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 #5 #, its base has sides of length #8 #, and its base has a corner with an angle of #(2 pi)/3 #. What is the pyramid's surface area?

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Answer 1 · 2018-07-04T10:16:48

#152.75\ \text{unit}^2#

Area of rombus base with each side #8# & an interior angle #{2\pi}/3#

#=8\cdot 8\sin({2\pi}/3)=55.426#

The rhombus base of pyramid has its semi-diagonals #8\cos(\pi/6)# & #8\sin(\pi/6)# i.e. #4\sqrt3# & #4#

Now, the sides of triangular lateral face of pyramid as given as

#\sqrt{5^2+(4\sqrt3)^2}=\sqrt{73}=8.544# &

#\sqrt{5^2+(4)^2}=\sqrt{41}=6.403#

Each of 4 identical lateral triangular faces of pyramid has the sides #8, 8.544# & #6.403#

semi-perimeter of triangle, #s={8+8.544+6.403}/2=11.4735#

Now, using heron's formula the area of lateral triangular face of pyramid

#=\sqrt{11.4735(11.4735-8)(11.4735-8.544)(11.4735-6.403)}#

#=24.331#

Hence, the total surface area of pyramid

#=4(\text{area of lateral triangular face})+\text{area of rhombus base}#

#=4(24.331)+55.426#

#=152.75\ \text{unit}^2#