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, \(\frac{AO}{CO}\) = \(\frac{BO}{DO}\) ⇒ \(\frac{4}{(4x – 2)}\) = \(\frac{(x +1)}{(2x + 4)}\) 4(2x + 4) = (4x – 2)(x +1) 8x + 16 = x(4x – 2) + 1(4x – 2) 8x + 16 = 4x² – 2x + 4x – 2 -4x² + 8x + 16 + 2 – 2x = 0 -4x² + 6x + 8 = 0 4x² – 6x – 18 = 0 4x² – 12x + 6x – 18 = 0 4x(x – 3) + 6(x – 3) = 0 (4x + 6) (x – 3) = 0 ∴ x = – \(\frac{6}{4}\) or x = 3

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,

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

⇒ \(\frac{4}{(4x – 2)}\) = \(\frac{(x +1)}{(2x + 4)}\)

4(2x + 4) = (4x – 2)(x +1)

8x + 16 = x(4x – 2) + 1(4x – 2)

8x + 16 = 4x² – 2x + 4x – 2

-4x² + 8x + 16 + 2 – 2x = 0

-4x² + 6x + 8 = 0

4x² – 6x – 18 = 0

4x² – 12x + 6x – 18 = 0

4x(x – 3) + 6(x – 3) = 0

(4x + 6) (x – 3) = 0

∴ x = – \(\frac{6}{4}\) or x = 3

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