| In the given figure\[DE||BC\]. If \[AD=x\]\[DB=x-2\], \[AE=x+2\] and \[EC=x-1\], then the value of x is |
 |
A) 9
B) 4
C) 4.5
D) 8
Correct Answer: B
Solution :
Given in \[\Delta ABC\,DE\,|\,\,|\,\,BC\], Now, \[\frac{AD}{DB}=\frac{AE}{EC}\] [by Thales theorem] \[\Rightarrow \,\,\,\,\frac{x}{x-2}=\frac{x+2}{x-1}\] \[\Rightarrow \,\,\,x\left( x-1 \right)=\left( x-2 \right)\left( x+2 \right)\] \[\Rightarrow \,\,\,{{x}^{2}}-x={{x}^{2}}-4\] \[\therefore \,\,\,\,x=4\]
You need to login to perform this action.
You will be redirected in
3 sec
