Get it on Google Play
1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
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
image
In \(\triangle{OO'A}\), we have,
\({AO'}^2 + {OO'}^2 = {3}^2 + {4}^2 = 9 + 16 = 25\)
i.e., \({OA}^2 = {5}^2 = {AO'}^2 + {OO'}^2\)
Therefore, OO'A is a right angled traingle and right angled at O'

Now, Area of \(\triangle{OO'A}\) = 1\2 × O'A × OO'
= 1\2 × 3 × 4
Area of \(\triangle{OO'A}\) = 6 sq.cm ...(i)
Also, Area of \(\triangle{OO'A}\) = 1\2 × OO' × AM
= 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.