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

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Joined: 02 Sep 2009
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The volume of the cube in the figure above is 64. If the vertices of ∆  [#permalink]

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30 Nov 2017, 23:48
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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 [ 6.23 KiB | Viewed 593 times ]

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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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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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01 Dec 2017, 09:43
Bunuel wrote:

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 [ 8.41 KiB | Viewed 358 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:
$$2\sqrt{2} + 2\sqrt{2} + 2\sqrt{2} = 6\sqrt{2}$$

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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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01 Dec 2017, 10:27
Bunuel wrote:

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

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Re: The volume of the cube in the figure above is 64. If the vertices of ∆ &nbs [#permalink] 01 Dec 2017, 10:27
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