Find area of trapezoid contained in circle 
The picture below represents a circumference of radius 5 and center at
  the origin of the referencial 0xy. In the picture is represented too a
  trapezoid [ABCD], with [AD] and [BC] parallel. A is the point of the
  circumference that belongs to the x axis. Points B and C belong to the
  circumference. D belongs to the 0x axis and has x-coordinate
  $-\frac{5}{2}$. Consider $AÔB = \alpha, \alpha \in ]0,\frac{\pi}{2}[$.
 
Show that the area of the trapezoid in function of $\alpha$ is given
  by:
$$A(\alpha) = 25\sin(\alpha)\cos(\alpha)+\frac{75}{4}\sin(\alpha)$$

First I tried to calculate the area of the whole trapezoid at once and so I got:
$$\frac{2 \cdot 5\cos(\alpha)+(5+\frac{5}{2})}{2}\cdot 5\sin(\alpha)$$
But when I put this one and the given one in the calculator they gave different values, so I tried dividing the main trapezoid in two smaller ones at the y axis and I got:
\begin{align} 25\sin(\alpha) \cdot \frac{\cos(\alpha)+1}{2}+25\sin(\alpha) \cdot \frac{2\cos(\alpha)+1}{4 }= \\
25\sin(\alpha)\Bigg(\frac{\cos(\alpha)+1}{2}+\cdot \frac{2\cos(\alpha)+1}{4}\Bigg) = \\
25\sin(\alpha)\Bigg(\frac{4(\cos(\alpha)+1)+2(2\cos(\alpha)+1)}{8}\Bigg) = \\
25\sin(\alpha)\Bigg(\frac{\cos(\alpha)+1+4\cos(\alpha)+2}{4}\Bigg) = \\
25\sin(\alpha)\Bigg(\frac{5\cos(\alpha)+3}{4}\Bigg) = ??? \end{align}
I have two questions:


*

*Why didn't my first attempt work?

*How do I solve this, continuing with my second one?

 A: You were correct but you did not continue with the first formula you derived: $$ {2⋅5cos\alpha+(5+5/2)\over 2}⋅5sin\alpha = 5{10\over 2}cos\alpha sin\alpha + 5{7.5\over 2}sin\alpha ⇒ \\ A(\alpha) = 25sin\alpha cos\alpha +{75\over 4}sin\alpha$$
Here's an alternative way:
We have to notice that as the angle a changes the point B changes and in order for the trapezoid to remain a trapezoid that means point C changes in order to stay paraller with AD. So: $$ A\hat OB = C\hat OD $$
Because BD and AD are parallel that means: $$ O\hat BC = O\hat CB $$
If we calculate the area of OAB triangle in terms of the angle α we got: $$ OA = 5\\ sin\alpha = h/5 \Rightarrow h = 5sin\alpha \\ AreaOfOAB = {25\over 2}sin\alpha$$
If we calculate the area of OCD triangle in terms of the angle α we got: $$ OD = 2.5\\ sin\alpha = h/5 \Rightarrow h = 5sin\alpha \\ AreaOfOCD = {12.5\over 2}sin\alpha$$
Now let's look at the OCB triangle: $$ cos\alpha = {(BC/2)\over 5}={BC\over 10} \Rightarrow BC = 10cos\alpha \\sinα=h/5⇒h=5sinα\\ AreaOfOCB = 25sin\alpha cos\alpha$$
Now if we add them we got: $$ A(\alpha) = 25sin\alpha cos\alpha +{75\over 4}sin\alpha$$
