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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
I thought it was B>A, but now I think A>B... Sandy, can you give the solution?
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
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The result of cutting this cube is two prisms as shown below.

Attachment:
prism.png.jpg
prism.png.jpg [ 833.33 KiB | Viewed 6351 times ]


Now the two triangles on the A and B form a square of side 3 feet. The base is a rectangle with one side 3 feet long and other \(3\sqrt{2}\) long (diagonal of a square is side times square root of 2).

So we have effectively 3 squares of 3 feet side length and one rectangle of sides 3 and \(3\sqrt{2}\).

Total area = 39.72.

Quantity A is greater.
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
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Sandy's solution is correct, although I think he meant to answer A, not B. Here's how I would do it in a less math-intensive way.
Each side of the block is 9 sq ft. After it's been cut in half, we can just count sides of the block we've got. If we choose the block closest to us, then we've got the front side, the left side, and then the top and bottom sides. I'll ignore the cut through the middle for now. Since the top and bottom sides are each exactly half of a side, they can be thought of as on more side together. So since we've got 3 sides of 9 sq ft each, so far we've got 27 sq ft.
What about that last side? Well we know that quantity B is 36, so if the last side is 9, the quantities would be equal. However, we know it's bigger than 9 since its dimensions are going to be 3 by whatever the diagonal of the side is. I know the diagonal of a 3x3 square will be longer than 3, so the sides of the block must be bigger than 36 sq ft. Thus, A is the answer.

Now, we don't actually need to know the size of the last side to get the answer, but if you know your special triangles and square roots it would be pretty fast. Any square cut across its diagonal will form 2 right isosceles triangles with side ratios of 1-1-√2. So the diagonal of this particular square should be 3√2 and the dimensions of that central side would be 3x3√2. Since we know (or we should) that √2 is about 1.4, then we again know this last side will be bigger than 9.
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
sandy wrote:
The result of cutting this cube is two prisms as shown below.

Attachment:
prism.png.jpg


Now the two triangles on the A and B form a square of side 3 feet. The base is a rectangle with one side 3 feet long and other \(3\sqrt{2}\) long (diagonal of a square is side times square root of 2).

So we have effectively 3 squares of 3 feet side length and one rectangle of sides 3 and \(3\sqrt{2}\).

Total area = 39.72.

Quantity B is greater.


I think you are wrong. The answer is A (39.72 square feet) which is bigger than B (36 square feet)
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
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I meant A. Sorry fixed it.
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
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Think it about this way:

Each side of the cube has an area or 9 ft^2. Once cut as in the picture, you are left with two whole sides and two half sides (the half upper and half lower), which means that you are left with three whole parts, adding an area of 27.
Then just estimate the area of the "cut" side (shaded), which has an height of 3 and a base of 3√2 (since all the diagonals of a square are x*√2, being x the side of the square).

Therefore the shaded area is 3*3√2=9√2.

The full area is 27 + 9√2. Since 9√2>9 then 27 + 9√2> 27 + 9 =36.

Quantity A is greater.
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
Thank you guys....
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Re: GRE Math Challenge #72- A solid cubical block of wood [#permalink]
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