A right triangular pyramid XYZB is cut from corner of a cube as shown in the figure. The side of the cube is 16 cm. X, Y, Z are mid points of the edges of the cube. What is the total surface area of this pyramid? 

32(√3+ 3) cm²

XB = YB =ZB = 8 cm

XY = YZ = XZ = \(\sqrt {8^2 + 8^2}\) = 8\(\sqrt{2}\) cm

Area of base of pyramid = \(\frac{\sqrt {3}}{4}\) × (side)²

                                   = \(\frac{\sqrt {3}}{4}\) × (8\(\sqrt{2}\))² = 32\(\sqrt{3}\) cm²

Slant height of pyramid = \(\sqrt {8² -(4\sqrt{2})²}\)

                                 = \(\sqrt {64 - 32}\) = 4\(\sqrt{2}\) cm

Lateral surface area = \(\frac{1}{2}\) × perimeter × slant height

                            = \(\frac{1}{2}\) × 3 × 8\(\sqrt{2}\) × 4\(\sqrt{2}\) = 96 cm²

T.S.A. = 96 + 32\(\sqrt{3}\) = 32 (3 + \(\sqrt {3}\)) cm²