Found the area of TrapeziumABCD whose parallel side are 14cm and 24cm.its non parallel sides are 8cm and 6cm
Make a sketch, dropping perpendiculars from the shorter parallel line to the longer
Let that height be h
You now have a rectangle 24 by h, and two triangles.
One has hypotenuse 6, height of h, let its base be x
The other has hypotenuse 8, height of h, and base of 10-x
from the first triangle:
x^2 + h^2 = 36
from the 2nd:
(10-x)^2 + h^2 = 64
subtract them:
(10-x)^2 - x^2 = 28
100 - 20x + x^2 - x^2 = 28
20x = 72
x = 18/5
then (18/5)^2 + h^2 = 36
h^2 = 36-324/25 = 576/25
h = ± 24/5 , ignoring the negative height
so now you have x = 18/5 and y = 24/5
so find the area of the two triangles, and add that to the area of the rectangle