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Kaplan Geometry

by djkvakin » Thu Jan 21, 2010 12:45 pm
If each curved portion of the boundary of the figure attached is formed from the circumference of two semicircles, each with a radius of 2 and each of the parallel sides have a length of 4, what is the area of the region?
16
32
16-8Pi
32-8Pi
32-4Pi

The official answer is B. However, I think there is an error and the answer should be 16Pi
Here is my rationale:
[spoiler]It's a rectangle if you "unfold" the figure. One side is 4, and the other side equals the sum of lengths of 2 semicircles (which would be (2Pi*2)/2 +(2Pi*2)/2=) 4Pi. So our area of the rectangle should be 4*4Pi=16Pi. I do not see this answer among the choices.
[/spoiler]
Please explain why I am wrong.
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by Brent@GMATPrepNow » Thu Jan 21, 2010 12:57 pm
djkvakin wrote:If each curved portion of the boundary of the figure attached is formed from the circumference of two semicircles, each with a radius of 2 and each of the parallel sides have a length of 4, what is the area of the region?
16
32
16-8Pi
32-8Pi
32-4Pi

The official answer is B. However, I think there is an error and the answer should be 16Pi
Here is my rationale:
[spoiler]It's a rectangle if you "unfold" the figure. One side is 4, and the other side equals the sum of lengths of 2 semicircles (which would be (2Pi*2)/2 +(2Pi*2)/2=) 4Pi. So our area of the rectangle should be 4*4Pi=16Pi. I do not see this answer among the choices.
[/spoiler]
Please explain why I am wrong.
Image
I'm assuming the figure here is the same as the curved portion of a cylinder. If so, I'd have to concur with your solution.
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by djkvakin » Thu Jan 21, 2010 1:15 pm
I'm assuming the figure here is the same as the curved portion of a cylinder. If so, I'd have to concur with your solution.
Exactly. Sorry for poor artistry, I was just trying to draw a copy of what is drawn in the book.
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by KapTeacherEli » Thu Jan 21, 2010 9:46 pm
I agree with both of you of the logic--essentailly, we can slice and dice the semicircles to make this an easy-to-solve figure. However, doing the math myself, it seems this problem is unsolvable. Stretching the figure out as you suggested, djkvaking, will get us a parallelogram, but we don't know the angle measure to prove that it's a rectangle.

What's the source on this problem? And, is there any way for you to scan the original diagram?
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by Testluv » Fri Jan 22, 2010 3:10 am
@djkvakin: Since the radius of each semicircle is equal, as Brent points out, we can re-imagine the shape as a right circular cylinder, and the question seems to be asking for the area of the curved portion of this cylinder (i.e., the surface area of the cylinder without the top and the bottom). In that case, your solution is certainly correct. The area of the curved portion of any right cylinder is 2*pi*r*h, and here the radius is 2 while the height is 4, so it would be 16pi, which isn't among the answer choices.

I wonder about two things:

--what is the specific source of this question?

-- is it possible that some information from the figure got lost in translation?
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by sanju09 » Fri Jan 22, 2010 4:54 am
The term 'circumference or perimeter' is limited to closed figures only, hence while talking about the circumference or perimeter of a semicircle of radius r, we must take it as π r + 2 r; and if you want it to be π r only, then consider calling it semicircumference or semiperimeter of circle of radius r. Could that take us to anywhere from here in this problem?

Let's wait n watch!!
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by Silver711 » Fri Jan 22, 2010 7:17 am
Apologies in advance if I am thinking of a different question. However, I believe I recall this question so I've recreated the image to hopefully guide the discussion.

If the curved portion of each of the boundaries forms the circumference of two semicircles, then the area extended from the figure (A) and the area carved into the figure (B) will be equivalent. Therefore, the area of the figure is equal to a rectangle with a length of 8 (2 x the diameter of the circles) and a width of 4.
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by djkvakin » Fri Jan 22, 2010 7:50 am
The source is Kaplan GMAT 800 2008-2009 edition page 327 question 9. The picture is attached.
I too, re imagined the figure as a cylinder, and the answer remains the same.
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Last edited by djkvakin on Fri Jan 22, 2010 8:42 am, edited 2 times in total.
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by djkvakin » Fri Jan 22, 2010 7:51 am
Sorry, multiple post.
Last edited by djkvakin on Fri Jan 22, 2010 8:37 am, edited 1 time in total.
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by djkvakin » Fri Jan 22, 2010 7:51 am
Sorry, multiple post
Last edited by djkvakin on Fri Jan 22, 2010 8:38 am, edited 1 time in total.
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by djkvakin » Fri Jan 22, 2010 7:53 am
sorry for multiple postings - some sort of a glitch
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by djkvakin » Fri Jan 22, 2010 7:59 am
justin.silverman wrote:Apologies in advance if I am thinking of a different question. However, I believe I recall this question so I've recreated the image to hopefully guide the discussion.

If the curved portion of each of the boundaries forms the circumference of two semicircles, then the area extended from the figure (A) and the area carved into the figure (B) will be equivalent. Therefore, the area of the figure is equal to a rectangle with a length of 8 (2 x the diameter of the circles) and a width of 4.
So you are saying that the diameter of a circle equals the arc of a semicircle above it? I don't think it's true. For a circle with the diameter of 4 the arc of a semicircle above it has a length of 2PiR/2=2Pi. Which is approximately 6.28, and not 4.
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by Testluv » Fri Jan 22, 2010 12:24 pm
Okay, here's one way we can figure that the answer is 32:

We can say that the shaded figure looks like a rectangle that has two semicircles removed, AND two semicircles added.

As the two added semicircles cancel out the two removed semicircles, adopting this visual interpretation would mean that the area of the figure is just the area of the rectangle. So, it would be just length*width, where the width is 4 units, and the length is the 4 radii along the horizontal side. Each radius is 2 units long, which makes the length 8 units. 8*4 is 32.

(I think this may have been the solution Justin had in mind when he wrote his post).

I guess this question has a couple of different reasonable visual interpretations, so I wouldn't worry about this particular question too much (as official GMAT questions have had all ambiguity painstakingly removed).
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by djkvakin » Sat Jan 23, 2010 9:01 am
Testluv wrote:Okay, here's one way we can figure that the answer is 32:

We can say that the shaded figure looks like a rectangle that has two semicircles removed, AND two semicircles added.

As the two added semicircles cancel out the two removed semicircles, adopting this visual interpretation would mean that the area of the figure is just the area of the rectangle. So, it would be just length*width, where the width is 4 units, and the length is the 4 radii along the horizontal side. Each radius is 2 units long, which makes the length 8 units. 8*4 is 32.

(I think this may have been the solution Justin had in mind when he wrote his post).

I guess this question has a couple of different reasonable visual interpretations, so I wouldn't worry about this particular question too much (as official GMAT questions have had all ambiguity painstakingly removed).
Testluv -
in the gmat geometry area or length or perimeter cannot be negative, wouldn't you agree? How is it possible to "remove" a semicircle. I would have to disagree with this interpretation entirely. What we have here is a cylinder (and we don't even know if it's a right cylinder, but assuming it is we can solve for the area.)
In my opinion this question needs to be redesigned.
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