Question

What value(s) of x will make DE || AB in the given figure?
AD = 8x + 9, CD = x + 3 BE = 3x + 4, CE = x.

Answer 1

hope this helps you....

Answer 2

Given :

In ∆ABC , DE//AB

AD = 8x + 9 , CD = x + 3 ,

BE = 3x + 4 , CE = x ;

Solution :

In ∆ABC , DE // BC

CD/DA = CE/BE

[ Thales theorem ]

( x + 3 )/( 8x + 9 ) = x/( 3x + 4 )

=> ( x+3 )( 3x+4 ) = x( 8x+9 )

=> 3x² + 4x + 9x + 12 = 8x² + 9x

=> 3x² + 13x + 12 = 8x² + 9x

=> 5x² - 4x - 12 = 0

=> 5x² - 10x + 6x - 12 = 0

=> 5x( x - 2 ) + 6 ( x - 2 ) = 0

=> ( x - 2 )( 5x + 6 ) = 0

x - 2 = 0 or 5x + 6 = 0

Therefore ,

x = 2 or x = -6/5