Question

In Fig., AB || DC. Find the value of x.


For maths expert ​

Answer 1

Given: In quadrilateral ABCD , AB | | CD

AO = 3x - 19 , OC = x - 5 , BO = x - 3 , OD = 3

To Find out: Find the value of x .

Solution: AB | | CD ( Quadrilateral ABCD is a trapezium. )

Since, the diagonals of a trapezium divide each other proportionally.

Therefore,

AO/OC = BO/OD

⇒ 3x - 19 / x - 5 = x - 3 / 3

⇒ 3 ( 3x - 19 ) = ( x - 5 ) ( x - 3 )

⇒ 9x - 57 = x² - 8x + 15

⇒ x² - 17x + 72 = 0

⇒ x² - 9x - 8x + 72 = 0

⇒ x ( x - 9 ) - 8 ( x - 9 ) = 0

⇒ ( x - 8 ) ( x - 9 ) = 0

Thus, x = 8 units or x = 9 units

Answer 2

$\large\underline\mathrm{Given:-}$

• In quadrilateral ABCD, AB//CD

$\implies$ AO = 3x - 19

$\implies$ OC = x - 5

$\implies$ BO = x - 3

$\implies$ OD = 3

$\large\underline\mathrm{To \: find}$

• Find the value of x.

$\large\underline\mathrm{Solution}$

• AB//CD (Quadrilateral ABCD is a trapezium.)

$\implies$ AO/OC = BO/OD

$\implies$ (3x - 19)/x - 5) = (x - 3)/3

$\implies$ 3(3x - 19) = (x - 5)(x - 3)

$\implies$ 9x - 57 = x² - 8x + 15

$\implies$ x² - 9x - 8x + 72 = 0

$\implies$ x(x - 9) - 8(x - 9) = 0

$\implies$ (x - 8) (x - 9) = 0

$\implies$ x = 8, or 9.