Question

In the given figure, if AB || DC, AO=x+5, BO = x-1, CO = x+3, and DO = x - 2 then find
the value of x

Answer 1

Step-by-step explanation:

Given,

ABCDABCD is trapezium where diagonals intersect at OO ,AO=x+5AO=x+5 ,OC=x+3OC=x+3 ,DO=x-2DO=x−2 , BO=x-1BO=x−1 and AB\parallel DCAB∥DC

From figure,

\triangle AOB△AOB and \triangle DOC△DOC

\angle AOB=\angle DOC∠AOB=∠DOC (vertically opposite angle)

\angle OAB=\angle OCD∠OAB=∠OCD (AB\parallel DCAB∥DC and alternate angle)

\angle OBA=\angle ODC∠OBA=∠ODC (AB\parallel DCAB∥DC and alternate angle)

∴ \triangle AOB△AOB ≅\triangle DOC△DOC

So, We can write,

\frac{AO}{OC}=\frac{BO}{OD}

OC

AO

=

OD

BO

Plug the value in above equation,

⇒ \frac{x+5}{x+3}=\frac{x-1}{x-2}

x+3

x+5

=

x−2

x−1

⇒x^{2} -2x+5x-10=x^{2} -x+3x-3x

2

−2x+5x−10=x

2

−x+3x−3

⇒3x-2x=-3+103x−2x=−3+10

⇒x=7x=7

∴ The value of xx is 77

Hope it helps you