A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is #3 #, its base has sides of length #2 #, and its base has a corner with an angle of #(3 pi)/4 #. What is the pyramid's surface area?

Question

A pyramid has a base in the shape of a rhombus and a peak directly above the base's center.  The pyramid's height is #3  #, its base has sides of length #2 #, and its base has a corner with an angle of #(3 pi)/4   #.  What is the pyramid's surface area?

Violet Aldrich · Accepted Answer

Total Surface Area of the pyramid with rhombus base

#T S A = color (purple)(12.8136)#

Area of base #A_b = a^2 sin ((theta) = 2^2 sin ((3pi)/4) =color(blue)(0.1644)#

Lateral surface area #A_l#= area of the four triangles.

Area of triangle #A_t = (1/2) a H# where H is the slant height of the lateral surface.

#H = sqrt(h^2 + (a/2)^2) = sqrt(3^2 + 1^2) = 3.1623#

L S A #A_l = 4 * (1/2) a * H = 4 * (1/2) * 2 * 3.1623 = color (blue)(12.6492)#

#T S A = A_b + A_l = 0.1644 + 12.6492 = color(purple)(12.8136)#

Violet Aldrich · Answer

undefined To find the surface area of the pyramid, we need to calculate the area of each of its faces and then sum them up.

1. The area of the base:
   The area of a rhombus can be calculated using the formula: $ \text{Area} = \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 $
   Given that the side length of the rhombus base is 2 and one of its angles is $ \frac{3\pi}{4} $ radians, we can find the diagonals using trigonometry.
   Each diagonal of the rhombus can be calculated as: $ \text{diagonal} = 2 \times \sin(\frac{\theta}{2}) $, where $ \theta $ is the angle opposite the given side.
   Substituting the values, we get the diagonals of the rhombus as $ \sqrt{8} $.

Therefore, the area of the base is $ \frac{1}{2} \times \sqrt{8} \times \sqrt{8} = 4 $ square units.

2. The area of each triangular face:
   Each triangular face is an isosceles triangle with two sides equal to the slant height of the pyramid and one side equal to the side length of the base. The slant height can be found using the Pythagorean theorem as: $ \text{slant height} = \sqrt{\text{height}^2 + (\frac{\text{side}}{2})^2} $.
   Substituting the given values, we find the slant height as $ \sqrt{9 + 1} = \sqrt{10} $ units.

The area of each triangular face is $ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times \sqrt{10} = \sqrt{10} $ square units.

There are four triangular faces in total.

Thus, the total surface area of the pyramid is $ 4 \times \sqrt{10} + 4 = 4\sqrt{10} + 4 $ square units.