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 #4 #, its base's sides have lengths of #8 #, and its base has a corner with an angle of #(7 pi)/8 #. 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 #4  #, its base's sides have lengths of #8 #, and its base has a corner with an angle of #(7 pi)/8 #.  What is the pyramid's surface area?

Ella Armstrong · Accepted Answer

T S A = 115.0021

AB = BC = CD = DA = a = 8
Height OE = h = 4
OF = a/2 = 8/2 = 4
# EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(4^2+4^2) = color(red)(5.6569)#

Area of #DCE = (1/2)*a*EF = (1/2)*8*5.6569 = color(red)(22.6276)#
Lateral surface area #= 4*Delta DCE = 4*22.6276 = color(blue)(90.5104)#

#/_C = (pi) - ((7pi)/8) = (pi)/8#
Area of base ABCD #= a* a * sin /_C = 8^2 sin (pi/8) = 24.4917#

T S A #= Lateral surface area + Base area#
T S A # =90.5104 + 24.4917 = color(purple)(115.0021)#

Autumn Armstrong · Answer

undefined To find the surface area of the pyramid:

1. Calculate the area of the rhombus base.
2. Determine the area of each triangular face.
3. Add the areas of all faces together.

First, find the area of the rhombus base using its diagonals $d_1$ and $d_2$:

$$ A_{\text{base}} = \frac{d_1 \times d_2}{2} $$

Since it's a rhombus, $d_1 = d_2 = 8$ (because all sides are equal).

$$ A_{\text{base}} = \frac{8 \times 8}{2} = 32 \, \text{square units} $$

Now, calculate the area of one triangular face. Each triangular face is an isosceles triangle with base $b$ (which is a side of the rhombus) and height $h$.

$$ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} $$

Given that the base side length is $8$ and the height of the pyramid is $4$, the area of one triangular face is:

$$ A_{\text{triangle}} = \frac{1}{2} \times 8 \times 4 = 16 \, \text{square units} $$

Since there are four identical triangular faces, the total area of all four triangular faces is $4 \times 16 = 64$ square units.

The total surface area of the pyramid is the sum of the base area and the area of the four triangular faces:

$$ \text{Surface Area} = A_{\text{base}} + 4 \times A_{\text{triangle}} $$

$$ \text{Surface Area} = 32 + 4 \times 16 = 32 + 64 = 96 \, \text{square units} $$