Question

D and E are points on the sides AB and AC respectively of a ΔABC such that DE || BC. Find the value of x, when AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm.

Answer 1

In Δ ABC, it is given that DE || BC.
Applying Thales’ theorem, we have :

`(AD)/(DB) = (AE)/(EC)`

⟹  `(7x - 4)/(3x +4) = (5x -2)/(3x)`

⟹ 3x (7x-4) = (5x-2) (3x+4)

⟹  `21x^2 – 12x = 15x^2 + 14x-8`

⟹ `6𝑥^2 – 26x + 8 = 0`

⟹ (x-4) (6x-2) =0

⟹ `x = 4,1/2`

∵ `x ≠ 1/3` (as if x = `1/3` 𝑡ℎ𝑒𝑛 𝐴𝐸 𝑤𝑖𝑙𝑙 𝑏𝑒𝑐𝑜𝑚𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)

∴ x = 4 cm