As shown in the figure, in the isosceles triangle ABCD, AB is parallel to CD, AC = 10 cm, CE is perpendicular to AB, e, CE = 6, find the area of trapezoid ABCD

As shown in the figure, in the isosceles triangle ABCD, AB is parallel to CD, AC = 10 cm, CE is perpendicular to AB, e, CE = 6, find the area of trapezoid ABCD

Isosceles trapezoid ABCD, right?
Make AF perpendicular to CD and extend CD to F
AE squared = AC squared - ce squared
So AE = 8
Because AB is parallel to CD
So angle FDA = angle DAB = angle B
Because angle f = angle CEB = 90, AF = CE
So △ AFD is equal to △ CEB
The area of ladder shaped ABCD = △ AEC + △ ACD + △ CEB
=△AEC+△ACD+△AFD
=Δ AEC + △ AFC = 6 times 8 = 48