Surface Area and Volume Help (page 2)
Introduction to Polyhedrons
In Euclidean three-space, geometric solids with straight edges have flat faces, also called facets, each of which forms a plane polygon. An object of this sort is known as a polyhedron .
A polyhedron in 3D must have at least four faces. A four-faced polyhedron is called a tetrahedron. Each of the four faces of a tetrahedron is a triangle. There are four vertices. Any four specific points, not all in a single plane, form a unique tetrahedron.
Surface Area Of Tetrahedron
Figure 8-1 shows a tetrahedron. The surface area is found by adding up the interior areas of all four triangular faces. In the case of a regular tetrahedron , all six edges have the same length, and therefore all four faces are equilateral triangles. If the length of each edge of a regular tetrahedron is equal to s units, then the surface area, B , of the whole four-faced regular tetrahedron is given by:
B = 3 1/2 s 2
where 3 1/2 represents the square root of 3, or approximately 1.732. This also happens to be twice the sine of 60°, which is the angle between any two edges of the figure.
Volume Of Tetrahedron
Imagine a tetrahedron whose base is a triangle with area A , and whose height is h as shown in Fig. 8-1. The volume, V , of the figure is given by:
V = Ah /3
Surface Area and Volume of a Pyramid
Figure 8-2 illustrates a pyramid . This figure has a square or rectangular base and four slanted faces. If the base is a square and the apex (the top of the pyramid) lies directly above a point at the center of the base, then the figure is a regular pyramid , and all of the slanted faces are isosceles triangles.
Surface Area Of Pyramid
The surface area of a pyramid is found by adding up the areas of all five of its faces (the four slanted faces plus the base). In the case of a regular pyramid where the length of each slanted edge, called the slant height , is equal to s units and the length of each edge of the base is equal to t units, the surface area, B , is given by:
B = t 2 + 2 t ( s 2 − t 2 /4) 1/2
In the case of an irregular pyramid , the problem of finding the surface area is more complicated, because it involves individually calculating the area of the base and each slanted face, and then adding all the areas up.
Volume Of Pyramid
Imagine a pyramid whose base is a square with area A , and whose height is h as shown in Fig. 8-2. The volume, V , of the pyramid is given by:
V = Ah /3
This holds true whether the pyramid is regular or irregular.
Surface Area and Volume of a Cube
Figure 8-3 illustrates a cube . This is a regular hexahedron (six-sided polyhedron). It has 12 edges, each of which is of the same length. Each of the six faces is a square.
Surface Area Of Cube
Imagine a cube whose edges each have length s , as shown in Fig. 8-3. The surface area, A , of the cube is given by:
A = 6 s 2
Volume Of Cube
Imagine a cube as defined above and in Fig. 8-3. The volume, V , of the solid enclosed by the cube is given by:
V = s 3
Surface Area and Volume of a Rectangular Prism
The Rectangular Prism
Figure 8-4 illustrates a rectangular prism . This is a hexahedron, each of whose six faces is a rectangle. The figure has 12 edges, but they are not necessarily all equally long.
Surface Area Of Rectangular Prism
Imagine a rectangular prism whose edges have lengths s 1 , s 2 , and s 3 as shown in Fig. 8-4. The surface area, A , of the prism is given by:
A = 2 s 1 s 2 + 2 s 1 s 3 + 2 s 2 s 3
Volume Of Rectangular Prism
Imagine a rectangular prism as defined above and in Fig. 8-4. The volume, V , of the enclosed solid is given by:
V = s 1 s 2 s 3
Surface Area and Volume of a Parallelepiped
A parallelepiped is a six-faced polyhedron in which each face is a parallelogram, and opposite pairs of faces are congruent. The figure has 12 edges. The acute angles between the pairs of edges are x , y , and z , as shown in Fig. 8-5.
Surface Area Of Parallelepiped
Imagine a parallelepiped with edges of lengths s 1 , s 2 , and s 3 . Suppose the angles between pairs of edges are x , y , and z as shown in Fig. 8-5. The surface area, A , of the parallelepiped is given by:
A = 2 s 1 s 2 sin x + 2 s 1 s 3 sin y + 2 s 2 s 3 sin z
where sin x represents the sine of angle x , sin y represents the sine of angle y , and sin z represents the sine of angle z .
Volume Of Parallelepiped
Imagine a parallelepiped whose edges have lengths s 1 , s 2 , and s 3 , and that has angles between edges of x , y , and z as shown in Fig. 8-5. Suppose further that the height of the parallelepiped, as measured along a line normal to the base, is equal to h . The volume, V , of the enclosed solid is equal to the product of the base area and the height:
V = hs 1 s 3 sin y
Straight-Edged Objects Practice Problems
Suppose you want to paint the interior walls of a room in a house. The room is shaped like a rectangular prism. The ceiling is exactly 3.0 meters above the floor. The floor and the ceiling both measure exactly 4.2 meters by 5.5 meters. There are two windows, the outer frames of which both measure 1.5 meters high by 1.0 meter wide. There is one doorway, the outer frame of which measures 2.5 meters high by 1.0 meter wide. With two coats of paint (which you intend to apply), one liter of paint can be expected to cover exactly 20 square meters of wall area. How much paint, in liters, will you need to completely do the job?
It is necessary to find the amount of wall area that this room has. Based on the information given, we can conclude that the rectangular prism formed by the edges between walls, floor, and ceiling measures 3.0 meters high by 4.2 meters wide by 5.5 meters deep. So we can let s 1 = 3.0, s 2 = 4.2, and s 3 = 5.5 (with all units assumed to be in meters) to find the surface area A of the rectangular prism, in square meters, neglecting the area subtracted by the windows and doorway. Using the formula:
A = 2 s 1 s 2 + 2 s 1 s 3 + 2 s 2 s 3
= (2 ×; 3.0 × 4.2) + (2 × 3.0 × 5.5) + (2 × 4.2 × 5.5)
= 25.2 + 33.0 + 46.2
= 104.4 square meters
There are two windows measuring 1.5 meters by 1.0 meter; each of these therefore takes away 1.5 × 1.0 = 1.5 square meters of area. The doorway measures 2.5 meters by 1.0 meter, so it takes away 2.5 × 1.0 = 2.5 square meters. Thus the windows and doorway combined take away 1.5 + 1.5 + 2.5 = 5.5 square meters of wall space. Then we must also take away the areas of the floor and ceiling. This is the final factor in the above equation, 2 s 2 s 3 = 46.2. The wall area to be painted, call it A w , is therefore:
A w = (104.4 − 5.5) − 46.2
= 52.7 square meters
A liter of paint can be expected to cover 20 square meters. Therefore, we will need 52.7/20, or 2.635, liters of paint to do this job.
Practice problems for these concepts can be found at: Surface Area And Volume Practice Test.
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