A) 10/3cm
B) 9/4cm
C) 5/3cm
D) 20/3cm
Correct Answer: D
Solution :
\[\frac{AD}{BC}=\frac{Perimerer\,of\,\Delta ADE}{Perimeter\,of\,\Delta BCE}\] \[\frac{AD}{5}=\frac{4}{3}\] \[AD=20/3\,cm\]
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