The base of a pyramid is the regular polygon ABCDEFGH, which has 14 inch sides. All eight of the pyramid's lateral edges, VA, VB,...,VH, are 25 inches long.

Question

To the nearest tenth of an inch, calculate the height of the pyramid VABCDEFGH.

 

 

 

2. (Continuation) To the nearest tenth of a degree, calculate the size of the dihedral angle formed by the octagonal base and the triangular face VAB.

Answer 1

Moxa,

This may be a challenge to follow without a diagram, but here it goes.

 

The lateral face ΔVAB is in fact an Isosceles triangle with a base of length 14 and legs of length 25. The altitude of this triangle necessarily bisects angle V and the base, creating a right triangle with a hypotenuse of length 25 and the short leg of length 7. Using Pythagorean theorem we can solve for the altitude, which is also the slant height of the pyramid.

 

s.h. = alt. = √(25² - 7²) = 24

 

Now we can use this slant height as the hypotenuse of another right triangle by combining with the height of the pyramid and 1/2 the base. To find half the base, we look at a regular octagon. You'll see that 4 of the sides are vertical or horizontal, while the other 4 sides are at 45º to the first four. So the distance between parallel sides of a regular octagon would be = s + 2s/√2, or in this case:  14 + 28/√2 = 14 + 14√2

 

So half of the base would = 7 + 7√2

 

This right triangle will give us both the height of the pyramid and the dihedral angle with the base.

 

h = √(24² - (7 + 7√2)²) = 17.0"

 

θ = cos^-1((7 + 7√2)/24) = 45.2º

 

I know it's a challenge. Try drawing out the figures as you follow my description.

Good Luck

Bryan