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The figure above shows a side view of the insert and the four componen

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The figure above shows a side view of the insert and the four componen [#permalink]

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New post 30 Jul 2016, 05:16
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The figure above shows a side view of the insert and the four components of a box designed to package four lightbulbs. The package is a rectangular box, whose base is a square with a side of 8 inches and whose height is 12 inches. A single rectangular insert is folded twice to form two identical diagonal planes, one at each end, and a horizontal plane in the middle, so that the box is divided into four identical compartments. What is the area of the insert in square inches?

A) 192

B) 224

C) \(64+4\sqrt{52}\)

D) \(64+96\sqrt{2}\)

E) \(64+128\sqrt{2}\)

[Reveal] Spoiler:
Attachment:
Figure.jpg
Figure.jpg [ 4.78 KiB | Viewed 2506 times ]
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Jul 2016, 06:14, edited 1 time in total.
Edited the question.

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Re: The figure above shows a side view of the insert and the four componen [#permalink]

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New post 30 Jul 2016, 12:19
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Hi NandishSS,

While this question looks complex, it's actually built around some fairly simple Geometry. It might help to break this calculation down into 'pieces' and think about the rules involved in this 3-dimensional shape.

To start, the 'insert' is a rectangle that's been folded in 2 spots. By definition, it has a length and a width; once we figure out those two dimensions, we can figure out its area.

We're meant to assume that the insert 'touches' the sides, so the width of the insert has to match the width of the box. Since the base of the box is an 8-inch square, the width of the insert is also 8 inches.

Next, we'll work on the 'middle' horizontal piece (it's the easiest part) - since it's horizontal, then it has the same length as the box (which is also 8 inches). Thus, that 'middle piece' is an 8x8 square = 64 in^2

The two diagonal pieces are the same length, so once we figure out one, we can double it and get the total area of those 2 'pieces.' You should notice how a bunch of right triangles are formed. The base of each of those triangles is 8 and the height is 6. This is a classic 3/4/5 right triangle that's been doubled to become a 6/8/10. Thus, the two diagonal pieces are 8x10 rectangles = 80 in^2 each.

64+80+80 = 224

Final Answer:
[Reveal] Spoiler:
B


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Re: The figure above shows a side view of the insert and the four componen [#permalink]

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New post 24 Aug 2016, 07:38
NandishSS wrote:
Image
The figure above shows a side view of the insert and the four components of a box designed to package four lightbulbs. The package is a rectangular box, whose base is a square with a side of 8 inches and whose height is 12 inches. A single rectangular insert is folded twice to form two identical diagonal planes, one at each end, and a horizontal plane in the middle, so that the box is divided into four identical compartments. What is the area of the insert in square inches?

A) 192

B) 224

C) \(64+4\sqrt{52}\)

D) \(64+96\sqrt{2}\)

E) \(64+128\sqrt{2}\)

[Reveal] Spoiler:
Attachment:
Figure.jpg


VeritasPrepKarishma, Abhishek009, Engr2012, Skywalker18, Bunuel Any other method to approach this problem :-)
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Re: The figure above shows a side view of the insert and the four componen [#permalink]

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New post 04 Sep 2016, 04:15
NandishSS wrote:
NandishSS wrote:
Image
The figure above shows a side view of the insert and the four components of a box designed to package four lightbulbs. The package is a rectangular box, whose base is a square with a side of 8 inches and whose height is 12 inches. A single rectangular insert is folded twice to form two identical diagonal planes, one at each end, and a horizontal plane in the middle, so that the box is divided into four identical compartments. What is the area of the insert in square inches?

A) 192

B) 224

C) \(64+4\sqrt{52}\)

D) \(64+96\sqrt{2}\)

E) \(64+128\sqrt{2}\)

[Reveal] Spoiler:
Attachment:
Figure.jpg


VeritasPrepKarishma, Abhishek009, Engr2012, Skywalker18, Bunuel Any other method to approach this problem :-)


The solution above is pretty straightforward.

