question_answer

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm then the distance of AB from the centre of the circle is ____.

A)  17 cm                         

B)         8 cm               

C)         4 cm                           

D)         15 cm             

Correct Answer: B

Solution :

Given, AD = 34 cm,  AB = 30  cm \[\therefore \]\[AO=\frac{1}{2}AD=\frac{1}{2}(34)=17\,cm\] Draw \[OP\bot AB.\] Since, perpendicular drawn center to the chord bisects the chord. \[\therefore \]      \[AP=\frac{1}{2}AB=15\,cm\] Now, in right angled \[\Delta APO\] \[{{(OP)}^{2}}={{(17)}^{2}}-{{(15)}^{2}}\] \[=289-225=64\] \[\Rightarrow \]\[OP=8\,cm\] \[\therefore \]Distance of AB from the center of the circle is 8 cm.