Question

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?

Answer 1

`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.