Frustum of Pyramid Problems, 2

Category: Solid Geometry

"Published in Newark, California, USA"

A block of granite is in the form of the frustum of a regular square pyramid whose upper and lower base edges are 3 ft. and 7 ft., respectively. If each of the lateral faces is inclined at an angle of 62°30' to the base, find the volume of granite in the block.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

Since the length of the edges of the upper and lower bases are given, then we can solve for the area of the bases. For the upper base, the area is

and for the lower base, the area is

The altitude of the frustum of a regular square pyramid is not given in the problem. If the angle of inclination of each lateral faces to the lower base is given, then we can solve for the altitude by isolating and label further the vertical section of the frustum of a regular square pyramid as follows

Photo by Math Principles in Everyday Life

There are two equal right triangles in the vertical section of the frustum of a regular square pyramid. This section is an isosceles trapezoid. By using simple trigonometric function, the altitude of the section which is also the altitude of the frustum of a regular square pyramid is

Therefore, the volume of the frustum of a regular square pyramid which is the volume of a block of granite is

More Pyramid Problems

Category: Solid Geometry

"Published in Newark, California, USA"

What fraction of the volume of a pyramid must be cut off by a plane parallel to the base if the pyramid thus formed has a lateral area equal to one half of the lateral area of the original pyramid?

Solution:

To illustrate the problem, it is better to draw a pyramid with any base as follows

Photo by Math Principles in Everyday Life

If B₁ is parallel to B₂, then the two pyramids are similar. We can relate their lateral areas with their altitudes using the principles of ratio and proportion as follows

where S₁ and S₂ are the lateral area of the two pyramids, respectively. If S₂ = 2S₁, then the above equation becomes

Take the square root on both sides of the equation, we have

We can relate also their volumes with their altitudes using the principles of ratio and proportion as follows

but

Finally, their volume ratio is

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