triangle ABC and triangle BCD have vertices A(-6 -7),B(-6,4),C(2,-7),and D(8,4).what is the area,in square units,of trapezoid ABDC which is formed by the two triangles?
area = 121u^2 .... area = (Parallel side one + parallel side two)/2 X height
Pls find attached file.
and u just need to find the distance by formula
\(\sqrt{(y2-y1)^2+(x2-x1)^2}\)
where x1,x2,y1,y2 are the cordinate of ABCD trapeziod