Mathematics
Volume of Pyramid
The volume of a pyramid is the measure of space enclosed by its three-dimensional shape. It is calculated using the formula V = (1/3) * base area * height, where the base area is the area of the pyramid's base and the height is the perpendicular distance from the base to the apex. This measurement is important in geometry and engineering for determining the capacity or space occupied by pyramidal structures.
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11 Key excerpts on "Volume of Pyramid"
No longer available |Learn more- Daniel C. Alexander, Geralyn M. Koeberlein, , , Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Figure 9.20 VOLUME OF A PYRAMID The final theorem in this section is presented without any attempt to construct the proof. In an advanced course such as calculus, the statement can be proved. The factor “one-third” in the formula for the volume of a pyramid provides exact results. This formula can be applied to any pyramid, even one that is not regular; in Figure 9.20(b), the length of the altitude is the perpendicular distance from the vertex to the plane of the square base. Read the Discover activity in the margin at the left before considering Theorem 9.2.4 and its applications. (a) Regular square pyramid (b) Square pyramid 9.2 ■ Pyramids, Area, and Volume 403 Unless otherwise noted, all content on this page is © Cengage Learning. D iscover There are kits that contain a hollow pyramid and a hollow prism that have congruent bases and the same altitude. Using a kit, fill the pyramid with water and then empty the water into the prism. a) How many times did you have to empty the pyramid in order to fill the prism? b) As a fraction, the volume of the pyramid is what part of the volume of the prism? ANSWERS (a) Three times (b) 1 3 THEOREM 9.2.4 The volume V of a pyramid having a base area B and an altitude of length h is given by V 1 3 Bh EXAMPLE 4 Find the volume of the regular square pyramid with height and base edges of length (This was the pyramid in Example 1.) SOLUTION The area of the square base is or . Because , the formula becomes To find the volume of a pyramid by using the formula , it is often necessary to determine B or h from other information that has been provided. In Example 5, calculating the length of the altitude h is a challenge! In Example 6, the difficulty lies in finding the area of the base.
No longer available |Learn more- Tom Bassarear, Meg Moss(Authors)
- 2015(Publication Date)
- Cengage Learning EMEA(Publisher)
In this triangular prism, the base is the triangle and the height is the distance between the triangular bases (25 cm). 7 cm 10 cm 25 cm 10 cm 10 cm Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Unless otherwise noted, all content on this page is © Cengage Learning Section 9.3 Surface Area and Volume 545 Volume = (Area of Base)(Height) Since we have a triangular base, we use the formula Area of Base = 1 2 (base)(height). Note here that we are using the words “base” and “height” to mean different things. We have the Base and Height of the prism, and the base and height of the triangle. Using capital letters when we are talking about the three-dimensional figure (the prism) and lowercase letters when we are talking about the two-dimensional figures (the triangle) can help us differentiate them. Volume 5 (Area of Base)(Height) 5 ( 1 2 bh )(Height) 5 ( 1 2 3 10 3 10)(25) 5 1250 cubic centimeters UNDERSTANDING THE VOLUMES OF PYRAMIDS AND CONES We will illustrate the formula of a pyramid using a cube. It so happens that three congruent pyramids will fit inside the cube. Most people need to see this to believe it, and Figure 9.24 provides the necessary information to make three congruent pyramids that can be joined together to make a cube. Two-dimensional drawings of three-dimensional objects can be difficult to visualize. Geosolids are manipulatives that enable you to demonstrate concepts related to volume. You can fill a geosolid pyramid with water and see that it takes three of them to fill up a geosolid cube with the same height and same size base.
