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

Answer: (C)

Given, a cuboid has all edges are integers and base is square.

Let the length, breadth and height of cuboid is $x,x,y$.



Sum of all edges of cuboid $=4x+4x+4y$

Sum of area of all faces $=2x^2+2xy+2xy$

Given,

Sum of all edges of cuboid = Sum of area of all faces

$\therefore 4x+4x+4y=2(x^2+xy+xy)$

$\Rightarrow 4(2x+y)=2(x^2+2xy)$

$\Rightarrow x^2+2xy-4x-2y=0$

$\Rightarrow x^2+2x(y-2)-2y=0$

$\Rightarrow x=\frac{-2\left(y-2\right)\pm \sqrt{4{\left(y-2\right)}^2+4\left(2y\right)}}{2}$

$\Rightarrow x=y-2\pm \sqrt{y^2-2y+4}$

$x$ is integer, when $y=2$

$\therefore y=2,x=2$

Hence, sum of edges $=8x+4y=16+8=24$