Question

In the given figure, DE || BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer 1

Given: AD = x, DB = (3x + 4), AE = (x + 3), EC = (3x + 19). DE || BC.

By the Basic Proportionality Theorem (since DE || BC), `(AD)/(DB) = (AE)/(EC)`.

So, `x/(3x + 4) = (x + 3)/(3x + 19)`.

Cross-multiply: x(3x + 19) = (x + 3)(3x + 4).

Expand: 3x² + 19x = 3x² + 13x + 12.

Subtract 3x²: 19x = 13x + 12

⇒ 6x = 12

⇒ x = 2