+2862 Given trapezoid ABCD with the measure of \(\angle{D}\) (in degrees) equals 45, AD = \(8\sqrt{2}\), AB = 4, BC = 10. Find the area of the trapezoid.
(image is not to scale)
I used trigonometry, and I got the answer of 88 (please check). However, this problem shouldn't use trig and shouldn't use the calculator. Can someone tell me how to solve it?
+9466 Draw a line segment from A perpendicular to DC that meets DC at point E , and
draw a line segment from B perpendicular to DC that meets DC at point F , like this:
The sum of the interior angles in △ADE = 180°
45° + 90° + m∠DAE = 180°
m∠DAE = 180° - 45° - 90°
m∠DAE = 45°
△ADE is an isosceles triangle and the sides opposite the base angles are congruent.
DE = AE
And if DE = b then AE = b
By the Pythagorean Theorem,
DE2 + AE2 = (8√2)2
b2 + b2 = (8√2)2
2b2 = 128
b2 = 64
b = 8
ABFE is a rectangle so
BF = AE = 8
EF = AB = 4
By the Pythagorean Theorem,
BF2 + FC2 = 102
82 + FC2 = 102
FC2 = 102 - 82
FC2 = 36
FC = 6
Now we know:
AB = 4
So we can say base1 = 4
DE = 8 and EF = 4 and FC = 6
So we can say base2 = 8 + 4 + 6 = 18
AE = 8
So we can say height = 8
area of trapezoid = (1/2)(base1 + base2)(height)
area of trapezoid = (1/2)(4 + 18)(8)
area of trapezoid = 88
If you have the 45-45-90 triangle memorized, you can immediately recognize that DE and AE must be 8,
but if not you can always figure it out this way. 
+2862 Thank you! This should've been easy, I will have to remember about isosceles triangles and Pythagorean theorem.
