Question

ABCD is trapezium, and the diagonals intersect at O... AO = x +5 , OC= x +3 , DO = x-2 and BO = x-1.
Find x if AB is parallel to DC.

Answer 1

I hope this help you

Answer 2

The value of $x$ is $7$

Step-by-step explanation:

Given,

$ABCD$ is trapezium where diagonals intersect at $O$,$AO=x+5$,$OC=x+3$,$DO=x-2$, $BO=x-1$ and $AB\parallel DC$

From figure,

$\triangle AOB$ and $\triangle DOC$

• $\angle AOB=\angle DOC$ (vertically opposite angle)
• $\angle OAB=\angle OCD$ ($AB\parallel DC$ and alternate angle)

• $\angle OBA=\angle ODC$  ($AB\parallel DC$ and alternate angle)  

∴ $\triangle AOB$≅$\triangle DOC$

So, We can write,

 $\frac{AO}{OC}=\frac{BO}{OD}$

Plug the value in above equation,

⇒ $\frac{x+5}{x+3}=\frac{x-1}{x-2}$

⇒$x^{2} -2x+5x-10=x^{2} -x+3x-3$

⇒$3x-2x=-3+10$

⇒$x=7$

∴ The value of $x$ is $7$