Question

In the figure, AB II DC, find the value of x.​

Answer 1

BD

= 6x-1=2x+1

= 6x-2x=1+1

= 4x=2

= x=2/4

= x=1/2

AC

= 3x-1=5x-1

= 3x-5x=-1+1

= -2x=0

= x=0/-2

= x=0

Therefore, BD= 1/2. and AC= 0

Answer 2

       $\underline{\bold{Solution:}}$

       $\text{It is given that }AB \parallel CD$

       $\text{In }\triangle{COD}\text{ and }\triangle{AOB}$

       $\angle{ODC} = \angle{OBA} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\text{Angles on opposite sides of the transversal are equal})$

       $\angle{OCD} = \angle{OAB} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\text{Angles on opposite sides of the transversal are equal})$

$\implies \text{By AA similarity }\triangle{COD}\text{ and }\triangle{AOB}\text{ are similar}$

$\implies \dfrac{AO}{CO}= \dfrac{BO}{DO}\text{ ------ (i)}$

       $\text{From the figure, we know that:}$

       $AO = 3x-1$

       $CO = 5x-1$

       $BO = 2x-1$

       $DO = 6x-1$

       $\text{Substituting in (i)}$

$\implies \dfrac{3x-1}{5x-1}= \dfrac{2x-1}{6x-1}$

       $\text{Cross multiplying...}$

$\implies (3x-1)(6x-1)= (2x-1)(5x-1)$

       $\text{Simplifying...}$

$\implies 18x^{2} - 3x - 6x + 1= 10x^{2} - 2x - 5x + 1$

       $\text{Simplifying...}$

$\implies 18x^{2} - 9x + 1= 10x^{2} -7x + 1$

       $\text{Transposing terms of R.H.S to L.H.S}$

$\implies 18x^{2} - 9x + 1- 10x^{2} +7x - 1 = 0$

       $\text{Simplifying...}$

$\implies 8x^{2} - 2x = 0$

       $\text{Taking }2x \text{ common}$

$\implies 2x(4x- 1) = 0$

$\implies x = 0 , \dfrac14$

       $\text{Putting }x = 0 \text{ we see that all lengths become negative}$

       $\text{We know that length cannot be negative. Hence, }x=0 \text{ is eliminated}$

 $\therefore \text{ }\text{ }\boxed{x = \dfrac{1}{4}}$