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

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Answer 1 · 2018-07-15T21:08:43

#100.683\ \text{unit}^2#

Area of rhombus base with each side #8# & an interior angle #{\pi}/4#

#=8\cdot 8\sin({\pi}/4)=45.255#

The rhombus shaped base of pyramid has its semi-diagonals

#8\cos({\pi}/8)=7.391# &

#8\sin({\pi}/8)=3.0615#

Now, two unequal lateral edges of each triangular lateral face of pyramid are given as

#\sqrt{2^2+(7.391)^2}=7.657# &

#\sqrt{2^2+(3.0615)^2}=3.657#

There are four identical triangular lateral faces of pyramid each has sides #8, 7.657# & #3.657#

Area of each of four identical triangular lateral faces with sides #8, 7.657# & #3.657#

semi-perimeter of triangle, #s={8+7.657+3.657}/2=9.657#

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

#=\sqrt{9.657(9.657-8)(9.657-7.657)(9.657-3.657)}#

#=13.857#

Hence, the total surface area of pyramid (including area of base)

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

#=4(13.857)+45.255#

#=100.683\ \text{unit}^2#

Note:
Height of pyramid is taken as vertical height of pyramid