2001 CEMC Gauss (Grade 8) Problems/Problem 15

Problem

The surface area of a cube is $24 \text { cm}^{2}$. The volume of this cube is

$\text{ (A) }\ 4 \text { cm}^{3} \qquad\text{ (B) }\ 24 \text { cm}^{3} \qquad\text{ (C) }\ 8 \text { cm}^{3} \qquad\text{ (D) }\ 27 \text { cm}^{3} \qquad\text{ (E) }\ 64 \text { cm}^{3}$

Solution 1

Let $l$ be the edge length of the cube and $V$ be the volume of the cube. Using the surface area, we have:

$6l^2 = 24$

$l^2 = 4$

$l$ must be positive because it represents a cube's edge length, so $l = 2 \text { cm}$.

We have $V = l^3$. Plugging in $l = 2 \text { cm}$, we have:

$V = 2^3 = \boxed {\textbf {(C) } 8 \text { cm}^{3}}$

~anabel.disher

2001 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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CEMC Gauss (Grade 8)