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Q. 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

KVPYKVPY 2016

Solution:

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

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

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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$