In Fig. 4.71, if AB ∥ CD, find the value of x.

Given: AB ∥ CD

To find: The value of x.

Theorem Used:

The diagonals of a trapezium divide each other proportionally.

Explanantion:

⇒ (3x-1) (6x-5) = (2x+1) (5x-3)

⇒ 18x² - 15x - 6x + 5 = 10x² - 6x + 5x - 3

⇒ 18x² – 21 x + 5 = 10x² – x -3

⇒ 18x² - 21x + 5 - 10x² + x + 3 = 0

⇒ 8x² - 20x + 8 = 0

⇒ 4(2x² - 5x + 2) = 0

⇒ 2x² - 5x + 2 = 0

⇒ 2x² - 4x – x + 2 = 0

⇒2x(x-2) -1(x-2) =0

⇒ (x-2) (2x-1) =0

⇒ x-2=0

x=2

Or, 2x-1=0

2x=1

⇒ x=1/2

But x=1/2 is not possible because it will make following lengths negative.

= 3 – 5

= -2

So, x=2