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Re: The figure above shows a side view of the insert and the four componen [#permalink]

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New post 13 Nov 2017, 02:00
EMPOWERgmatRichC wrote:
Hi NandishSS,

While this question looks complex, it's actually built around some fairly simple Geometry. It might help to break this calculation down into 'pieces' and think about the rules involved in this 3-dimensional shape.

To start, the 'insert' is a rectangle that's been folded in 2 spots. By definition, it has a length and a width; once we figure out those two dimensions, we can figure out its area.

We're meant to assume that the insert 'touches' the sides, so the width of the insert has to match the width of the box. Since the base of the box is an 8-inch square, the width of the insert is also 8 inches.

Next, we'll work on the 'middle' horizontal piece (it's the easiest part) - since it's horizontal, then it has the same length as the box (which is also 8 inches). Thus, that 'middle piece' is an 8x8 square = 64 in^2

The two diagonal pieces are the same length, so once we figure out one, we can double it and get the total area of those 2 'pieces.' You should notice how a bunch of right triangles are formed. The base of each of those triangles is 8 and the height is 6. This is a classic 3/4/5 right triangle that's been doubled to become a 6/8/10. Thus, the two diagonal pieces are 8x10 rectangles = 80 in^2 each.

64+80+80 = 224

Final Answer:
[Reveal] Spoiler:
B


GMAT assassins aren't born, they're made,
Rich


Hi. Please help, I don't even understand the language of the question. Please break the question down as I cant connect the language to the given diagram.
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Re: The figure above shows a side view of the insert and the four componen [#permalink]

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New post 13 Nov 2017, 03:47
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NandishSS wrote:
Image
The figure above shows a side view of the insert and the four components of a box designed to package four lightbulbs. The package is a rectangular box, whose base is a square with a side of 8 inches and whose height is 12 inches. A single rectangular insert is folded twice to form two identical diagonal planes, one at each end, and a horizontal plane in the middle, so that the box is divided into four identical compartments. What is the area of the insert in square inches?

A) 192

B) 224

C) \(64+4\sqrt{52}\)

D) \(64+96\sqrt{2}\)

E) \(64+128\sqrt{2}\)

[Reveal] Spoiler:
Attachment:
Figure.jpg


Too much of confusing explanation is given in the question which makes me think that the diagram tells it all. That is pretty much what the first sentence of the question says too.

We see the 8 x 12 rectangular box as shown in the figure.
A single rectangular insert is folded twice to make four identical compartments. We see in the figure that we have 4 identical right triangular compartments. So the insert is just the lightening shape shown in the figure. We need its area. It is actually a rectangle which is folded into this shape. If we open it up, we will get a rectangle and its area will be the total height * Width (which is 8 since the box has a square base of side 8)

Now all we need is the height. Since the compartments are equal, height of each compartment is 6. So each diagonal in the figure is 10 (using multiple of pythagorean triplet 3-4-5).
Total height of the rectangular insert = 10 + 8 + 10 = 28

Its area = 28*8 = 224
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Re: The figure above shows a side view of the insert and the four componen [#permalink]

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New post 13 Nov 2017, 18:35
NandishSS wrote:
The figure above shows a side view of the insert and the four components of a box designed to package four lightbulbs. The package is a rectangular box, whose base is a square with a side of 8 inches and whose height is 12 inches. A single rectangular insert is folded twice to form two identical diagonal planes, one at each end, and a horizontal plane in the middle, so that the box is divided into four identical compartments. What is the area of the insert in square inches?

A) 192

B) 224

C) \(64+4\sqrt{52}\)

D) \(64+96\sqrt{2}\)

E) \(64+128\sqrt{2}\)

[Reveal] Spoiler:
Attachment:
Figure.jpg



I'm surprised this is an official GMAT question, it's pretty badly worded. I took me a good 10 minutes to understand that we're dealing with a 3D shape, and that the base (not shown) is 8x8...

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Re: The figure above shows a side view of the insert and the four componen   [#permalink] 13 Nov 2017, 18:35
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The figure above shows a side view of the insert and the four componen

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