A) 4 cm and 3 cm
B) 5 cm and 3 cm
C) 2 cm and 4 cm
D) 4 cm and 5 cm
Correct Answer: A
Solution :
| [a] In \[\Delta ADE\] and \[\Delta ABC,\] |
| \[\angle ADE=\angle ABC\] |
 |
| [\[\because DE\parallel BC\] and AB is a transversal i.e. corresponding angle] |
| \[\angle AED=\angle ACB\] |
| [\[\because DE\parallel BC\] and AB is a transversal i.e. corresponding angle] |
| \[\angle A=\angle A\] [common] |
| \[\therefore \] \[\Delta ADE\sim \Delta ABC\] |
| So, \[\frac{AD}{AB}=\frac{AE}{AC}\]\[\Rightarrow \]\[\frac{2.5}{7.5}=\frac{2}{AC}\] |
| \[\Rightarrow \] \[AC=\frac{2\times 75}{25}=6\] |
| \[\therefore \] \[EC=AC-AE=6-2=4\,cm\] |
| and \[\frac{AD}{AB}=\frac{DE}{BC}\]\[\Rightarrow \]\[\frac{2.5}{7.5}=\frac{DE}{9}\] |
| \[\Rightarrow \] \[DE=9\times \frac{25}{75}=3\,cm.\] |
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