Problem:
A right pyramid has regular octagon ABCDEFGH with side length 1 as its base and apex V. Segments AV and DV are perpendicular. What is the square of the height of the pyramid?
Answer Choices:
A. 1
B. 21+2
C. 2
D. 23
E. 32+2
Solution:
Let O be the center of the octagon, and let r=AO. As can be seen from the figure below, AD=1+2.
Because △AVD is an isosceles right triangle,
AV=22⋅AD=22+2
Applying the Law of Cosines to △AOH yields
r2+r2=12+2r2cos45∘=1+r22
Thus r2(2−2)=1 and
r2=2−21=22+2
The Pythagorean Theorem applied to △VOA gives the requested square of the height of the pyramid:
VO2=AV2−r2=(22+2)2−22+2=(B)21+2
OR
Place the figure in a three-dimensional coordinate system with the center O of the base of the pyramid at the origin, the octagon in the x−y plane with positive x coordinates for A,B,C, and D, the y-axis parallel to AD, and apex V(0,0,h) on the positive z-axis. See the figure.
Then consider vectors
OV=⟨0,0,h⟩,OA=⟨21,−21+2,0⟩, and OD=⟨21,21+2,0⟩
It follows that
AV=OV−OA=⟨−21,21+2,h⟩
and
DV=OV−OD=⟨−21,−21+2,h⟩
Because AV and DV are perpendicular, their dot product is zero. Therefore
