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 Post subject: geometryPosted: Fri Jul 02, 2010 10:07 pm

Joined: Sun Apr 04, 2010 2:11 pm
Posts: 8
Can you please explain the problem number 23 in the Geometry VI book.

thanks,
Bandita

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 Post subject: Re: geometry book page 53, number 23Posted: Sat Jul 03, 2010 5:26 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
This one may be easiest if you use a flat sheet of paper to visualize it.
If we were to fold up a little of each side, we'd have an open box with height x (that's the "little bit" you folded from each side.)
Now, in this case, we fold up x from each side, leaving a width of 15-2x (It's 2x instead of x because we fold up x from both sides instead of just one side) and a length of 12-2x (Again, the 2x because we are folding from both sides.)

The volume of this open box is now w*L*H which is (15-2x)(12-2x)(x)
We can multiply this together, or we can start with our answer choices. I'd do the latter, because multipying this will result in something a little unwieldy.

The most efficient path here may be to make a chart: (We'll see if it comes through well in this format...)
(We're using height = x, width=15-2x, and length = 12-2x

Height Width Length Volume
A 5 5 2 50
B 4 7 4 112
C 3 9 6 162
D 2 11 8 176
E 1 13 10 130

(You may be tempted, as you see a pattern, to stop calculating at C or D when you see that the volumes are getting bigger -- if you do this, you'll miss noting that it gets smaller again when the height is just 1... )

Also note that you can probably estimate on these -- no need to be terribly specific here...

Hope this helps,

Veritas Help

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 Post subject: Re: geometryPosted: Sat Oct 01, 2011 12:26 pm

Joined: Fri Sep 16, 2011 8:30 am
Posts: 1
Can anyone help me with this?

 Attachments: problem.jpg [ 132.58 KB | Viewed 1615 times ]
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 Post subject: Re: geometryPosted: Tue Oct 04, 2011 5:20 pm

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Hey Gy,

Love this question! The key to this one comes in recognizing that, if the triangle is an equilateral triangle, then:

-All angles are 60 degrees
-The square splits it into three sub-triangles to the left, top, and right of the square. And since the bottom of the square runs parallel to the base of the triangle, then:

-The left and right triangles are right triangles with the 60-degree angle of the equilateral triangle remaining unchanged in each
-the top triangle is another equilateral triangle with all angles 60 (the top of the square runs parallel to the base of the triangle, so the angles don't change)

So, knowing that the sides of the square are all 12, we know that the top triangle has all sides of 12. And with the vertical sides of the square creating the 60-degree side of a 30-60-90 triangle, then solving for the sides of the left/right triangles we know that they're in the ratio:

x, x*sqrt 3, 2x (with 2x being the hypotenuse)

So x*sqrt 3 = 12 (the middle side)
And therefore x = 12/(sqrt 3). But the GMAT doesn't let you use radicals in the deonimntor, so we'll multiply top and bottom by sqrt3 to rationalize the denominator:

12/sqrt3 * sqrt3/sqrt3 = 12sqrt3 / 3 = 4sqrt 3

So we know that x = 4sqrt 3, but that's the small side of the right/left triangles. The hypotenuses are double that, so 8sqrt 3. So any one side of the large triangle is comprised of 12 (from the top triangle) + 8sqrt 3 (from the right or left triangle) so the perimeter will be 3 (the number of sides) * 12 + 8sqrt 3 (the length of one side) = 36 + 24sqrt 3.

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