MCQ
In the given figure, $ABCD$ is a trapezium whose diagonals $AC$ and $BD$ intersect at $O$ such that $OA = (3x - 1)cm, OB = (2x + 1)cm, OC = (5x - 3)cm$ and $OD (6x - 5)cm$. Then $x =?$
- ✓$2$
- B$3$
- C$2.5$
- D$4$
✓
Answer
Correct option: A.
$2$
The diagonals of a trapezium divide each other proportinally.
$\Rightarrow\frac{\text{OD}}{\text{OB}}=\frac{\text{OC}}{\text{OA}}$
$\Rightarrow\frac{6\text{x}-5}{2\text{x}+1}=\frac{5\text{x}-3}{3\text{x}-1}$
$\Rightarrow18\text{x}^2-21\text{x}+5=10\text{x}^2-\text{x}-3$
$\Rightarrow8\text{x}^2-20\text{x}+8=0$
$\Rightarrow2\text{x}^2-5\text{x}+2=0$
$\Rightarrow(\text{x}-2)(2\text{x}-1)=0$
$\Rightarrow\text{x}=2$ or $\text{x}=\frac{1}{2}$
if $\text{x}=\frac{1}{2},$ then $\text{OD}=6\text{x}-5=6\Big(\frac{1}{2}\Big)-5=-2$
RThis is not possible since length cannot be negative.
$\Rightarrow x = 2$
$\Rightarrow\frac{\text{OD}}{\text{OB}}=\frac{\text{OC}}{\text{OA}}$
$\Rightarrow\frac{6\text{x}-5}{2\text{x}+1}=\frac{5\text{x}-3}{3\text{x}-1}$
$\Rightarrow18\text{x}^2-21\text{x}+5=10\text{x}^2-\text{x}-3$
$\Rightarrow8\text{x}^2-20\text{x}+8=0$
$\Rightarrow2\text{x}^2-5\text{x}+2=0$
$\Rightarrow(\text{x}-2)(2\text{x}-1)=0$
$\Rightarrow\text{x}=2$ or $\text{x}=\frac{1}{2}$
if $\text{x}=\frac{1}{2},$ then $\text{OD}=6\text{x}-5=6\Big(\frac{1}{2}\Big)-5=-2$
RThis is not possible since length cannot be negative.
$\Rightarrow x = 2$
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