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In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.
If $AD\ =\ 4x\ –\ 3$, $AE\ =\ 8x\ –\ 7$, $BD\ =\ 3x\ –\ 1$, and $CE\ =\ 5x\ –\ 3$, find the value of $x$.

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Academic Mathematics NCERT Class 10

Given:

In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.

$AD\ =\ 4x\ –\ 3$, $AE\ =\ 8x\ –\ 7$, $BD\ =\ 3x\ –\ 1$, and $CE\ =\ 5x\ –\ 3$.

To do:

We have to find the value of $x$.

Solution:

$DE\ ||\ BC$ (given)

Therefore,

By Basic proportionality theorem,

$\frac{AD}{DB}\ =\ \frac{AE}{EC}$

$\frac{4x-3}{3x-1} =\frac{8x-7}{5x-3}$

$(5x-3)( 4x-3) =( 8x-7)( 3x-1)$

$5x( 4x-3) -3( 4x-3) =8x( 3x-1) -7( 3x-1)$

$20x^{2} -15x-12x+9=24x^{2} -8x-21x+7$

$( 24-20) x^{2} +( -8-21+12+15) x+7-9=0$

$4x^{2} -2x-2=0$

$2x^{2} -x-1=0$

$2x^{2} -2x+x-1=0$

$2x( x-1) +1( x-1) =0$

$( x-1)( 2x+1) =0$

$x-1=0$ or $2x+1=0$

$x=1$ or $2x=-1$

$x=1$ or $x=\frac{-1}{2}$

The side of a triangle cannot be negative. Therefore, the value of $x$ is $1 cm$.

Updated on: 10-Oct-2022

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