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# The figure above was cut out of paper and folded to form a pyramid wit

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Retired Moderator
Joined: 29 Apr 2015
Posts: 751
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Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
The figure above was cut out of paper and folded to form a pyramid wit  [#permalink]

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31 May 2015, 03:04
1
7
00:00

Difficulty:

75% (hard)

Question Stats:

58% (02:19) correct 42% (01:57) wrong based on 141 sessions

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T6254.png [ 5.4 KiB | Viewed 3262 times ]
The figure above was cut out of paper and folded to form a pyramid with a square base and four equilateral triangles. If the side of the pyramid's base is 1, what is the height of the pyramid?

A. 1/2
B. √2/2
C. (√3)/2
D. (√5)/2
E. √3

Attachment:

T6254c.png [ 41.26 KiB | Viewed 3200 times ]

Split one of the equilateral triangle at the center to mark the height, as shown in the figure. The split results in two 30, 60, 90 right triangles with sides of ratio 1, √3, 2. Given that the base of the square is 1, short leg of each of the 30:60:90 triangles is 0.5, and the height of the equilateral triangles is 0.5√3.

Now calculate the height of the pyramid through the triangle formed by connecting the center of the base, point x and the tip of the pyramid. The base and hypotenuse of this triangle are 0.5 and 0.5√3. Use the Pythagorean theorem to calculate the height:

(0.5√3)2 = 0.52 + h2

--> 0.52(√3)2 = 0.52 + h2

1/22 is 1/4, and (√3)2 is simply 3 (as the square root and power cancel each other out), so

--> 1/4⋅3 = 1/4 + h2

--> h2 = (3/4) - (1/4)

--> h2 = 1/2

--> h = 1/√2

To make the final transition to √2/2, multiple the top and bottom by √2:

(1/√2) × (√2/√2) = √2/(√2·√2) = √2/2
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4483
Re: The figure above was cut out of paper and folded to form a pyramid wit  [#permalink]

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01 Jun 2015, 15:53
1
reto wrote:
Attachment:
The attachment T6254.png is no longer available
The figure above was cut out of paper and folded to form a pyramid with a square base and four equilateral triangles. If the side of the pyramid's base is 1, what is the height of the pyramid?

A. 1/2
B. √2/2
C. (√3)/2
D. (√5)/2
E. √3

Dear reto,
This is a fun question, and I am happy to respond.
Attachment:

square-based pyramid, equilateral faces.JPG [ 16.09 KiB | Viewed 3112 times ]

Let M be the center of the square base, and let F be the midpoint of AB. We know FM = 1/2.

We know AEB is an equilateral triangle, so EFB is a 30-60-90 triangle. See this blog for more on these:
http://magoosh.com/gmat/2012/the-gmats- ... triangles/
We know that EB = 1 and that MF = 1/2, so
$$EF = \sqrt{3}/2$$

Now, look at right triangle EFM: we know the hypotenuse and horizontal leg, and we want the vertical leg.
$$(EM)^2 = (EF)^2 - (MF)^2 = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$$

Therefore, the height, EM, is the square root of 1/2:
$$EM = \sqrt{2}/2$$

Mike
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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10664
Location: Pune, India
Re: The figure above was cut out of paper and folded to form a pyramid wit  [#permalink]

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01 Jun 2015, 19:54
1
reto wrote:
Attachment:
T6254.png
The figure above was cut out of paper and folded to form a pyramid with a square base and four equilateral triangles. If the side of the pyramid's base is 1, what is the height of the pyramid?

A. 1/2
B. √2/2
C. (√3)/2
D. (√5)/2
E. √3

Attachment:
T6254c.png

Split one of the equilateral triangle at the center to mark the height, as shown in the figure. The split results in two 30, 60, 90 right triangles with sides of ratio 1, √3, 2. Given that the base of the square is 1, short leg of each of the 30:60:90 triangles is 0.5, and the height of the equilateral triangles is 0.5√3.

