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?

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?

Sadie Adams · Accepted Answer

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

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://tutor.hix.ai), 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.

Sadie Allen · Answer

undefined To find the surface area of the pyramid, you first need to calculate the area of the base and then add the areas of the four triangular faces.

1. Area of the base (parallelogram):
   Area = base × height
   Area = 7 × 3 = 21 square units

2. To find the area of each triangular face, you'll need to calculate the base and height of each triangle. The base is the side length of the base of the pyramid (either 3 or 7), and the height is the slant height of the pyramid, which can be found using the Pythagorean theorem.

3. Calculate the slant height (l) of the pyramid:
   l = √(h^2 + (b/2)^2)
   l = √(8^2 + (3/2)^2)  [Since the base is a parallelogram, the diagonal length is used as the height (h)]
   l ≈ √(64 + 2.25)
   l ≈ √66

4. Calculate the area of one triangular face:
   Area = 0.5 × base × height
   Area = 0.5 × 3 × √66

5. Calculate the area of all four triangular faces and add them together:
   Total triangular area = 4 × (0.5 × 3 × √66)

6. Finally, add the base area and the total triangular area to find the total surface area of the pyramid:
   Surface area = Base area + Total triangular area