(Q13) What value(s) of x will make DE || AB, in the given figure?

AD = 8x + 9, CD = x + 3, BE = 3x + 4, CE = x.

Solution :

ABCDE

Given : In AABC, DE ∥ AB AD = 8x + 9, CD = x + 3, BE = 3x + 4, CE = x

By Basic proportionality theorem,

If DE ∥ AB then we should have

CD
DA
 =
CE
EB
=

⇒ (x + 3) (3x + 4) = x (8x + 9)

⇒ x (3x + 4) + 3 (3x + 4) - 8x2 + 9x

⇒ 3x2 + 4x + 9x + 12 = 8x2 + 9x

⇒ 8x2 + 9x - 3x2 - 4x - 9x -12 = 0

⇒ x2 - x - = 0

⇒ 5x2 - 10x + 6x - 12 = 0

⇒ 5x (x - 2) + 6 (x - 2) = 0

⇒ (5x + 6) (x - 2) = 0

⇒ 5x + 6 = 0 or x - 2 = 0

⇒ x =
or x = ;

x cannot be negative.

∴ The value x = 2 will make DE ∥ AB.