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In a circle of radius 10 cm, with centre O, PQ and PR are two chords each of length 12 cm. PO intersects chord QR at the points S. The length of OS is:
Answer & Solution
Correct Answer:
Option
A

QS2 = 122 - (10 - x)2 = 102 - x2
122 - 102 = (10 - x)2 - x2
22 × 2 = 10(10 - 2x)
4.4 = 10 - 2x
2x = 5.6
x = 2.8 cm
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