The value of x is 6 and y is 3.
Given that,
ABCD is a parallelogram in which sides of parallelogram are,
AB = 8x- 34, BC = 6x- 4
CD = 9x-40, AD = 4y+20
We have to find,
The value of x and y in a quadrilateral ABCD ?
According to the question,
By the property of a parallelogram,
The opposite sides of a parallelogram are equal and parallel to each other.
AD = BC
CD = AB
Substitute the values of AD and BC and solve for the value of x and y,
[tex]\rm 4y +20 = 6x -4 \\\\4y - 6x = -4 -20\\\\4y -6x = -24\\\\ Divide \ by \ 2 \ on \ both \ the \ sides\\\\ 2y - 3x = -12[/tex]
And
[tex]CD = AB \\\\9x-40=8x-34\\\\9x-8x = 40-34\\\\x = 6[/tex]
Substitute the value of x in equation 1,
[tex]2y-3x = -12\\\\2y - 3(6) = -12\\\\2y = -18=-12\\\\2y = -12+18\\\\2y = 6\\\\y = \dfrac{6}{2}\\\\y = 3[/tex]
Hence, The value of x is 6 and y is 3.
For more details refer to the link.
https://brainly.com/question/20216437
