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# GMAT: Math 32

163-166, pg. 174. Created by Sal Khan.

## Want to join the conversation?

• if the length of the longest rod that can be placed in a cubical room is 4*2^0.5 m
,then the area of the four walls of the room is
(1 vote)
• Well, the dimensions of a rectangular room are length, width and height (l, w, h) so
area will be the sum of the areas of each wall.
For a cubical room, if all we want is the area of the 4 walls, this simplifies to the sum of 4 squares of side s
which is 4s²
So A₄ = 4s²
For the rectangular room, the distance of a diagonal is √(l² + w² + h²)

For a cubical room, we can use s = w = l = h
so the length of the diagonal is √(s² + s² + s²) = √(3 s²) which simplifies to s√3
Now you were given a length of 4√2, so we set that equal to s√3 to solve for s
s√3 = 4√2
We COULD divide both sides by √3 and then rationalize the denominator, but,
since we need s², in order to solve the final part, I am going to square both sides
(s√3)² = (4√2)²
s²∙3 = 16 ∙ 2
s² = 16 ∙ 2/3
s²=32/3
4s²=4 ∙ 32/3
=128/3 m² or 42 2/3 m²
(1 vote)