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Find the value of x for which DE || AB in Fig.

 

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Answer : [x=2]

Given:  DE || AB

According to basic proportionality theorem. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, then the other two sides are divided in the same ratio.

\\\therefore \frac{CD}{AD}=\frac{CE}{BE}\\\frac{x+3}{3x+19}=\frac{x}{3x+4}\\\left ( x+3 \right )\left ( 3x+4 \right )=x\left ( 3x+19 \right )

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

19x - 13x = 12

6x = 12

x=\frac{12}{6}=2

Hence the required value of x is 2.

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