question_answer

O is the centre of the circle having radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. If OA meets BC at P, then OP = ____.     

A)  3.6 cm                        

B)         1.4 cm                        

C)         2 cm                           

D)         3 cm

Correct Answer: B

Solution :

We have, AB = AC = cm \[OB=OC=5\,cm\] \[\therefore \]ABOC is a kite. Since, diaconates of kite are perpendicular of one another. \[\angle OPB=\angle OPC={{90}^{o}}\] Let \[OP=x\]\[\therefore \]\[AP=OA-OP=5-x\]       ?(i) In \[\Delta {\mathrm O}PB,{{(OB)}^{2}}={{(OP)}^{2}}+{{(PB)}^{2}}\] \[\Rightarrow \]\[{{(5)}^{2}}={{x}^{2}}+{{(PB)}^{2}}\Rightarrow {{(PB)}^{2}}=25-{{x}^{2}}\]     ?(ii) In \[\Delta \Alpha PB,{{(AB)}^{2}}={{(AP)}^{2}}+{{(PB)}^{2}}\] \[\Rightarrow \]\[{{(6)}^{2}}={{(5-x)}^{2}}+{{(PB)}^{2}}\] \[\Rightarrow \]\[{{(PB)}^{2}}={{(6)}^{2}}-{{(5-x)}^{2}}\]                             ?(iii) From (ii) and (iii), we have \[25-{{x}^{2}}={{(6)}^{2}}-{{(5-x)}^{2}}\] \[\Rightarrow \]\[25-{{x}^{2}}=36-25-{{x}^{2}}+10x\] \[\Rightarrow \]\[14=10x\Rightarrow x=1.4\,cm\]