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 \[25-{{x}^{2}}={{(6)}^{2}}-{{(5-x)}^{2}}\] \[\Rightarrow \]\[25-{{x}^{2}}=36-25-{{x}^{2}}+10x\] \[\Rightarrow \]\[14=10x\Rightarrow x=1.4\,cm\]You need to login to perform this action.
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