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

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

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

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

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

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

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

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

$\therefore y = 2, x = 2$

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