Geometry - really hard trapezium problem. Can anyone explain me how to find the area of this trapezium if we have a base and radius of the circle?

 A: So, you know a little more than just the top base and the radius.

Using the letters in the plot that you attached in the comment, you also know


*

*that the circle is inscribed in the trapezium 

*that the angles in D and C are right angles 

*that the angle AIF is right too 

*that the height DC is 2r 

*that the segment HF=1, because it is $CF-r=3-2$


Since the circle is inscribed in the polygon, the two angles $\widehat{FAI}$ and $\widehat{HAF}$ are equal and so are the two triangles, since you also know that $AH=AI$ and $AF$ is shared. So, since the two triangles are equal, the two segments $HF$ and $FI$ are qual too, giving $IF=b-r=1$.
Now, name $x$ the piece of the lower base that we need to find, $x=GE$. If you draw the perpendicular form A to the side DE and, as before, draw also the bisector of the angle $\widehat{GEF}$ you will see (with an argument identical to the previous one) that $IE=x+b-r=x+1$, so that the hypotenuse $FE=x+2(b-r)=x+2$.
Using Pythagora's theorem for the triangle GEF you will have:
$$x^2+4^2=(x+2)^2$$
or in more general terms
$$x^2+(2r)^2=(x+2(b-r))^2$$
This means 
$$x=\frac{r^2-(b-r)^2}{b-r}=3$$
which is in agreement with what obtained with the Pitot theorem, mentioned by others.
Finally, your area is
$$A=2r\frac{b+(b+x)}{2}=18$$
in agreement with Mann's comment.
A: Theorem
For every pair of tangents meeting at a point $A$, $AB=AB'$, where $B,B'$ are the tangential points.
Problem
Let $GC=a$ and $G'E=u$. Applying the theorem we have $BG=BH=3-a$, $GC=G'C=a$,
$G'E=H'E=u$ and $HD=H'D=a+1$. Notice $BH=GA\implies a=1$.
Observe $CE^2=(GH')^2+(u-a)^2\implies(u+1)^2=4^2+(u-1)^2\implies u=4$.
The area is $\displaystyle \frac{4[3+(a+1+u)]}{2}=18$

A: Sorry for the crummy drawing (I wouldn't object if someone replaced it with a better one!), but hopefully it gets the point across:

The key triangles are similar, so
$${L\over r}={r\over b-r}$$
This gives $L=r^2/(b-r)$, so the area of the trapezium is 
$${1\over2}(2r)(b+(r+L))=r\left(b+r+{r^2\over b-r}\right)={rb^2\over b-r}$$
A: area of the trapezium is 
$$=(r)(2r)+(r)(b-r)+(r)\left(\frac{r^2}{\sqrt{b^2+r^2-2br}}\right)=r^2+br+\frac{r^3}{b-r}=\frac{rb^2}{b-r}$$
Now, setting the values $r=2$ & $b=3$, we get area of trapezium
$$=\frac{2(3)^2}{3-2}=18 \space sq. units$$ 
A: Similar triangles, similar triangles...

Drawing lines out from the centre of the circle $O$ as shown, we see that 


*

*$\triangle OEB$ and $\triangle OBF$ are congruent right triangles and also that 

*$\triangle OFC$ and $\triangle OCG$ are congruent right triangles. 


We can also see that $2\alpha + 2\beta$ forms a straight line, so - the key insight for me 


*

*$\alpha + \beta$ is a right angle. 


This means that $\triangle OBF, \triangle OBC$ and $\triangle OFC$ are all similar, and thus, since $|OE| = 2|EB|, \: |CG| = 2|OG| = 4$ and $|CD|=6$, giving the area of the trapezium:
$$\text{Area} = |AD|\frac{|AB|+|CD|}{2}= 4\frac{3+6}{2} = 18$$
A: 
By Pythagoras’ Theorem, $$
\begin{aligned}
\because 4^{2}+(x-1)^{2}=(x+1)^{2} & \Leftrightarrow x=4 \\
\therefore \text { Area of the trapezium } &=\frac{1}{2}(4)(3+3) =18
\end{aligned}
$$
