Question

let ABCD.A'B'C'D' be a cube with AC' =√3. Find the totsl surface area of this cube

Answer 1

We can easily prove that : \(AC'^2=2.A'B'^2\Rightarrow3=2.AB^2\Rightarrow AB=\sqrt{\frac{3}{2}}\) ( AB > 0 )

We also have \(AB\)is the side of this cube. So its total surface area is \(AB.AB.6=\frac{3}{2}.6=9\)

Answer 2

i DOnt understand