Right Circular Cylinder Problems, 14

Category: Solid Geometry

"Published in Newark, California, USA"

Pass a plane through a cube of edge 6 in. so that the section formed will be a regular hexagon. Find the volume of a right circular cylinder 8 in. long, (a) whose base circumscribed this hexagon, (b) whose base is inscribed in this hexagon.

Solution:

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

Photo by Math Principles in Everyday Life

If a cube is cut by a plane that passes through the midpoints of two adjacent sides from the upper base to the opposite lower base with the midpoints of two adjacent sides, then the intersection is a regular hexagon. By Pythagorean Theorem, the length of the sides of a regular hexagon is

 
Let's consider the section of a cube which is a regular hexagon as follows

Photo by Math Principles in Everyday Life

There are six equal equilateral triangles in a regular hexagon in which their common vertex is a center of a regular hexagon. All sides of equilateral triangles are all equal which are 3√2 in.

(a) If the base of a right circular cylinder circumscribes the regular hexagon, then the radius is equal to 3√2 in. A circle contains all the vertices of a regular hexagon. Therefore, the volume of a right circular cylinder is
 

 

 

 

(b) If the base of a right circular cylinder inscribes the regular hexagon, then the radius is tangent to all the sides of a regular hexagon. The radius of a right circular cylinder is also an apothem of a regular hexagon and an altitude of an equilateral triangle. The altitude of an equilateral triangle bisects its base. By Pythagorean Theorem, the radius of a right circular cylinder is

Therefore, the volume of a right circular cylinder is 

 

 

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

Right Circular Cylinder Problems, 13

Category: Solid Geometry

"Published in Newark, California, USA"

A wedge ABCDEF (see figure) is cut from a right circular cylindrical block of altitude 10 in. and radius 4 in. The dihedral angle of the wedge is 42°30'. Calculate its volume and total surface.

Photo by Math Principles in Everyday Life

Solution:

To analyze more the problem, it is better to label further the given figure as follows 

Photo by Math Principles in Everyday Life

In this problem, we will use the given angle in solving for the length of an arc as well as the area of a circular sector. The given angle should be expressed in radians because it is a unitless value. 

The area of a circular sector DEF is 

Therefore, the volume of wedge ABCDEF is 

The length of arc EF is 

Therefore, the total surface area of wedge ABCDEF is