Two circles of radii 7 cm and 5 cm intersect each other at A and B, and the distance between their centres is 10 cm. The length (in cm) of the common chord AB is:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e (4√(66))/5.
| Understanding/Application |
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AP = 5 cm, AQ = 7 cm, PQ = 10 cm Let CQ be x cm. PC = (10 - x) Triangle ACQ is a right-angled triangle. AQ2 = AC2 + CQ2 ⇒ 72 = AC2 + x2 ⇒ AC2 = 49 - x2 .....(1) Triangle APC is a right-angled triangle. AP2 = AC2 + PC2 ⇒ 52 = AC2 + (10 - x)2 ⇒ AC2 = 25 - (10 - x)2 .....(2) |
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From eq 1 and 2, we get - ⇒ 49 - x2 = 25 - (10 - x)2 ⇒ 24 - x2 = - (10 - x)2 ⇒ x2 = 24 + 100 + x2 - 20x ⇒ 20x = 124 ⇒ x = 6.2 cm From eq (1), we get - AC2 = 49 - (6.2)2 ⇒ AC2 = 49 - 38.44 ⇒ AC2 = 10.56 = 1056/100 = 264/25 ⇒ AC = (2√66)/5 Chord AB = 2(2√66)/5 = (4√66)/5 |
