Two circles of radii $5cm$ and $3cm$ intersect at two points and the distance between their centres is $4cm$. Find the length of common chord.

The correct option is (B): $6cm$

We know that if two circles intersect each other at $2$ points, then the line joining their centres is the perpendicular bisector to their common chord.

In right-angled $\Delta ADC$, applying Pythagoras Theorem, we have

${AC}^2={AD}^2+{CD}^2$

or, $25=x^2+p^2\Rightarrow p^2=25-x^2---------------\left(1\right)$

In right-angled $\Delta CDB,$ we have

${BC}^2={CD}^2+{BD}^2$

or, $9=p^2+{(4-x)}^2---------------\left(2\right)$

From equation (1), we have

$\Rightarrow 9=25-x^2+16+x^2-8x$ [using equation (1)]

$\Rightarrow 8x=41-9$

$\Rightarrow 8x=32$

$\Rightarrow x=\frac{32}{8}$

$\Rightarrow x=4$

Now again from equation (1), we have

$p^2=25-x^2$

$\Rightarrow p^2=25-4^2$

$\Rightarrow p^2=25-16$

$\Rightarrow p^2=9$

$\Rightarrow p=3$

Hence, length of the common chord $=2p=2\times 3cm=6cm$.