Question

In the given figure, DE || BC.

AD = x, DB = x – 2

AE = x + 2 , EC = x – 1

What is the value of x?

(a) 3

(b) 4

(c) 5

(d) 6

Answer 1

Correct option is (b) 4

In ΔABC, DE ∥ BC

\(\therefore \frac{AD}{DB} = \frac{AE}{EC}\)

(By basic proportionality theorem)

⇒ \(\frac x{x - 2} = \frac{x - 2}{x - 1}\)

⇒ \(x(x - 1) = (x + 2) (x - 2)\)

⇒ \(x^2 - x = x^2 - 4\)

⇒ \(x =4\)