A) 5/3
B) 4
C) 81/625
D) 16
Correct Answer: D
Solution :
\[\frac{{{l}_{\max }}}{{{l}_{\min }}}=\frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}=\frac{25}{9}\] \[\therefore \,\,\,\,\frac{a+b}{a-b}=\frac{5}{3}\,\,or\,\,5a-5b=3a+3b\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,a=4b\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{(4b)}^{2}}}{{{b}^{2}}}=16\]You need to login to perform this action.
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