circle.
Answers
Answer:
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Step-by-step explanation:
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. Draw 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
OA^2 = OP^2 + AP^2 and OC^2 = OQ^2 + CQ^2
⇒ r^2 = (x + 3)^2 + 3^2 and r^2 = x^2 + 6^2
[∵ AP = ½ AB = 3 cm and CQ = ½ CD = 6 cm
⇒ (x + 3)^2 + 3^2 = x^2 + 6^2 (on equating the value of r2)
⇒ x^2 + 6x + 9 + 9 = x^2 + 36
⇒ 6x = 18 ⇒ x = 3 cm
Putting the values of x in r^2 = x^2 + 6^2, we get
r^2 = 3^2 + 6^2 = 4^5
⇒ r = √45 cm = 6.7 cm
Hence, the radius of the circle is 6.7 cm.
