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:
Show answer & explanation
Correct answer: D
Explanation
Correct answer is option 4 i.e. 2.8 cm
| Concept |
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Radius always bisect the chord at 90° angle. |
| Understanding |
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Radius OQ = OP = 10 cm, PQ = PR = 12 cm Hence, PS will bisect QR. So, ∠PSQ = ∠OSQ = 90° Let PS = x then OS = 10 - x |
| Application |
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According to Pythagoras theorem, 122 - x2 = 102 - (10 - x)2 ⇒ 144 - x2 = 100 – 100 – x2 + 20x ⇒ 20x = 144 ⇒ x = 144/20 ⇒ x = 7.2 OS = 10 – 7.2 = 2.8 cm |
