Question

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

A) 48[(√3) + 1] B) 24[4 + (√3)] C) 28[6 + (√3)] D) 32[3 + (√3)]

Answer 1

Given: A right triangular pyramid XYZB is cut from cube as shown in figure. The side of cube is 16 cm. X, Y and Z are mid points of the edges of the cube.

To find:  What is the total surface area (in sq.cm) of the pyramid?

Solution:

• Now we have given side of cube is 16 cm. X, Y and Z are mid points of the edges of the cube so XB = BY = BZ = 8 cm.

• Now as per the figure, we have:

                   XY^2 = XB^2 + BY^2

                    XY^2 = 8^2 + 8^2

                   XY^2 = 64 + 64

                    XY = 8√2 cm

• Similarly, YZ = ZX = 8√2 cm

• Now for base area of the triangular pyramid, we have:

                   A = √3/4 x (8√2)^2

                   A = √3/4 x 128

                   A = 32√3 cm^2

• Now consider ΔXBY, we have:

                   A1 = 1/2 x base x height

                   A1 = 1/2 x 8 x 8

                   A1 = 32 cm^2

• Area of 3 surfaces of pyramid will be:

                   = 3 (A1) = 3(32) cm^2

• Total surface area = 32(3) + 32√3 = 32(3 + √3) cm^2

Answer:

           So the area of the pyramid is 32(3 + √3) cm^2

Answer 2

Answer:

32(3+$\sqrt{3$)

Step-by-step explanation: