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 399 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
_________________
The only thing more dangerous than ignorance is arrogance.
-- Albert Einstein
close
81% (01:10) correct 
