MCQ
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$18$
- ✓$24$
- D$36$
✓
Answer
Correct option: C.
$24$
c
(c)
(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 $=4 x+4 x+4 y$
Sum of area of all faces $=2 x^2+2 x y+2 x y$
Given,
Sum of all edges of cuboid = Sum of area of all faces
$\therefore \quad 4 x+4 x+4 y=2\left(x^2+x y+x y\right)$
$\Rightarrow \quad 4(2 x+y)=2\left(x^2+2 x y\right)$
$\Rightarrow \quad x^2+2 x y-4 x-2 y=0$
$\Rightarrow \quad x^2+2 x(y-2)-2 y=0$
$\Rightarrow \quad x=\frac{-2(y-2) \pm \sqrt{4(y-2)^2+4(2 y)}}{2}$
$\Rightarrow \quad x=y-2 \pm \sqrt{y^2-2 y+4}$
$x \text { is integer, when } y=2$
$\therefore \quad \quad y=2 x=2$
Hence, sum of edges $=8 x+4 y=16+8=24$
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