BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Geometry: Strange Figure

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Fri Feb 27, 2009 8:38 pm

Geometry: Strange Figure

by TryHarder » Wed May 20, 2009 9:47 am
Source: Kaplan 800
(Figure is attached)

If each curved portion of the boundary of the figure above is formed from the circumferences of two semicircles, each with a radius of 2, and each of the parallel sides has length 4, what is the area of the shaded figure?

The official anwers is 32
The explaination is that the figure can be visualized as a rectangle with one side as 4 (which is ok!) and the other side as 4x2=8 (which does not make sense to me)

I think the correct answer is 2pirh = 2pi x 2 x 4 = 16pi

Here is how I think I can prove that the answer is wrong. Let us assume that 32 is infact the right answer for a moment. Then it should be the surface area of the hollow right circular cylinder (with no circular lids on either side). We know that this part of the surface area is 2pi rh

Hence if 2pi rh = 32
then 2pi r x 4 = 32 (4 is the undisputed height)
then r = 4/pi but we know that r = 2

Bottomline, I think that the curvature of the side has been overlooked!
Attachments
GeoFig.JPG
Referenced figure
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 19, 2009 10:48 am
Location: Chicago
Thanked: 1 times
GMAT Score:680

by Katrusya » Wed May 20, 2009 10:24 am
Look at the upper line.
Since both semicircles have same radius, you can cut out the upper semicircle and put it next to it. You'll get the upper line consisted of 2 diameters of 2 semicircles. (D=2R--> 2D=8)
Same you could do with bottom line.
Area =4*8=32
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Wed May 20, 2009 10:36 am
If you visualize it conceptually, you will notice that if you bring the two semicircles together, the figure becomes a cylinder (you could use a paper as an example to demonstrate this.

Height of a cylinder = 4

radius is still 2

area of a cylinder is (pi*r^2) * 2 * h = 32pi
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 19, 2009 10:48 am
Location: Chicago
Thanked: 1 times
GMAT Score:680

by Katrusya » Wed May 20, 2009 10:43 am
Culinder is 3D figure. Formula that you use is for Volume of Culinder and has nothing to do with area. You are given 2D figure.
You can divide this figure on some simple figures and calculate the areas of those and then sum it up. But here I can't see such a separation on simple figures. Therefore,you need to transform it to the shape that will allow you to find an area.
Top
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Fri Feb 27, 2009 8:38 pm

Strange figure

by TryHarder » Wed May 20, 2009 10:50 am
Katrusya,
Help me understand this - If you move the first circular area over the second, you get a hollow cylinder. Then why would the surface area not be 2pi*rh?
Second, I agree you can flatten the cylinder to form a rectangle and find the area that way - but then the length would not be equal to 4 times the radius, it would be more. Remember you are flattening two sem-circumference which is equal to 2pi*r. Thus even with this approach the answer should be lxb = 2pi*r*h
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 19, 2009 10:48 am
Location: Chicago
Thanked: 1 times
GMAT Score:680

by Katrusya » Wed May 20, 2009 11:17 am
Take the semicircle 1 and put it instead of room of semicircle 2. They are the same size. Do the same for the bottom line.
The length of new figure - rectangular - will be not the lenght of curcomference (wich is 2pi*r), but simply 2 Diametres of 2 semicircles.

Yes, you are right. If you flatten cylinder, then the length of rect will be equal to length of curcomference. But this is not the case here.

I hope my picture will help!
Attachments
geofig_727-1.JPG
Top
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Fri Feb 27, 2009 8:38 pm

Curved surface

by TryHarder » Wed May 20, 2009 11:43 am
Katrusya,
Thanks for taking the time, but I am not convinced, and in fact I plan to email this to Kaplan. My issue is simply this, the author is asking us to get the area of a curved surface which to me means that we do either of the following
Note: The question says "what is the area of the curved surface?"
a. Visualize this as a cylinder by covering the cross section and use 2pi*rh
OR
b. Visualize this as a rectangle but ensuring that we flatten out the curvature (each of which is a semicircle i.e. pi *r) and then use l x b where l = 2pi *r and b = 4)

Any others who would like to chime in?
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 19, 2009 10:48 am
Location: Chicago
Thanked: 1 times
GMAT Score:680

by Katrusya » Wed May 20, 2009 11:57 am
According to my pic, you take figure 1 (imagine that you cut off that semicircle), turn it clockwise and place it instead of figure2. Can you see small "r"s that stand there for Radius.
I wish I were near by you to explain it better.
Top
Master | Next Rank: 500 Posts
Posts: 101
Joined: Tue Apr 07, 2009 12:30 am
Thanked: 2 times
GMAT Score:750

Re: Curved surface

by Svedankae » Thu May 21, 2009 11:23 am
TryHarder wrote:Katrusya,
Thanks for taking the time, but I am not convinced, and in fact I plan to email this to Kaplan.
come on, dont make a fool out of yourself.

Just dont see it as a 3d figure. because it aint one. its a rectangle with 4*8
Top
Post new topic Post Reply
• Page 1 of 1