In the figure, $D E ∥ B C$ . Find the value of $x$ .

\sqrt{5}

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\sqrt{6}

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\sqrt{3}

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\sqrt{7}

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Solution

The correct option is D $\sqrt{7}$
From the figure, it is given that DE║BC and $A D = (x + 4) c m, D B = (x + 3) c m,$
$A E = (2 x - 1) c m, E C = (x + 1) c m$

In $Δ A D E$ and $Δ A B C$ ,we have

$∠ A = ∠ A$ ( common)

$∠ E D A = ∠ C B A$ (Corresponding angles)

Hence, by AA similarity,

$Δ A D E$ is similar to $Δ A B C$

Therefore, by similarity of triangles,

$\frac{A D}{A B} = \frac{D E}{B C} = \frac{A E}{A C}$

$\frac{A D}{A D + B D} = \frac{A E}{A E + E C}$

$\frac{A D + B D}{A D} = \frac{A E + E C}{A E}$

$1 + \frac{B D}{A D} = 1 + \frac{E C}{A E}$

$\frac{B D}{A D} = \frac{E C}{A E}$

Substituting the values, we have

$\frac{x + 4}{2 x + 7} = \frac{2 x - 1}{3 x}$

$\frac{B D}{A D} = \frac{E C}{A E} \Rightarrow \frac{x + 3}{x + 4} = \frac{x + 1}{2 x - 1} \Rightarrow 2 x^{2} + 5 x - 3 = x^{2} + 5 x + 4 \Rightarrow x^{2} = 7 \Rightarrow x = \sqrt{7}$

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