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?
Answer & explanation
Correct answer: option 1
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