question_answer

In a circle of radius 4 cm with centre at O, AB and CD are two diameters perpendicular to each other. The length of chord BD is

A) \[4\,cm\]

B) \[4\sqrt{2}\,cm\]

C) \[4\sqrt{3}\]

D) None of these

Correct Answer: B

Solution :

[b] \[\Delta BOD\]is a right angled triangle as. \[\angle BOD=90{}^\circ \] So,       \[DO=BO=4\] and \[B{{D}^{2}}=D{{O}^{2}}+B{{O}^{2}}\] \[BD=\sqrt{{{4}^{2}}+{{4}^{2}}}=4\sqrt{2}\,cm\]