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) |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2x-1}{x-3}=\frac{2x+5}{x-1}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,(2x-1)\,\,(x-1)=(2x+5)\,(x-3)\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{x}^{2}}-2x-x+1=2{{x}^{2}}+5x-6x-15\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,2x=16\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,x=8\] |
You need to login to perform this action.
You will be redirected in
3 sec