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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles3 Marks
Question
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides $AB$ and $AC$ respectively such that $DE || BC$.
If $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$, find the value of $x$.
✓

Answer

In the figure, $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$

$\because$ $DE || BC$
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\Rightarrow\frac{\text{x}}{\text{x}-2}=\frac{\text{x}+2}{\text{x}-1}$
$\Rightarrow x(x - 1) = (x + 2)(x - 2)$
$ \Rightarrow x^2 - x = x^2 - 4 $
$\Rightarrow x^2 - x - x^2 = -4 $
$\Rightarrow -x = -4 $
$\Rightarrow x = 4$
​​​​​​​$\therefore$ $x = 4$

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