Two circles of radii 5 cm and 3 cm intersect each other at A and B, and the distance between their centres is 6 cm. The length (in cm) of the common chord AB is:
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 
| Understanding | Application |
| Tricks |
According to question
Here Oo = 6 cm OB = 5 oB = 3 & let ot = x cm |
| Now |
When AB intersects the Oo it makes of 90 From the right angle triangle At2 = AO2 - Ot2 .....i) & At2 = Ao2 - ot2 .........ii) |
| Hence |
AO2 - Ot2 = Ao2 - ot2 {from eq i) and eq ii)} ⇒ 52 - (6 - x)2 = 32 - x2 ⇒ 25 - (6 - x)2 = 9 - x2 ⇒ 25 - 36 + 12x - x2 = 9 - x2 ⇒ -11 + 12x = 9 ⇒ 12x = 9 + 11 ⇒ x = 20/12 ⇒ x = 5/3 |
| Hence |
At2 = 9 - x2 {from eq ii)} ⇒ At2 = 9 - 25/9 ⇒ At2 = (81 - 25)/9 = 56/9 So AB = 56 × 2/9 = 112/9 So
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