- Kevin Corner, Leslie Jackson, William Embleton(Authors)
- 2013(Publication Date)
- Thomas Reed(Publisher)
238 • Mathematics Therefore, the volume of a cone is one-third of the volume of a solid cylinder having the same height and base diameter as the cone. Therefore, volume of a cone = π · r 2 · h 3 . Similarly, the volume of a square pyramid is one-third the volume of a bar of square base with height equal to that of the pyramid. In all cases: Volume of Pyramid = 1 3 (Area of base × Perpendicular height) Oblique Prisms and Pyramids If the prism or pyramid be imagined as being made up of a number of discs or lamina-tions and pushed over to one side, it can be seen by reference to Figure 10.4 that the same rule for finding the volumes of regular prisms or regular pyramids can be applied provided the perpendicular height is used. Figure 10.4 Volume – Mass, Centre of Gravity, Moment • 239 Frustums A frustum of a cone or pyramid is the bottom piece left, after a portion has been sliced off the top (Figure 10.5). The volume can be found by subtracting the volume of the sliced-off top part from the volume of the complete cone. h R r L Figure 10.5 Example Find the volume of a frustum of a cone if the radius of the base is 5 cm, the radius of the top surface is 3 cm and the frustum has a height of 10 cm. 10 cm 3 cm 10 cm 240 • Mathematics Looking at the frustum in cross-section gives: 3 5 10 It is necessary to project the sides to form the original cone and then find the height of the projected section, h . 3 5 10 h By similar triangles h 3 = h + 10 5 So 5 h = 3 ( h + 10 ) ⇒ 3 h = 5 h + 30 ⇒ 2 h = 30 Therefore h = 15 cm The volume of the whole cone is V W = π · r 2 · h 3 = π × 5 2 × ( 15 + 10 ) 3 = 654. 50 cm 3 The volume of the top cone is V T = π · r 2 · h 3 = π × 3 2 × 15 3 = 141. 37 cm 3 So the volume of the frustum is 654. 50 − 141. 37 = 513. 13 cm 3
eBook - PDFMathematics for Elementary Teachers
A Contemporary Approach
- Gary L. Musser, Blake E. Peterson, William F. Burger(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
The height of one equilateral triangle is found by using the Pythagorean theorem. h h h 2 2 2 2 4 8 64 16 48 48 4 3 + = = − = = = Thus the area of the hexagon is 6 8 4 3 96 3 1 2 ( ) ¥ ( ) = square units. We now need to find the height of the pyramid in order to find the volume. Once again the Pythagorean theorem will be applied. Figure 13.72 shows the right triangle under consideration. The height of the pyramid is found as follows: h h h 2 2 2 2 8 17 289 64 225 225 15 + = = − = = = Putting all the pieces together, the volume of the pyramid is ■ V = = 1 3 96 3 15 480 3 ¥ ¥ cubic units. 17 8 Figure 13.70 h 4 8 Figure 13.71 h 17 8 Figure 13.72 Check for Understanding: Exercise/Problem Set A #6–7 ✔ Cones We can determine the volume of a cone in a similar manner by considering a sequence of pyramids with increasing numbers of sides in the bases (Figure 13.73). Since the 702 Chapter 13 Measurement Spheres In Section 13.3 it was stated that Archimedes observed that the volume of a sphere is two-thirds the volume of the smallest cylinder containing the sphere. Figure 13.74 shows this situation. The volume of the cylinder is V r r r = = ( ) . p p 2 3 2 2 Hence the volume of the sphere is 2 3 2 3 ( ), p r or 4 3 3 p r . Thus we have derived a formula for the volume of a sphere. Figure 13.73 volume of each pyramid is one-third of the volume of the smallest prism containing it, we would expect the volume of a cone to be one-third of the volume of the small- est cylinder containing it. This is, in fact, the case. That is, the volume of a cone is one-third of the product of the area of its base and its height. This property holds for right and oblique cones. Volume of a Cone The volume V of a cone whose base has area A and whose height is h is V Ah = 1 3 . h r A Common Core – Grade 7 Know the formulas for the vol- umes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.
eBook - PDF- Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
- 2019(Publication Date)
- Cengage Learning EMEA(Publisher)
(a) Regular square pyramid (b) Square pyramid Figure 9.20 Discover There are kits that contain a hollow pyramid and a hollow prism that have congruent bases and the same altitude. Using a kit, fill the pyramid with water and then empty the water into the prism. a) How many times did you have to empty the pyramid in order to fill the prism? b) As a fraction, the volume of the pyramid is what part of the volume of the prism? ANSWERS (a) Three times (b) 1 3 TABLE 9.1 Type of Measure Geometric Measure Type of Unit Linear Length of segment, such as length of slant height in., cm, ft Area Amount of plane region enclosed, such as area of lateral face in 2 , cm 2 , ft 2 Volume Amount of space enclosed, such as volume of a pyramid in 3 , cm 3 , ft 3 Theorem 9.2.4 The volume V of a pyramid having a base area B and an altitude of length h is given by V 5 1 3 Bh VOLUME OF A PYRAMID The final theorem in this section is presented without any attempt to construct the proof. In an advanced course such as calculus, the statement can be proved. The factor “one-third” in the formula for the volume of a pyramid provides exact results. This formula can be applied to any pyramid, even one that is not regular; in Figure 9.20(b), the length of the altitude is the perpendicular distance from the vertex to the plane of the square base. Read the Discover activity in the margin at the left before considering Theorem 9.2.4 and its applications. 414 CHAPTER 9 ■ SURFACES AND SOLIDS In Example 5, we apply Theorem 9.2.5. This application of the Pythagorean Theorem relates the lengths of the lateral edge, the radius of the base, and the altitude of a regular pyramid. Figure 9.21(c) provides a visual interpretation of the theorem. Theorem 9.2.5 In a regular pyramid, the lengths of altitude h, radius r of the base, and lateral edge e satisfy the Pythagorean Theorem; that is, e 2 5 h 2 1 r 2 . EXAMPLE 5 Find the volume of the regular square pyramid in Figure 9.21(a).
