A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #3 # and the pyramid's height is #6 #. If one of the base's corners has an angle of #pi/3#, what is the pyramid's surface area?

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Answer 1 · 2016-09-13T02:02:37

#82.319#

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Total surface area = base area (a1) + side area 1 (a2) + side area 2 (a3).

1) base area (a1) = parallelogram area = BxxH
#=7xx3xxsin60 = 18.186# , (#H=3sin60#)

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2) side area 1 (a2) #=(1/2)xxB1xxS1#, (#B1=7#)

#S1=sqrt((6^2+1.5^2))=6.185#
#a2=(1/2)xx7xx6.185xx2=43.295#, (2 sides)

3) side area 2 (a3) #=(1/2)xxB2xxS2#, (#B2=3#)

#S2=sqrt((6^2+3.5^2))=6.946#
#a3=(1/2)xx3xx6.946xx2=20.838#, (2 sides)

Fianlly, total surface area = a1+a2+a3

#=18.186+43.295+20.838#
#=82.319#