Question

In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA = (3x – 1) cm, OB = (2x + 1) cm, OC = (5x – 3) cm and OD = (6x – 5) cm. Then, x = ?

विकल्प

• 2

• 3

• 2.5

• 4

Answer 1

2

Explanation:

We know that the diagonals of a trapezium divide each other proportionally.

∴ `(OA)/(OC) = (OB)/(OD)`

⇒ `(3x - 1)/(5x - 3) = (2x + 1)/(6x - 5)`

⇒ (3x – 1)(6x – 5) = (5x – 3)(2x + 1)

⇒ 18x² – 21x + 5 = 10x² – x – 3

⇒ 8x² – 20x + 8 = 0

⇒ 2x² – 5x + 2 = 0

⇒ 2x² – 4x – x + 2 = 0

⇒ 2x(x – 2) – (x – 2) = 0

⇒ (x – 2)(2x – 1) = 0

⇒ x = 2 or x = `1/2`

But, x = `1/2` gives (6x – 5) < 0 and the distance cannot be negative.

∴ x = 2