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# Surface Area and Volume Help (page 2)

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### 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

Fig. 8-4 . A rectangular prism has six rectangular faces and 12 edges.

### 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

### The 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.

Fig. 8-5 . A parallelepiped has six faces, all of which are parallelograms, and 12 edges.

### 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

#### PROBLEM 1

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?

#### SOLUTION 1

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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