Two light rods \[AB=a+b\]and \[CD=a-b\]symmetrically lying on a horizontal. There are kept intact by two strings AC and BD. The perpendicular distance between rods is a. The length of AC is given by |
A) \[{{a}^{2}}+{{b}^{2}}\]
B) \[{{a}^{2}}-{{b}^{2}}\]
C) \[\sqrt{{{a}^{2}}-{{b}^{2}}}\]
D) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]
Correct Answer: D
Solution :
Since they are symmetrically lying on horizontal plane. |
\[\therefore \] \[AC=BD\] |
\[\therefore \] \[AE=BF=x\] |
Now, \[AB=(a-b)+2x\] |
i.e. \[a+b=a-b+2x\]\[\Rightarrow \]\[2b=2x\] |
\[\therefore \] \[x=b\] |
Now, in \[\Delta ACE,\]\[{{x}^{2}}+{{a}^{2}}=A{{C}^{2}}\] |
\[\Rightarrow \] \[A{{C}^{2}}={{b}^{2}}+{{a}^{2}}\] |
\[\therefore \] \[AC=\sqrt{{{b}^{2}}+{{a}^{2}}}\] |
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