Question

In Δ DEF, AB || EF. The value of x is ______.

विकल्प

• 0, 2

• 2 only

• −2

• 1

Answer 1

In Δ DEF, AB || EF. The value of x is 2 only.

Explanation:

DE = DA + AE

= 2x + 3x + 1

= 5x + 1

DF = DB + BF

= `x + 2x - 1/2`

= `3x - 1/2`

Δ DEF, AB || EF

`(DA)/(DE) = (DB)/(DF)`

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

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

`6x^2 - (2x)/2 = 5x^2 + 5`

6x² − x = 5x² + x

6x² − x − 5x² − x = 0

x² − 2x = 0

x(x − 2) = 0

x = 0 or x = 2

∴ x = 2