SOLUTION: Please help me solve this question tried my best but couldn't answer it! Question: In quadrilateral ABCD, m/A=(x-10), m/B=(2x+10), m/C=(2x-70) m/D(3x-50).What kind of quadrilate

Algebra ->  Parallelograms -> SOLUTION: Please help me solve this question tried my best but couldn't answer it! Question: In quadrilateral ABCD, m/A=(x-10), m/B=(2x+10), m/C=(2x-70) m/D(3x-50).What kind of quadrilate      Log On


   

Question 911466: Please help me solve this question tried my best but couldn't answer it!
Question: In quadrilateral ABCD, m/A=(x-10), m/B=(2x+10), m/C=(2x-70) m/D(3x-50).What kind of quadrilateral ia ABCD? Why?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
In quadrilateral ABCD, m/A=(x-10), m/B=(2x+10), m/C=(2x-70) m/D(3x-50)
Sum of angles of quadrilateral = 360 deg
(x-10)+(2x+10)+(2x-70)+(3x-50)=360
8x-120=360
6x=360+120
8x=480
x=60
m/A=(x-10),=50
m/B=(2x+10),130
m/C=(2x-70) 50
m/D(3x-50)130
The opposite angles are equal
Hence its a parallelogram