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 + √2 sq. units 

(b) 2 + √3 sq. units 

(c) 3 √ 2 sq. units 

(d) 3 sq. units

Answer 1

(a) (3+\(\sqrt2\)) sq. units

Total surface area of the cube = 6a² = 6 × 1² = 6 sq. units

DB = \(\sqrt{AD^2+AD^2}\) = \(\sqrt{1+1}\) = \(\sqrt2\)

Surface area of the section DBFH 

= DB × BF = \(\sqrt2\) × 1 sq. units

= \(\sqrt2\) sq. units

∴ Total surface area of one part = \(\frac62+\sqrt2\)

= (3+\(\sqrt2\)) sq. units