A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #8 # and the pyramid's height is #7 #. If one of the base's corners has an angle of #(5pi)/6#, 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 #4 # and #8  # and the pyramid's height is #7 #.  If one of the base's corners has an angle of #(5pi)/6#, what is the pyramid's surface area?

Evelyn Arboleda · Accepted Answer

Total surface area # = 74.2409#

#CH = 4*sin ((5pi)/6) = 2#
Area of parallelogram base #= a* b1 = 8*2=16#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+4^2)= sqrt65#
Area of #Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4*sqrt65=2sqrt65#

#EG = h_2 = sqrt(h^2+(b/2)^2 = sqrt(7^2+2^2) = sqrt53#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*sqrt53=4sqrt53#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#= 2*2sqrt65 + 2*4sqrt53= 32.249 + 58.2409#

Total surface area =Area of parallelogram base + Lateral surface area # = 16 + 58.2409 = 74.2409#

Aaliyah Anderson · Answer

undefined To find the surface area of the pyramid, we need to calculate the areas of the base and the four triangular faces.

1. **Area of the Base (Parallelogram):**
   The area of a parallelogram can be calculated using the formula: $ \text{Area} = \text{base} \times \text{height} $.
   Given that one side of the base is 8 and the corresponding height is 7, the area of the base is $ 8 \times 7 = 56 $ square units.

2. **Area of the Triangular Faces:**
   Since the pyramid has four triangular faces, we need to calculate the area of one triangular face and then multiply by 4.
   Each triangular face is an isosceles triangle with two sides equal to the slant height of the pyramid and one side equal to the length of a side of the base.
   To find the slant height, we use the Pythagorean theorem. The slant height (l) can be found using the formula: $ l = \sqrt{h^2 + (\frac{b}{2})^2} $, where $ h $ is the height of the pyramid and $ b $ is the length of the base side.
   $ l = \sqrt{7^2 + (\frac{4}{2})^2} = \sqrt{49 + 4} = \sqrt{53} $.
   Now, we can find the area of one triangular face using the formula for the area of a triangle: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $.
   $ \text{Area} = \frac{1}{2} \times 8 \times \sqrt{53} = 4\sqrt{53} $ square units.
   Since there are four identical triangular faces, the total area of all four faces is $ 4 \times 4\sqrt{53} = 16\sqrt{53} $ square units.

3. **Total Surface Area:**
   The total surface area of the pyramid is the sum of the area of the base and the area of the four triangular faces.
   Total Surface Area $ = 56 + 16\sqrt{53} $ square units.

Therefore, the surface area of the pyramid is $ 56 + 16\sqrt{53} $ square units.