+1725 In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.
+129771 Angle OCD = 45
Angle COD = 90
So....Angle ODC = 45
So triangle DCO is 45 - 45 - 90
So OD = 10/sqrt 2 = OC = 5sqrt 2
Area of triangle DCO = (1/2) (5sqrt 2)(5sqrt 2) = 25
Height of triangle DCO can be found as
25 = (1/2)CD * height
50 = 10 * height
50/10 = height = 5
Triangle ABO similar to Triangle DCO
And the scale factor from triangle ABO to Triangle DCO = 4/10 = 2/5
So the height of triangle ABO = (2/5) (height of tragle DCO) = (2/5) (5) = 2
So the trapezoid has a height of 5 + 2 = 7
And the bases are 10 and 4
Area = (1/2) (7) (10 + 4) = 49
