Consider a cuboid where all edges are integer | vedclass

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(C) Let the side length of the square base be $x$ and the height of the cuboid be $y$,where $x, y \in \mathbb{Z}^+$.The sum of all edges of the cuboid

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) Let the side length of the square base be $x$ and the height of the cuboid be $y$,where $x, y \in \mathbb{Z}^+$.The sum of all edges of the cuboid is $4x + 4x + 4y = 8x + 4y$.The sum of the areas of all six faces is $2x^2 + 4xy$.According to the problem,the sum of edges equals the sum of areas:$8x + 4y = 2x^2 + 4xy$Dividing by $2$:$4x + 2y = x^2 + 2xy$Rearranging to solve for $y$:$2y - 2xy = x^2 - 4x$$2y(1 - x) = x(x - 4)$$y = \frac{x(x - 4)}{2(1 - x)} = \frac{x(4 - x)}{2(x - 1)}$Since $y > 0$,we must have $1 If $x = 2$,$y = \frac{2(4 - 2)}{2(2 - 1)} = \frac{4}{2} = 2$.Sum of edges $= 8(2) + 4(2) = 16 + 8 = 24$.If $x = 3$,$y = \frac{3(4 - 3)}{2(3 - 1)} = \frac{3}{4}$,which is not an integer.Thus,the only integer solution is $x = 2, y = 2$,and the sum of edges is $24$.

Consider a cuboid where all edges are integers and the 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

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