# Express Area as a function of r

by rocomath
Tags: express, function
 P: 1,756 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? $$A=A_{circle}+A_{rectange}\rightarrow A=\pi r^2 + 2rL$$
 Sci Advisor HW Helper Thanks P: 25,251 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?
 Sci Advisor HW Helper P: 2,482 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.
P: 1,756
Express Area as a function of r

 Quote by 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.
Yes, that is the correct picture! :-]

So it's not? eek.
 Sci Advisor HW Helper P: 2,482 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."
P: 1,756
 Quote by 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."
AHH!!! Yes, very true.

Thanks a lot :-]
P: 1,756
 Quote by 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."
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.
 Sci Advisor HW Helper Thanks P: 25,251 Then I guess they tricked you. What was the exact phrasing of the problem?
P: 1,756
 Quote by Dick Then I guess they tricked you. What was the exact phrasing of the problem?
Sorry, I don't have the book, I will post it tomorrow. >:-[
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,338 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.
P: 1,756
 Quote by HallsofIvy 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.
$$\frac 1 4=2\pi r+2l \rightarrow l=\frac{1-8\pi r}{8}$$

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

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

$$A=\frac r 8$$

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.
P: 182
 ... and the area of the rectangle is lr
Wait a minute guys! Is not the area of the rectangle = $(2r)l$
?

This would make

$$A=\pi r^2+2lr$$
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,338 Your right. I have no idea why I wrote rl!

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