| In the adjoining figure, \[DE\left\| BC \right.\]. The value of x is: |
 |
A) 4
B) 6
C) 8
D) 10
Correct Answer: C
Solution :
| [c] In \[\Delta ABC,\] \[DE||BC\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{AD}{DB}=\frac{AE}{EC}\] (By Thales theorem) |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2x-1}{x-3}=\frac{2x+5}{x-1}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,(2x-1)\,\,(x-1)=(2x+5)\,(x-3)\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{x}^{2}}-2x-x+1=2{{x}^{2}}+5x-6x-15\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,2x=16\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,x=8\] |
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