In the given figure, $D E | | B C$ such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

(3 Marks)

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Solution

Triangle ABC, DE || BC, so by Basic Proportionality theorem,
$\frac{A D}{D B}$ = $\frac{A E}{E C}$ (1 Mark)

Substituting the values, we get,

$\frac{x}{3 x + 4}$ = $\frac{x + 3}{3 x + 19}$ (1 Mark)

By cross multiplying, we get,

$x (3 x + 19) = (x + 3) (3 x + 4)$

$3 x^{2} + 19 x = 3 x^{2} + 4 x + 9 x + 12$

$19 x - 13 x = 12$

$6 x = 12$
$x = 2$ (1 Mark)

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In the given figure, $D E | | B C$ such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

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