MCQ
In $\triangle\text{ABC},\text{DE }\|\text{ BC}$ so that $AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm$ and $EC = 3x \ cm.$ Then, we have:
- A$x = 3$
- B$x = 5$
- ✓$x = 4$
- D$x = 2.5$
✓
Answer
Correct option: C.
$x = 4$
In $\triangle\text{ABC},\text{DE }\|\text{ BC}$
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}}$
$\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8$
$\Rightarrow6\text{x}^2-26\text{x}+8=0$
$\Rightarrow3\text{x}^2-13\text{x}+4=0$
$\Rightarrow(\text{x}-4)(3\text{x}-1)=0$
$\Rightarrow\text{x}=4$ or $\text{x}=\frac{1}{3}$
If $\text{x}=\frac{1}{3},$ then $\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0$
This is not possible since length cannot be negative.
$\Rightarrow x = 4$
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}}$
$\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8$
$\Rightarrow6\text{x}^2-26\text{x}+8=0$
$\Rightarrow3\text{x}^2-13\text{x}+4=0$
$\Rightarrow(\text{x}-4)(3\text{x}-1)=0$
$\Rightarrow\text{x}=4$ or $\text{x}=\frac{1}{3}$
If $\text{x}=\frac{1}{3},$ then $\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0$
This is not possible since length cannot be negative.
$\Rightarrow x = 4$
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