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

1.	In fig, if AB∥CD, find the value of x.
Answer» It’s given that AB∥CD. Required to find the value of x. We know that, Diagonals of a parallelogram bisect each other So, \(\frac{AO}{CO}\) = \(\frac{BO}{DO}\) ⇒ \(\frac{(6x – 5)}{(2x + 1)}\) = \(\frac{(5x – 3)}{(3x – 1)}\) (6x – 5)(3x – 1) = (2x + 1)(5x – 3) 3x(6x – 5) – 1(6x – 5) = 2x(5x – 3) + 1(5x – 3) 18x² – 10x² – 21x + 5 + x +3 = 0 8x² – 16x – 4x + 8 = 0 8x(x – 2) – 4(x – 2) = 0 (8x – 4)(x – 2) = 0 x = \(\frac{4}{8}\) = \(\frac{1}{2}\) or x = -2 ∴ x = \(\frac{1}{2}\)

Answer»

It’s given that AB∥CD.

Required to find the value of x.

We know that,

Diagonals of a parallelogram bisect each other

So,

\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)

⇒ \(\frac{(6x – 5)}{(2x + 1)}\) = \(\frac{(5x – 3)}{(3x – 1)}\)

(6x – 5)(3x – 1) = (2x + 1)(5x – 3)

3x(6x – 5) – 1(6x – 5) = 2x(5x – 3) + 1(5x – 3)

18x² – 10x² – 21x + 5 + x +3 = 0

8x² – 16x – 4x + 8 = 0

8x(x – 2) – 4(x – 2) = 0

(8x – 4)(x – 2) = 0

x = \(\frac{4}{8}\) = \(\frac{1}{2}\) or x = -2

∴ x = \(\frac{1}{2}\)

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