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 #6 #, its base's sides have lengths of #5 #, and its base has a corner with an angle of #( pi)/6 #. What is the pyramid's surface area?

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Answer 1 · 2018-06-28T18:34:23

#73.788\ \text{unit}^2#

Area of rombus base with each side #5# & an interior angle #\pi/6#

#=5\cdot 5\sin(\pi/6)=12.5#

The rhombus base of pyramid has its semi-diagonals #5\cos(\pi/12)# & #5\sin(\pi/12)#

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

#\sqrt{6^2+(5\cos(\pi/12))^2}=7.702# &

#\sqrt{6^2+(5\sin(\pi/12))^2}=6.138#

Each of 4 identical lateral triangular faces of pyramid has the sides #5, 7.702 \ & \ 6.138#

semi-perimeter of triangle, #s={5+7.702+6.138}/2=9.42#

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

#=\sqrt{9.42(9.42-5)(9.42-7.702)(9.42-6.138)}#

#=15.322#

Hence, the total surface area of pyramid

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

#=4(15.322)+12.5#

#=73.788\ \text{unit}^2#