| In figure, \[LM\left\| AB \right.\]. If \[AL=x-3,\]\[AC=2x,\]\[BM=x-2\]and \[\text{BC}=\text{2x}+\text{3},\]find the value of x. |
 |
A) 2
B) \[9.5\]
C) 9
D) \[8.5\]
Correct Answer: C
Solution :
| [c] In \[\Delta ABC,\] \[LM||AB\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\frac{AL}{LC}=\frac{BM}{MC}\] (By Thales theorem) |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{AL}{AC-AL}=\frac{BM}{BC-BM}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{x-3}{2x-(x-3)}=\frac{x-2}{(2x+3)-(x-2)}\Rightarrow \frac{x-3}{x+3}=\frac{x-2}{x+5}\]\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(x-3)\,\,(x+5)=(x-2)\,(x+3)\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+2x-15={{x}^{2}}+x-6\,\,\,\,\,\,\,\,\,\Rightarrow \,\,x=9\] |
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