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Answer :

Let O and O' be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

Given: OA = 5cm, O'A = 3cm and OO' = 4cm

In \(\triangle{OO'A}\), we have,

\({AO'}^2 + {OO'}^2 = {3}^2 + {4}^2 = 9 + 16 = 25\)

\(\Rightarrow \) \({OA}^2 = {5}^2 = {AO'}^2 + {OO'}^2\)

Therefore, OO'A is a right angled traingle and right angled at O'

Now, Area of \(\triangle{OO'A}\) = \(\frac{1}{2} \) × O'A × OO'

= \(\frac{1}{2} \) × 3 × 4

Area of \(\triangle{OO'A}\) = 6 sq.cm ...(i)

Also, Area of \(\triangle{OO'A}\) = \(\frac{1}{2} \) × OO' × AM

= \(\frac{1}{2} \) × 4 × AM

Area of \(\triangle{OO'A}\) = 2 × AM ...(ii)

From eq. (i) and (ii), we get,

2 × AM = 6

Therefore, AM = 3 cm.

Since, when two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord.

Thus, AB= 2 × AM = 2 × 3 = 6cm.

Therefore, the length of the common chord is 6cm.

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