Question

Consider a cuboid all of whose edges are integers and whose base is 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

The correct option is C $24$

$\text{Let sides are}a,a\&b.\text{Sum of the edges}=8a+4b\text{Sum of the areas of six faces}=2a^2+4ab\Rightarrow 8a+4b=2a^2+4ab\Rightarrow b=\frac{a(4-a)}{2(a-1)}\text{Since,}a,b\in Z^{+}\Rightarrow a>1\&a<4\Rightarrow a=2,3\text{For}a=2\Rightarrow b=2\text{For}a=3\Rightarrow b=\frac{3}{4}\notin Z\therefore \text{Sum of all the edges}=8\times 2+4\times 2=24$