Question

Consider a cuboid all of whose edges are integers and whose base is a square. Suppose the sum of all its edges is numerically equal to the sum of the areas of all its six faces. Then the sum of all its edges is

• A. 12
• B. 18
• C. 24
• D. 36

Answer 1

The correct option is C 24

The base of the cuboid is a square. Hence, the two of the sides of the cuboid are equal. Let the sides of the cuboid be given by $(a,a,b)$, where $b$ is the height when the base is the square.
Sum of all edges $=8a+4b$
Sum of areas of all faces $=2a^2+4ab$
Both are equal
$\Rightarrow 8a+4b=2a^2+4ab$
$\Rightarrow 4a+2b=a^2+2ab$
$\Rightarrow a^2-4a+2ab-2b=0$
$\Rightarrow (a-2{)}^2+2ab-2b-4=0$ (eqn $1$)
$(a-b{)}^2\ge 0\Rightarrow 2ab-2b-4\le 0$
$\Rightarrow b(a-1)\le 2$
Now, keeping in mind $a\ge 1$ and $b\ge 1$(both being integers), the only possibe solutions to the above equation are $a=3,b=1$ or $a=2,b=2$.
$a=3,b=1$ doesn't satisfy eqn $1$ while $a=2,b=2$ does.
So, sum of all its edges $=24$