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

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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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