Now calculate the height of the pyramid through the triangle formed by connecting the center of the base, point x and the tip of the pyramid. The base and hypotenuse of this triangle are 0.5 and 0.5√3. Use the Pythagorean theorem to calculate the height:

(0.5√3)2 = 0.52 + h2

--> 0.52(√3)2 = 0.52 + h2

1/22 is 1/4, and (√3)2 is simply 3 (as the square root and power cancel each other out), so

--> 1/4⋅3 = 1/4 + h2

--> h2 = (3/4) - (1/4)

--> h2 = 1/2

--> h = 1/√2

To make the final transition to √2/2, multiple the top and bottom by √2:

(1/√2) × (√2/√2) = √2/(√2·√2) = √2/2

The question needs you to use your imagination. The height of the pyramid will be given by a line drawn from the centre of the square up to the point in air where vertices of all 4 equilateral triangles meet. This height forms the leg of a right triangle - the other leg is half the length of a side of square (formed by joining the centre of the square to the mid point of any one of its sides). Its length will be 1/2. The hypotenuse of this triangle will be the altitude of the equilateral triangle. Its length will be $$\sqrt{3}/2$$ because sides of the equilateral triangles are 1 each (they share a side with the square).

If this is hard to understand, try cutting a piece of paper in this shape. Draw a point at the centre of the square and join it to the mid point of any one side of the square. Also draw the altitude of that triangle. Now join all four triangles and see what happens.

Using pythagorean theorem, you get

$$Height^2 + (1/2)^2 = (\sqrt{3}/2)^2$$
$$Height = \sqrt{2}/2$$

_________________
Karishma
Veritas Prep GMAT Instructor

Retired Moderator
Joined: 29 Apr 2015
Posts: 751
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
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The figure above was cut out of paper and folded to form a pyramid wit  [#permalink]

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05 Sep 2015, 00:01
Thanks for your solutions. It is very tricky to first identify the two 30:60:90 triangles (ECF) and (DEF) and then conclude that EF is $$0.5\sqrt{3}$$. If you in a rush, you will focus on EMF and asking you how to figure out its altitude (Let's call the midpoint of the square below M):

Attachment:

T6254c.png [ 41.26 KiB | Viewed 2833 times ]
Manager
Joined: 29 Jul 2015
Posts: 150
Re: The figure above was cut out of paper and folded to form a pyramid wit  [#permalink]

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05 Sep 2015, 10:48
reto wrote:
Attachment:
T6254.png
The figure above was cut out of paper and folded to form a pyramid with a square base and four equilateral triangles. If the side of the pyramid's base is 1, what is the height of the pyramid?

A. 1/2
B. √2/2
C. (√3)/2
D. (√5)/2
E. √3

Attachment:
T6254c.png

Split one of the equilateral triangle at the center to mark the height, as shown in the figure. The split results in two 30, 60, 90 right triangles with sides of ratio 1, √3, 2. Given that the base of the square is 1, short leg of each of the 30:60:90 triangles is 0.5, and the height of the equilateral triangles is 0.5√3.

Now calculate the height of the pyramid through the triangle formed by connecting the center of the base, point x and the tip of the pyramid. The base and hypotenuse of this triangle are 0.5 and 0.5√3. Use the Pythagorean theorem to calculate the height:

(0.5√3)2 = 0.52 + h2

--> 0.52(√3)2 = 0.52 + h2

1/22 is 1/4, and (√3)2 is simply 3 (as the square root and power cancel each other out), so

--> 1/4⋅3 = 1/4 + h2

--> h2 = (3/4) - (1/4)

--> h2 = 1/2

--> h = 1/√2

To make the final transition to √2/2, multiple the top and bottom by √2:

(1/√2) × (√2/√2) = √2/(√2·√2) = √2/2

Diagonal of a square is given by$$\sqrt{2}$$ * side
so the diagonal of square base in the given figure will be $$\sqrt{2}$$
The triangles in the figure are equilateral. So, each side will be 1.
Let h be the height of pyramid.

Now,
Imagine the right angled triangle formed by 1/2 of the diagonal of square as its base, one of the sides of the triangle as its hypotenuse and and height of the pyramid as its perpendicular.
then by Pythagoras theorem,

$$1^2$$ = $$(\sqrt{2}*\frac{1}{2})^2$$ + $$h^2$$

or $$h^2$$ = $$1 - \frac{1}{2}$$ = $$\frac{1}{2}$$

or $$h =$$ $$\frac{1}{\sqrt{2}}$$

= $$\sqrt{2}*\frac{1}{2}$$

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Re: The figure above was cut out of paper and folded to form a pyramid wit  [#permalink]

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17 Apr 2020, 04:30
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Re: The figure above was cut out of paper and folded to form a pyramid wit   [#permalink] 17 Apr 2020, 04:30