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.