Isosceles trapezoid with semicircle I have an isosceles trapezoid, with a semicircle in the middle.  I need to know the difference in area of the two shapes. Radius of the semicircle is $6$cm, and the longest base is $14$cm.
 A: Assuming that the base of the circle lies along the longest base, here is a way to find the length of the shorter base (which along with the length of the longer base, 14, and the height of the trapezoid, 6, readily yields its area):
Label the vertices of the trapezoid ABCD, with AD being the longer base of length 14. Let the smaller of the bases have length $x$, and let the center of the circle be at $O$.
Area of trapezoid= $$\frac{6}{2}(x+14)=3x+42$$ by the area of a trapezoid formula. We can calculate this in a different manner by doing area= 
$$A_{OAB}+A_{OBC}+A_{OCA}=\frac{1}{2}(6AB+6*x+6*CD)=6AB+3x$$
Therefore, $$3x+42=6AB+3x\Leftrightarrow AB=7$$
Drop the perpendicular Q from B to AD and applying Pythagorean Theorem, we get $$AB^2=AQ^2+BQ^2 \Leftrightarrow 49=AQ^2+36 \Leftrightarrow AQ=\sqrt{13}$$
By symmetry, $$AD=2AQ+BC$$, which gives $$BC=14-2\sqrt{13}$$
Cheers,
Rofler
A: Hint:  You have drawn a picture of course.  Draw the line from the center of $O$ of the semicircle to the "left" point $T$ of tangency.
Drop a perpendicular from the left end $L$ of the second base to the first base (the one of length $14$).  Let it meet the first base at $P$.
Let $A$ be the left end of the first base.  
Now look at $\triangle OTA$ and $\triangle LPA$.  What do you observe about them?  Of course they are similar, but even more is true.
By the Pythagorean Theorem, you can find the length of $AT$.  So now you know $AP$.  
That's all you need to find the second base.
A: Hint:  What formula do you know for the area of a trapezoid?  How does the semicircle help you evaluate the values that enter?  What is the area of the semicircle?
