Question
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides AB and AC respectively such that DE || BC.
If AD = 4, AE = 8, DB = x - 4, and EC = 3x - 19, find x.
If AD = 4, AE = 8, DB = x - 4, and EC = 3x - 19, find x.
✓
Answer
We have,DE || BC
therefore, by basic proportionally theorem,
We have $\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\frac{4}{\text{x}-4}=\frac{8}{3\text{x}-19}$
$\Rightarrow4(3\text{x}-19)=8(\text{x}-4)$
$\Rightarrow12\text{x}-76=8\text{x}-32$
$\Rightarrow12\text{x}-8\text{x}=-32+76$
$\Rightarrow4\text{x }=44$
$\Rightarrow\text{x}=\frac{44}{4}=11\text{cm}$
$\therefore\text{x}=11\text{cm}$
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