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The volume of the cube in the figure above is 64. If the vertices of ∆

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The volume of the cube in the figure above is 64. If the vertices of ∆  [#permalink]

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New post 30 Nov 2017, 23:48
1
2
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

69% (01:14) correct 31% (01:31) wrong based on 49 sessions

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Re: The volume of the cube in the figure above is 64. If the vertices of ∆  [#permalink]

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New post 01 Dec 2017, 04:03
side of the cube = 4
mid point is =2
now perimeter 2root2*3
= 6root2
hence B.
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The volume of the cube in the figure above is 64. If the vertices of ∆  [#permalink]

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New post 01 Dec 2017, 09:43
Bunuel wrote:
Image
The volume of the cube in the figure above is 64. If the vertices of ∆ PQR are midpoints of the cube's edges, what is the perimeter of ∆ PQR?

(A) 6
(B) 6√2
(C) 6√3
(D) 12
(E) 12√2

Attachment:
The attachment 2017-12-01_0948.png is no longer available

Attachment:
cubetri.png
cubetri.png [ 8.41 KiB | Viewed 782 times ]

Find side length
\(s^3 = 64\)
\(\sqrt[3]{s^3} = \sqrt[3]{64}\)
\(s = 4\)



Perimeter?

Three two-dimensional triangles make up ∆ PQR (see diagram: ∆ PRS, ∆ QRS, and ∆ PQS)
Side length of all three = \(2\) (vertices at midpoint of cube's edges)
All three are congruent right isosceles triangles (45-45-90) with corresponding side ratios in length \(x : x : x\sqrt{2}\)

\(x = 2\), so hyptenuse of each two-dimensional triangle = \(2\sqrt{2}\)

Those hypotenuses are the perimeter of ∆ PQR. that is
Side length of ∆ PQR = \(2\sqrt{2}\)

Perimeter:
\(2\sqrt{2} + 2\sqrt{2} + 2\sqrt{2} = 6\sqrt{2}\)

Answer B
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Re: The volume of the cube in the figure above is 64. If the vertices of ∆  [#permalink]

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New post 01 Dec 2017, 10:27
Bunuel wrote:
Image
The volume of the cube in the figure above is 64. If the vertices of ∆ PQR are midpoints of the cube's edges, what is the perimeter of ∆ PQR?

(A) 6
(B) 6√2
(C) 6√3
(D) 12
(E) 12√2

Attachment:
2017-12-01_0948.png


Volume of big cube = a^3 = 64 = 4^3

i.e. Side of big cube = 4

i.e. Base of the Cut pyramid = Equilateral triangle with sides of length = 2√2 . (Using property of 45-45-90 on one triangular face of the pyramid)

Perimeter = 3*2√2 = 6√2

Answer: option B
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Re: The volume of the cube in the figure above is 64. If the vertices of ∆  [#permalink]

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New post 16 Jan 2019, 07:51
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Re: The volume of the cube in the figure above is 64. If the vertices of ∆   [#permalink] 16 Jan 2019, 07:51
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