| In the given figure, if\[AB||CD\], then the value of x is |
 |
A) 6
B) 8
C) 3
D) 9
Correct Answer: C
Solution :
| Given, \[AB\,\,|\,\,|\,\,CD\] |
| \[\therefore \] Quadrilateral ABCD is a trapezium |
| \[\therefore \,\,\,\,\frac{AO}{CO}=\frac{BO}{DO}\] |
| [diagonals of trapezium divides each other proportionally] |
| \[\Rightarrow \,\,\,\,\frac{4}{4x-2}=\frac{x-1}{2x+4}\] |
| \[\Rightarrow \,\,\,\,4\left( 2x+4 \right)=\left( x+1 \right)\left( 4x-2 \right)\] |
| \[\Rightarrow \,\,\,\,8x+16=4{{x}^{2}}-2x+4x-2\] |
| \[\Rightarrow \,\,\,\,4{{x}^{2}}-6x-18=0\] |
| \[\Rightarrow \,\,\,2{{x}^{2}}-3x-9=0\] |
| \[\Rightarrow \,\,\,\,2{{x}^{2}}-6x+3x-9=0\] |
| \[\Rightarrow \,\,\,\,2x\left( x-3 \right)+3\left( x-3 \right)=0\] |
| \[\Rightarrow \,\,\,\,\,\left( 2x+3 \right)\left( x-3 \right)=0\] |
| \[\Rightarrow \,\,\,x=3\] and \[x=-\frac{3}{2}\] (not possible) |
| Hence, the value of x = 3. |
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