question_answer

Two equal circles of radii 6 cm intersect each other such that each passes through the centre of the other. The length of the common chord is -

A)  \[6\sqrt{3}\,cm\]                       

B)  \[6\,cm\]

C)  \[5\sqrt{3}\,cm\]                       

D)  \[5\,cm\]

Correct Answer: A

Solution :

Let two circles with centre A and B. Here \[\Delta \,AOC\cong \Delta BOC\] \[\therefore \,\,\,\,\,AO=BO\Rightarrow AO=BO=\frac{6}{2}=3\,cm\] \[AC=6\text{ }cm\] (Radius of circle) \[\therefore \,\,\,\,\,O{{C}^{2}}=A{{C}^{2}}-A{{O}^{3}}=36-9=27\] \[\therefore \,\,\,OC=\sqrt{27}=3\sqrt{3}\,cm\] \[\therefore \]   Length of common chord (CD) \[=2\times 3\sqrt{3}=\,\underline{\mathbf{6}\sqrt{\mathbf{3}}\,\mathbf{cm}}\]