Rectangular Parallelepiped Problem, 13

Category: Solid Geometry, Plane Geometry, Trigonometry

"Published in Newark, California, USA"

The figure represents a rectangular parallelepiped; AD = 20 in., AB = 10 in., AE = 15 in. (a) Find the number of degrees in the angles AFB, BFO, AFO, BOF, AOF, OFC. (b) Find the area of each of the triangles ABO, BOF, AOF. (c) Find the perpendicular distance from B to the plane AOF.

Photo by Math Principles in Everyday Life

Solution:

Since the given figure is a rectangular parallelepiped, then all sides are perpendicular to each other. We can solve for the length of line segments and area of triangles very well.

(a) By Pythagorean Theorem, the length of AF is

Therefore, ∠AFB is

or

By Pythagorean Theorem, the length of BD is

Since the lower base of a rectangular parallelepiped is a rectangle, then the diagonals are equal and bisect each other. In this case,

Hence, OB = OD = OA = OC = 11.1804 in.

Therefore, ∠BFO is

or

By Pythagorean Theorem, the length of OF is

Therefore, by Cosine Law, ∠AFO is

or

Therefore, ∠BOF is

or

Therefore, by Sine Law, ∠AOF is

or

By Pythagorean Theorem, the length of CF is

Therefore, by Sine Law, ∠OFC is

or

(b) The area of  ∆ABO is

The area of ∆BOF is

The semi-perimeter of ∆AOF is

Therefore, the area of ∆AOF by Heron's Formula is

(c) The volume of pyramid AFOB is

Therefore, if the base of the pyramid AFOB is ∆AOF, then the length of altitude from the vertex B to ∆AOF is