Question

A cube having each side of unit length is cut into two parts by a plane through two diagonals of two opposite faces. What is the total surface area of each of these parts?

• A. $3+\sqrt{2}$ sq.units
• B. $2+\sqrt{3}$ sq.units
• C. $3\sqrt{2}$ sq.units
• D. $3$ sq. units

Answer 1

The correct option is A $3+\sqrt{2}$ sq.units

Total surface area of the cube $=6a^2=6\times 1^2=6$ sq. units.

$DB=\sqrt{A{D}^2+A{B}^2}=\sqrt{1+1}=\sqrt{2}$

Surface area of the section $DBFH=DB\times BF$

$=\sqrt{2}\times 1$ $=\sqrt{2}$ sq. units.

$\therefore$ Total surface area of one part $=\frac{6}{2}+\sqrt{2}$ $=(3+\sqrt{2})$ sq. units.