A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #7 # and the pyramid's height is #8 #. If one of the base's corners has an angle of #(3pi)/8#, what is the pyramid's surface area?

1 Answer
Jun 13, 2018

#3/2sqrt(305)+7/4sqrt(1042+9sqrt(2))=83.0# to three significant figures

Explanation:

Work in the usual 3-dimensional Cartesian co-ordinate system #(x,y,z)#; let #(x,y,z)=(0,0,0)# at the centre of the parallelogram base and the #x#-axis run parallel to the long edge. Assume wlog that the smaller angle in the parallelogram is at positive #x# and positive #y# (or conversely negative both).

Then the apex of the pyramid is at #(0,0,8)# and the mid-points of the parallelogram's four sides are at #(0,+-7/2,0)# (short sides) and #(+-3/2sin((3pi)/8),0,0)# (long sides).

We calculate the area of the triangles on short and long sides by the usual triangle area formula: #1/2bh#. #b# is given in the question; #h# is the slant height, the straight-line distance between side mid-point and apex.

Short sides:

#h^2=(7/2)^2+8^2=49/4+256/4=305/4#
#h=sqrt(305/4)=sqrt(305)/2#
#A_{sh}=1/2bh=1/2*3*sqrt(305)/2=3/4sqrt(305)#

Long sides:

#h^2=(3/2)^2sin^2((3pi)/8)+8^2#
We may take that #sin((3pi)/8)=sqrt(2+sqrt(2))/2# (proof here: https://www.mathway.com/popular-problems/Precalculus/400495), and so
#h^2=9/4*(2+sqrt(2))/4+8^2=9/16(2+sqrt(2))+64=(1042+9sqrt(2))/16#
#h=sqrt(1042+9sqrt(2))/4#
#A_{lo}=1/2bh=1/2*7*sqrt(1042+9sqrt(2))/4=7/8sqrt(1042+9sqrt(2))#

Thus the total surface area of the pyramid:

#A=2A_{sh}+2A_{lo}=3/2sqrt(305)+7/4sqrt(1042+9sqrt(2))#

To three significant figures, this equals 83.0 square units. Note that I have derived an exact expression here, but it is possible that the question setter desires everything to be computed on a calculator instead, in which case we do not need to know the #sin# formula used.