eBook - ePub- W Bolton, W. Bolton(Authors)
- 2012(Publication Date)
- Routledge(Publisher)
cone is used.Figure 2.17 PyramidsThe term right pyramid is used when the apex is vertically above the centre of the base, otherwise the term oblique pyramid is used. If we think of a right pyramid as built up from a number of layers, then sliding successive layers over another generates an oblique pyramid will the same volume (Figure 2.18 ). A right pyramid and an oblique pyramid will have the same volumes if their base areas are the same and their vertical heights equal.Figure 2.18 Right and oblique pyramidsThe volume V of any pyramid is:V = × area of base × perrpendicular heightThis can be demonstrated by first considering a triangular-based prism (Figure 2.19 ). Diagonal planes can be used to divide the prism into three equal volume triangular-based pyramids ABCE, DEFC and BCDE. Thus, since the volume of the prism is the base area multiplied by the perpendicular height, the volume of the triangular prism is one third the base area multiplied by the perpendicular height. This argument applies to other prisms.Figure 2.19 A prism divided into pyramidsThe total surface area of a pyramid is the sum of the areas of the triangles forming the sides plus the area of the base.A cone with a base of radius r and vertical height h (Figure 2.17(c) ) has a base area of πr2 and thus the volume V of a cone is:V = πr2 hThe area of the curved surface of a cone may be obtained by imagining the cone surface to be a sheet of paper which is cut along the line OA and then unrolled onto a flat surface (Figure 2.20 ). The sheet is a segment of a circle with a radius equal to the slant height l of the cone and segment having a circumference equal to that of the cone, i.e. 2πr. The area of a circle of radius l is πl2 and its circumference is 2πl; thus the area of a segment with an arc length 2πr is (2πr/2πl) × πl2 = πrl
eBook - PDF- John Peterson, Robert Smith(Authors)
- 2019(Publication Date)
- Cengage Learning EMEA(Publisher)
The smaller base is the circle or polygon formed by the parallel cutting plane. The smaller base of the pyramid has the same shape as the larger base. The two bases are similar. The altitude is the perpendic-ular segment that joins the planes of the bases. The height is the length of the altitude. A frustum of a pyramid and a frustum of a cone with their parts identified are shown in Figure 63-3. Altitude Altitude Small Base Large Base ~ ~ Small Base ~ Large Base ~ FIGURE 63-3 VOLUMES OF FRUSTUMS OF REGULAR PYRAMIDS AND RIGHT CIRCULAR CONES The volume of the frustum of a pyramid or cone is computed from the following formula: V 5 1 3 h _ A B 1 A b 1 Ï A B A b + where V 5 volume of the frustum of a pyramid or cone h 5 height A B 5 area of larger base A b 5 area of smaller base The formula for the volume of a frustum of a right circular cone is expressed in this form: V 5 1 3 p h s R 2 1 r 2 1 Rr d where V 5 volume of the frustum of a right circular cone h 5 height R 5 radius of larger base r 5 radius of smaller base Example 1 A container is designed in the shape of a frustum of a pyramid with square bases as shown in Figure 63-4. a. Find the volume of the container in cubic feet. b. Compute the capacity (number of gallons) of liquid that the container can hold when full. Note: One cubic foot contains 7.5 gallons. Solutions a. Find the volume. Find the larger base area. A B 5 (14.0 in.) 2 5 196 sq in. Find the smaller base area. A b 5 (11.0 in.) 2 5 121 sq in. 14.0 in. 16.0 in. Height 11.0 in. FIGURE 63-4 UNIT 63 VOLUMES OF PYRAMIDS AND CONES 477 Find the volume. V 5 s 16.0 in. d f 196 sq in. 1 121 sq in. 1 Ïs 196 sq in. d s 121 sq in. d g 3 V 5 s 16.0 in. d s 196 sq in. 1 121 sq in. 1 154 sq in. d 3 V 5 2512 cu in. Express the volume in cubic feet. 2512 cu in. 4 1728 cu in./cu ft < 1.45 cu ft Ans b. Compute the capacity of the container in gallons. 1.45 cu ft 3 7.5 gal/cu ft < 10.9 gal Ans Example 2 A tapered shaft is shown in Figure 63-5.
