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# The U.S. Defense Department has decided that the Pentagon is

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Joined: 12 Aug 2011
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The U.S. Defense Department has decided that the Pentagon is  [#permalink]

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Updated on: 13 Jul 2013, 05:59
2
5
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Difficulty:

75% (hard)

Question Stats:

59% (02:55) correct 41% (02:47) wrong based on 115 sessions

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The U.S. Defense Department has decided that the Pentagon is an obsolete building and that it must be replaced with an upgraded version: the Hexagon. The Secretary of Defense wants a building that is exactly 70 feet high and 200 feet on a side, and that has a hexagonal bull's-eye cutout in the center (somewhat like the current one) that is 50 feet on a side. What will be the volume of the new building in cubic feet?

A. 3,937,500 cubic feet
B. 15,750 cubic feet
C. 11550√3 cubic feet
D. 15,750√3 cubic feet
E. 3,937,500√3 cubic feet

(Source: McGraw-Hill's GMAT: Graduate Management Admission Test)

Originally posted by Azizbek1983 on 11 Sep 2011, 11:12.
Last edited by Bunuel on 13 Jul 2013, 05:59, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Hexagon  [#permalink]

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11 Sep 2011, 12:39
1
Imagine a hexagonal building with base , side and roof as hexagon
Sum of angle of hexagon = 180(n-2) where n is number of side
= 180(6-2)=720
Each angle will be 720/number of sides = 720/6=120

Now imagine the base of building with six sides. You can form 6 triangle all converging at the center of hexagon. These 6 triangle each will be equilateral triangle with side 200 feet as that is the side of base of hexagon.
Area of 1 such equilateral triangle = $$\sqrt{3} / 4 * (side)^2$$
=$$\sqrt{3}/ 4 * (200)^2 = 10,000 \sqrt{3}$$

Area of 6 such triangle will be= $$6*10,000 \sqrt{3}= 60,000\sqrt{3}$$

Now we know area of base of hexagon. To find volume we need third dimension which is height
Hence volume of complete hexagon = area of hexagon * height
= $$60,000\sqrt{3} * 70 = 4,200,000\sqrt{3}$$

Now there is hexagon bull's eye cut out right through the building.
Proceeding in similar fashion to find out the volume of this hexagonal bull's eye
Volume of hexagon bull's eye = area of hexagonal bull's eye * height
= (area of 6 triangle with base 50) * height
=$$3750 \sqrt{3}* 70 =262500\sqrt{3}$$

Hence volume of new building= volume of complete building - volume of bull's eye
= $$4,200,000\sqrt{3}-262500\sqrt{3}$$
= $$3,937,500\sqrt{3}$$

OA E.

On a slightly different note, with 14 million unemployed, last thing Defense Department needs is new building.
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Re: Hexagon  [#permalink]

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11 Sep 2011, 17:15
Volume of the Hexagon with side 200 and height 70 = Area * height

= 6*(sqrt(3)/4)(200^2)(70)

Volume of the center bull's eye that is similar in shape of a hexagon but side 50

= 6 * (sqrt(3)/4)(50^2)*70

Volume of the building = 6*(sqrt(3)/4)(200^2)(70) - 6 * (sqrt(3)/4)(50^2)*70
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Re: Hexagon  [#permalink]

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12 Sep 2011, 01:05
Good question. The logic lies in picking up the regular hexagon principle and making it 6 equilateral triangles. Well explained Jamifahad.
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Re: The U.S. Defense Department has decided that the Pentagon is  [#permalink]

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30 Nov 2013, 04:13
6x70x$$\sqrt{3}$$/4$$[(200)^2-(50)^2]$$
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Re: The U.S. Defense Department has decided that the Pentagon is  [#permalink]

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12 Dec 2013, 09:25
A hexagon with six equal sides can be divided up into six, equal, equilateral triangles.

We know the "base" of each triangle is 200. We also know that you can divide an equilateral triangle up into two identical 30:60:90 triangles. This allows us to find the the height of each triangle. With this we know the area (or the area of the roof) is 10000√3. Then we multiply by the height 70 to get the volume --> 700000√3. Finally, we multiply by 6 to get the total area of 4200000√3. However, we need to account for the empty space in the middle which too is an equilateral hexagon. The same process we applied to the large hexagon can be applied to the small hexagon. The base of each triangle is 50 so the height is 50√3. The area is 1/2*50*50√3 = 1250√3. Then we multiply that by the height to get volume which is 87500√3. Finally we multiply by 6 to account for the four equilateral triangles --> 525000√3. We simply subtract 525000√3 from the total to get the answer.
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Re: The U.S. Defense Department has decided that the Pentagon is  [#permalink]

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11 Feb 2014, 06:08
madn800 wrote:
6x70x$$\sqrt{3}$$/4$$[(200)^2-(50)^2]$$

This is it, but I don't think this is a realistic question since the calculations needed to get to the final answer are long and tedious. The question is probably OK since it tests some basic geometry but answer choices could be somewhat summarized.

Cheers
J
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Re: The U.S. Defense Department has decided that the Pentagon is  [#permalink]

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12 Feb 2014, 04:50
2
jlgdr wrote:
madn800 wrote:
6x70x$$\sqrt{3}$$/4$$[(200)^2-(50)^2]$$

This is it, but I don't think this is a realistic question since the calculations needed to get to the final answer are long and tedious. The question is probably OK since it tests some basic geometry but answer choices could be somewhat summarized.

Cheers
J

Actually, you don't really need to calculate a lot because we don't need to get to the actual answer. The options are such that you can find the answer by setting up the numbers. Let me explain.

The volume is given as Surface Area * Height.
What we need to do is find the surface area of the large hexagon - surface area of the small hexagon.
How do you find the area of a hexagon? Let's split in into two equal trapezoids
Area of hexagon = 2* area of trapezoid = 2*(1/2)*Height * (Sum of bases) = Height * Sum of bases

Attachment:

Ques3.jpg [ 8.97 KiB | Viewed 1771 times ]

Since each internal angle of a hexagon is 120 degrees, we get a 30-60-90 triangle with sides in the ratio $$1:\sqrt{3}:2$$. Since hypotenuse is 200, the other two sides are 100 and $$100\sqrt{3}$$.

Area of large hexagon $$= 100\sqrt{3}* (200 + 400)$$
Similarly, we know that area of small hexagon $$= 25\sqrt{3}*(50 + 100)$$ (instead of 200, just put 50)

Surface area of the building $$= 100\sqrt{3}* (600) - 25\sqrt{3}*(150) = 10\sqrt{3}(6000 - 375)$$
Volume will be given by multiplying the surface area by 70 (the height)

So we will have $$\sqrt{3}$$ in the volume and two 0s. Only (E) satisfies these conditions.
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Re: The U.S. Defense Department has decided that the Pentagon is  [#permalink]

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10 Sep 2017, 06:52
Ans is E
3937500√3

See attached image for more clearity.
Attachments

IMG_1188.JPG [ 1.33 MiB | Viewed 555 times ]

Re: The U.S. Defense Department has decided that the Pentagon is   [#permalink] 10 Sep 2017, 06:52
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