| In the given figure, ABC is an equilateral triangles DE is parallel to BC such that area of quadrilateral DBCE is equal to one half the area of \[\Delta ABC\]. If \[BC=2cm,\] then \[DE=\] |
 |
A) \[1cm\]
B) \[1\frac{1}{2}cm\]
C) \[\sqrt{2}cm\]
D) None of these
Correct Answer: C
Solution :
| [c] Area of \[DBCE=\frac{1}{2}\times Area\,\,of\,\,\Delta ABC\] Area \[\therefore \] Area of \[\Delta ADE=\frac{1}{2}\times Area\,\,of\,\,\Delta ABC\] |
| \[DE||BC\] |
| \[\Rightarrow \,\,\,\,\frac{ar(\Delta ADE)}{ar(\Delta ABC)}=\frac{D{{E}^{2}}}{B{{C}^{2}}}\] |
| \[\Rightarrow \,\,\,\,\frac{1}{2}=\frac{D{{E}^{2}}}{4}\,\,\,\,\,\,\,\Rightarrow \,DE=\sqrt{2}\,cm.\] |
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