
#1
Feb608, 01:43 PM

P: 1,757

Ok, so I have this picture. A semicircle at the ends of a rectangle. It tells me that the perimeter is [tex]\frac 1 4[/tex]. So, isn't the total perimeter just the sum of the circle and rectange? [tex]P=2\pi r+2(L+W)[/tex]
[tex]r=radius[/tex] [tex]W=2r[/tex] [tex]P=2\pi r+2(L+2r)[/tex] And isn't the total area just the sum of the areas? [tex]A=A_{circle}+A_{rectange}\rightarrow A=\pi r^2 + 2rL[/tex] 



#2
Feb608, 01:47 PM

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Almost. It's only a semicircle. 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
Feb608, 01:49 PM

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P: 2,483

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
Feb608, 01:53 PM

P: 1,757

Express Area as a function of rSo it's not? eek. 



#5
Feb608, 01:58 PM

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




#7
Feb708, 10:18 PM

P: 1,757





#8
Feb708, 10:29 PM

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P: 25,178

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




#10
Feb808, 05:47 AM

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P: 38,904

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 semicircles totals [itex]\pi r^2[/itex] and the area of the rectangle is lr. The total area of figure is [itex]\pi r^2+ lr[/itex].
The perimeter of the figure is the distance around the two semicircles, [itex]2\pi r[/itex] and the two lengths, 2l: the perimeter is [itex]2\pi r+ 2l= 1/4[/itex]. You can solve that for l as a function of r and replace l by that in the area formula. 



#11
Feb808, 08:58 AM

P: 1,757

[tex]A=\pi r^2+lr[/tex] [tex]A=\pi r^2 +\left(\frac{18\pi r}{8}\right)r[/tex] [tex]A=\frac r 8[/tex] 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. 



#12
Feb808, 11:03 AM

P: 183

? This would make [tex]A=\pi r^2+2lr[/tex] 



#13
Feb808, 02:30 PM

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P: 38,904

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



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