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:

$\frac{4\sqrt{14}}{3}$

Given, AP = 5 cm, AQ = 3 cm and PQ = 6 cm

Let PM = x,

= MQ = (6 - x)

In \(\Delta \)AMP

\( {(PA) }^{2 } \) = \( {(AM) }^{2 } \) + \( {(PM) }^{2 } \)

= \( {5 }^{2 } \) = \( {(AM) }^{2 } \) + \( {(X }^{2 } \)

= \( {(AM) }^{2 } \) = 25 - \( {X }^{2 } \)      ..(1)

In \(\Delta \)AQM,

\( {(AQ) }^{2 } \) = \( {(AM) }^{2 } \) + \( {(6\;-\;X) }^{2 } \)

= \( {3 }^{2 } \) = \( {(AM) }^{2 } \) + 36 + \( {X }^{2 } \) - 12X

= \( {(AM) }^{2 } \) = 9 - 36 - \( {X }^{2 } \) + 12X      ..(2)

From eq(1) and eq(2)

25 - \( {X }^{2 } \) = 9 - 36 - \( {X }^{2 } \) + 12X

12X = 25 + 27

12X = 52

X = \(\frac{52}{12}\)

X = \(\frac{13}{3}\)

From equation (1)

\( {(AM) }^{2 } \) = 25 - \( {(\frac{13}{3}) }^{2 } \) = 25 - \(\frac{169}{9}\) = \(\frac{225\;-\;169}{9}\) = \(\frac{56}{9}\)

AM = \(\frac{\sqrt {56 }}{9}\) = \(\frac{2\sqrt {14 }}{3}\)

As we know,

AB = 2AM = 2 x  [\(\frac{2\sqrt {14 }}{3}\)] = \(\frac{4\sqrt {14 }}{3}\)

Therefore, AB is \(\frac{4\sqrt {14 }}{3}\).