# 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.

-
Does the semicircle have its diameter along the long base and is it tangent to the other three sides? I suspect so, as otherwise you don't have enough information. –  Ross Millikan Jun 14 '11 at 23:33

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

-
Thanks so much Rofler. I was asking cuz it was question on my math test today, and I couldn't figure it out. By chance is there another way as well, just as my teacher had mentioned something about finding the angles. –  Rachel Jun 15 '11 at 0:20
No worries, I figured it. :) thanx –  Rachel Jun 15 '11 at 0:27

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.

-

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?

-
Trapezoid - A =(a+b/2)h and the area of the semicircle is 56.55. And yes the height is the radius of a semicircle, but I do not have the second base to calculate the area of the trapezoid. –  Rachel Jun 14 '11 at 23:44
You might want to recheck the area of a trapezoid: $A = \frac 12\cdot h(a + b)$. Is that what you meant? Do you see how that is different than what you wrote? –  amWhy Jun 14 '11 at 23:57
The area of the semicircle is not 56.55, though it is approximately so. It is $18\pi$. –  Ross Millikan Jun 15 '11 at 0:10