A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #6 # and the pyramid's height is #9 #. If one of the base's corners has an angle of #(5pi)/12#, what is the pyramid's surface area?

Question

A pyramid has a parallelogram shaped base and a peak directly above its center.  Its base's sides have lengths of #8 # and #6 # and the pyramid's height is #9 #.  If one of the base's corners has an angle of #(5pi)/12#, what is the pyramid's surface area?

Evelyn Arboleda · Accepted Answer

Total surface area # = 168.7614#

#CH = 6*sin ((5pi)/12) = 5.7956#
Area of parallelogram base #= a* b1 = 6* 5.7956=33.7736#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(9^2 + 4^2) = sqrt97#
Area of #Delta AEF = BEC = (1/2)*b*h_1 = (1/2)*6*sqrt97=3sqrt97#

#EG = h_2 = sqrt(h^2+(b/2)^2 = sqrt(9^2 + 3^2 = sqrt90#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*sqrt90=4sqrt90#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#= 2*3sqrt97 + 2*4sqrt90 = 59.0931 + 75.8947 = 134.9878#

Total surface area =Area of parallelogram base + Lateral surface area # = 33.7736 + 134.9878 =168.7614#

Wyatt Anderson · Answer

undefined To find the surface area of the pyramid, we need to calculate the area of its base and the area of its lateral faces.

1. **Base Area (Parallelogram):**
   The area of a parallelogram can be found by multiplying the base by the height perpendicular to that base. In this case, the base has sides of lengths 8 and 6, and the height is not given directly. However, we can find the height using trigonometry.

Let's denote the angle at one of the corners of the base as $ \frac{5\pi}{12} $. Since the opposite side of this angle is the height, we can use trigonometric functions to find it. Using the given side lengths:
   $$ \tan\left(\frac{5\pi}{12}\right) = \frac{h}{6} $$
   Solving for $ h $:
   $$ h = 6 \tan\left(\frac{5\pi}{12}\right) $$

Now, we can find the area of the parallelogram:
   $$ \text{Area} = \text{base} \times \text{height} $$
   $$ \text{Area} = 8 \times 6 \tan\left(\frac{5\pi}{12}\right) $$

2. **Lateral Surface Area:**
   The lateral surface area of a pyramid is the sum of the areas of its triangular faces. Since it's a regular pyramid, all triangular faces are congruent. We can find the area of one of these triangles and then multiply it by the number of faces.

Each triangular face is an isosceles triangle with base equal to the side length of the base of the parallelogram (8) and equal legs, which are the slant heights of the pyramid. We can find the slant height using the Pythagorean theorem.

$$ \text{Slant height} = \sqrt{\text{height}^2 + \left(\frac{\text{base length}}{2}\right)^2} $$
   $$ \text{Slant height} = \sqrt{9^2 + \left(\frac{8}{2}\right)^2} $$
   $$ \text{Slant height} = \sqrt{81 + 16} $$
   $$ \text{Slant height} = \sqrt{97} $$

Now, we can find the area of one triangular face:
   $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$
   $$ \text{Area} = \frac{1}{2} \times 8 \times \sqrt{97} $$

Since there are four triangular faces, the total lateral surface area is:
   $$ 4 \times \frac{1}{2} \times 8 \times \sqrt{97} $$

Now, add the base area and the lateral surface area to find the total surface area of the pyramid.