# Express Area as a function of r

1. Feb 6, 2008

### rocomath

Ok, so I have this picture. A semi-circle at the ends of a rectangle. It tells me that the perimeter is $$\frac 1 4$$. So, isn't the total perimeter just the sum of the circle and rectange? $$P=2\pi r+2(L+W)$$

$$r=radius$$
$$W=2r$$

$$P=2\pi r+2(L+2r)$$

And isn't the total area just the sum of the areas? $$A=A_{circle}+A_{rectange}\rightarrow A=\pi r^2 + 2rL$$

2. Feb 6, 2008

### Dick

Almost. It's only a semi-circle. So the circumerence of that is just pi*r. And the perimeter of partial rectangle is just 2L+r. Why did you double everything?

3. Feb 6, 2008

### EnumaElish

Does it look like this:
.._____
(|____|)

or not? If it does look like the above then total perimeter is not the sum of the circle & rectangle.

4. Feb 6, 2008

### rocomath

Yes, that is the correct picture! :-]

So it's not? eek.

5. Feb 6, 2008

### EnumaElish

If by perimeter you mean "edges exposed to the outside" then it is the two half-circles plus the 2 long edges of the rectangle. The short edges are "internalized."

6. Feb 6, 2008

### rocomath

AHH!!! Yes, very true.

Thanks a lot :-]

7. Feb 7, 2008

### rocomath

So even after ignoring the width, I still was unable to solve it. The person I was helping me showed me the solution and the internal parts were included. Blah.

8. Feb 7, 2008

### Dick

Then I guess they tricked you. What was the exact phrasing of the problem?

9. Feb 7, 2008

### rocomath

Sorry, I don't have the book, I will post it tomorrow. >:-[

10. Feb 8, 2008

### HallsofIvy

Staff Emeritus
The wording you gave was clear. You have a rectangle with a semicircle at each end of radius r. That end of the rectangle, then, has length 2r. For the moment call the length of the other sides l. Then the area of the two semi-circles totals $\pi r^2$ and the area of the rectangle is lr. The total area of figure is $\pi r^2+ lr$.

The perimeter of the figure is the distance around the two semi-circles, $2\pi r$ and the two lengths, 2l: the perimeter is $2\pi r+ 2l= 1/4$. You can solve that for l as a function of r and replace l by that in the area formula.

11. Feb 8, 2008

### rocomath

$$\frac 1 4=2\pi r+2l \rightarrow l=\frac{1-8\pi r}{8}$$

$$A=\pi r^2+lr$$

$$A=\pi r^2 +\left(\frac{1-8\pi r}{8}\right)r$$

$$A=\frac r 8$$

That's still not the answer in the book! Is the book wrong? I will post the actual problem in a few hours, gotta go library.

Last edited: Feb 8, 2008
12. Feb 8, 2008

### TheoMcCloskey

Wait a minute guys! Is not the area of the rectangle = $(2r)l$
?

This would make

$$A=\pi r^2+2lr$$

13. Feb 8, 2008

### HallsofIvy

Staff Emeritus
Your right. I have no idea why I wrote rl!