A) \[\text{6}.\text{3 cm}\]
B) \[\text{7}\text{.2 cm}\]
C) \[\text{6}\text{.5 cm}\]
D) \[\text{7}\text{.6 cm}\]
Correct Answer: B
Solution :
| [b] In \[\Delta ABC,\] \[DE||BC\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{AD}{DB}=\frac{AE}{EC}\] (By Thales theorem) |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2}{3}=\frac{AE}{EC}\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\frac{3}{2}=\frac{EC}{AE}\] |
| Adding 1 on both the sides, we get |
| \[\frac{3}{2}+1=\frac{EC}{AE}+1\,\,\,\Rightarrow \frac{3+2}{2}=\frac{EC+AE}{AE}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{5}{2}=\frac{AC}{AE}\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\frac{5}{2}=\frac{18}{AE}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,5AE=36\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,AE=\frac{36}{5}\,\,\,\,\Rightarrow \,\,AE=7.2\,cm\] |
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