Mathematics
The adjoining figure shows a circle with centre O in which a diameter AB bisects the chord PQ at point R. If PR = RQ = 8 cm and RB = 4 cm, find the radius of the circle.
Circles
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Answer
Join OQ.
Radius = OQ = OB = x (let)
⇒ OR = OB - RB = (x - 4) cm.
We know that,
A straight line drawn from the center of a circle to bisect a chord, is perpendicular to the chord.
Since, PR = QR, thus R is mid-point of PQ.
∴ OR ⊥ PQ
In right-angled triangle OQR,
By pythagoras theorem,
⇒ Hypotenuse2 = Perpendicular2 + Base2
⇒ OQ2 = OR2 + QR2
⇒ x2 = (x - 4)2 + 82
⇒ x2 = x2 + 42 - 2 × x × 4 + 64
⇒ x2 = x2 + 16 - 8x + 64
⇒ x2 - x2 + 8x = 80
⇒ 8x = 80
⇒ x = = 10 cm.
Hence, radius of circle = 10 cm.
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