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

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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tkorzhan1995 I hope the figure attached will help you to understand this question better.

Note: If you could visualize a 3D figure based on the information's given and understand what area we need to find here, solving this question would be easier.

We have a rectangular box, whose base is a square with a side of 8 inches and whose height is 12 inches. A single rectangular sheet ( Yellow shaded region in the attached figure ) 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. We are asked to find the area of the rectangular insert.

Please refer to the figure attached for the 3D -view and we are asked to find the yellow shaded region in the figure.




Let's cut the entire rectangular insert into three parts: Sheets 1,2 and 3 as marked in the figure. The width of each sheet will be the same i.e 8 inches.

Since the box is divided into 4 identical compartments, the height of each compartment will be 12/2 = 6 inches.

The length of rectangular sheet 1 will be the hypotenuse of the right-angled triangle formed as shown in the figure with a base of 8 inches and a height of 6 inches.

Length of sheet 1 = Sq.root(\(8^2 + 6^2\)) = 10 inches

Area of Sheet 1 = 10 * 8 = 80 sq. inches

It's given in the question that Sheet 1 and Sheet 3 are two identical diagonal planes, hence their area will also be the same i.e 80 sq. inches.

Sheet 2 will be a square of base 8 cm, hence the area of sheet 2 = 8 * 8 = 64 sq. inches


Area of rectangular insert = Area of Sheet 1 + Area of sheet 2 + Area of sheet 3= 80 + 64 + 80 = 224 sq. inches.

Option B is the correct answer.

Thanks,
Clifin J Francis,
GMAT Mentor
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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:

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

VeritasPrepKarishma, Abhishek009, Engr2012, Skywalker18, Bunuel Any other method to approach this problem
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The solution above is pretty straightforward.

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.

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...

VeritasKarishma Thanks for such a nice explanation!

This was my first question in the practice test and I decided to move on by marking randomly after getting puzzled with the language.
If we just open up the sides we can get the height of the figure and we already know the base.
Maybe the text was just for distraction! Dugh!

I'm having trouble understanding why the area of the rectangular box wouldn't be: (10+8+6)*8? Isn't the right side of the triangle 6 in.? I'm missing where the other 10 in. comes from.

Hi pthet9801,

I assume that you're actually talking about the INSERT and not the BOX (since we're told the dimensions of the box - it's 8x8x12).

The insert can be broken down into 3 pieces: the two 'diagonal' sections and the middle 'horizontal' section. Since we're meant to assume that the insert 'touches' the sides, the width of the insert has to match the width of the box. The base of the box is an 8-inch square, so the width and length of the middle section is also 8x8 inches = 64 square inches.

The two diagonal pieces are the same length. You should notice how a bunch of right triangles are formed by those two pieces. The base of each of those triangles is 8 and since each 'pair' of triangles is HALF the height (re: half of 12 inches), the height of each triangle is 6 inches. This is a classic 3/4/5 right triangle that's been doubled to become a 6/8/10. Thus, the two diagonal pieces have a width of 8 inches and a length of 10 inches = 80 square inches each.

If you wanted to form a calculation in the way that you did in your post, then it would be (8+10+10)(8).

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

EMPOWERgmatRichC

Thank you for the explanation. I was completely off on what the insert actually looked like... For some reason I thought the insert was 4 different slots of a 2-fold insert... I'm not sure why I thought that, but I understand now that the insert starts on the top right of the box then diagonals down to middle left then horizontal to the right middle then diagonal down to bottom left. So you have 2 diagonals of 10 and a horizontal of 8 and a width of 8. Therfore, Length = 10+10+8 and width=8 A= 28*8=224. Thank you so much for the help.

Tricky visualization problem.....If you can visualize the Box in your head and realize that it is the SIDE view of a 3-D Rectangular Box, we can imagine what would happen if we pulled the Insert Out, Straightened the Diagonals out, and turned the Insert into a Rectangle. Then we can find the Surface Area of the Insert.

(1st) getting the Width of the Insert after Straightening it out into a Rectangle

Since the Bottom Base of the Box is a Square, the Insert must have a Breadth = 8 Inches. If we were to pull the Insert out and Straighten it into a Rectangle, it would have a Breadth into the 'background' of the picture = 8 inches.


(2nd) Finding the Length/Other Side of the Straightened Out Insert - Rectangle

we are told that the insert creates a "horizontal plane in the middle" of the 3-D Rectangular Box.

thus from the Side View we can say that the Line in the Middle of the figure is Parallel to the Bottom Side of 8 inches and Bisects the 12 Inch Height on the Sides of the figure.

The Insert has created 2 Right Triangles with Legs of 6 and 8 ------> which is a Multiple of a 3-4-5 Pythagorean Triplet Right Triangle.

Thus, the Hypotenuse of Each Triangle = 10

+

Middle Horizontal Plane that Stretches across the "middle" of the Rectangular Box = 8 inches
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Length = 10 + 10 + 8 = 28



The Area of the Insert, when stretched out to form a Rectangle, equals:

(Width of 8 Inches) * (Length of 28 Inches) = 8 * 28 = 224

-B-

GMATNinja, Bunuel, could you please help to understand what area we need to find here? (It will be really appreciated, if you can please highlight in the picture, what area should be found here).
I have hard time to understand why we are adding 8+10+10 to determine height. It may be due to the fact that I don't have a clear understanding what area should be found here.

GMATNinja, Bunuel, could you please help to understand what area we need to find here? (It will be really appreciated, if you can please highlight in the picture, what area should be found here).
I have hard time to understand why we are adding 8+10+10 to determine height. It may be due to the fact that I don't have a clear understanding what area should be found here.

CrackverbalGMAT is absolutely correct with the sketch.

I will add on to it a bit

INSERT is the cardboard which is placed inside ( Red in colour ).
Partitions are equal so ABCD gives us the length of the partition as the entire INSERT is made by folding the rectagular cardboard at two places - B and C.

AB=CD= hypotenuse with other two sides 6 and 8, so both are 10.

Thus the insert becomes a rectangular piece with sides 8 and 10+8+10 as shown in the third sketch.
Area = 8*28 = 224

B

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The question is all about finding the area of the insert, which has two surfaces in the form of slanting squares and one flat square.

The shape of the insert can be identified by reading the question carefully, i.e., the shape of the entire structure is a rectangle. Therefore, the part which isn't visible has to be a square as the base of the structure is square (mentioned in the question).

The area of the flat part of the insert = 8*8= 64
the area of the slanting part of the insert = 8* square root of 6^2 + 8^2 = 8*10 = 80
Total area = 2(80) + 64 = 224

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