Question

Two circle of radii 5 cm intersect of two point and the distance between their centre is 4 cm. Find the length of the common chord

Answer 1

$\cos B=\frac{(5{)}^2+(4{)}^2-(5{)}^2}{2\times 5\times 4}$ (cosine rule)

$\cos B=\frac{16}{2\times 5\times 4}$

$\Rightarrow \cos B=\frac{2}{5}$

$\Rightarrow \sin B=\frac{\sqrt{21}}{5}$

Now,

In $\Delta ABC$

$\cos B=\frac{BD}{AB}=\frac{BD}{5}$

$\Rightarrow \frac{2}{5}=\frac{BD}{5}$

$\Rightarrow BD=2$cm

$\Rightarrow \sin B=\frac{AD}{AB}$

$\Rightarrow \frac{\sqrt{21}}{5}=\frac{AD}{5}$

$\therefore AD=\sqrt{21}$cm

$\therefore$ Length of common chord$=2\times AD$

$=2\sqrt{21}$cm.