eBook - PDF- John Peterson, Robert Smith(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
The smaller base is the circle or polygon formed by the parallel cutting plane. The smaller base of the pyramid has the same shape as the larger base. The two bases are similar. The altitude is the perpendicular seg-ment that joins the planes of the bases. The height is the length of the altitude. A frustum of a pyramid and a frustum of a cone with their parts identified are shown in Figure 29–21. Figure 29–21 ALTITUDE ALTITUDE SMALL BASE SMALL BASE LARGE BASE LARGE BASE Volumes of Frustums of Regular Pyramids and Right Circular Cones The volume of the frustum of a pyramid or cone is computed from the formula V 5 1 3 h s A B 1 A b 1 Ï A B A b d where V 5 volume of the frustum of a pyramid or cone h 5 height A B 5 area of larger base A b 5 area of smaller base The formula for the volume of a frustum of a right circular cone is expressed in this form. V 5 1 3 p h s R 2 1 r 2 1 Rr d where V 5 volume of a right circular cone h 5 height R 5 radius of larger base r 5 radius of smaller base These examples illustrate the method of computing volumes of frustums of regular pyramids and right circular cones. EXAMPLES 1. A wastebasket is designed in the shape of a frustum of a pyramid with a square base, as shown in Figure 29–22. Find the volume of the basket in cubic feet. Find the larger base area. A B 5 s 14.0 in. d 2 5 196 sq in. Find the smaller base area. A b 5 s 11.0 in. d 2 5 121 sq in. Figure 29–22 16.0 in. HEIGHT 14.0 in. 11.0 in. Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Unit 29 PYRAMIDS AND CONES: VOLUMES, SURFACE AREAS, AND WEIGHTS 729 Find the volume.
eBook - PDF- John Bird(Author)
- 2017(Publication Date)
- Routledge(Publisher)
Determine the volume and the total surface area of the square pyramid shown in Fig. 20.3 if its perpendicular height is 12 cm. 5 cm 5 cm D C E A B Figure 20.3 Volume of Pyramid = 1 3 (area of base) × (perpendicular height) = 1 3 (5 × 5) × 12 = 100 cm 3 The total surface area consists of a square base and 4 equal triangles. Area of triangle ADE = 1 2 × base × perpendicular height = 1 2 × 5 × AC The length AC may be calculated using Pythagoras’ theorem on triangle ABC, where AB = 12 cm, BC = 1 2 × 5 = 2.5 cm Hence, AC = AB 2 + BC 2 = 12 2 + 2.5 2 = 12.26 cm Hence area of triangle ADE = 1 2 × 5 × 12.26 = 30.65 cm 2 Total surface area of pyramid = (5 × 5) + 4(30.65) = 147.6 cm 2 Problem 6. Determine the volume and total surface area of a cone of radius 5 cm and perpendicular height 12 cm The cone is shown in Fig. 20.4. r 5 5 cm l h 5 12 cm Figure 20.4 Volume of cone = 1 3 πr 2 h = 1 3 × π × 5 2 × 12 = 314.2 cm 3 Total surface area = curved surface area + area of base = πrl + πr 2 From Fig. 20.4, slant height l may be calculated using Pythagoras’ theorem l = 12 2 + 5 2 = 13 cm Volumes and surface areas of common solids 185 Hence total surface area = (π × 5 × 13) + (π × 5 2 ) = 282.7 cm 2 Problem 7. Find the volume and surface area of a sphere of diameter 8 cm Since diameter = 8 cm, then radius, r = 4 cm. Volume of sphere = 4 3 πr 3 = 4 3 × π × 4 3 = 268.1 cm 3 Surface area of sphere = 4πr 2 = 4 × π × 4 2 = 201.1 cm 2 Now try the following Practice Exercise Practice Exercise 82 Volumes and surface areas of regular solids (Answers on page 681) 1. A rectangular block of metal has dimensions of 40 mm by 25 mm by 15 mm. Determine its volume. Find also its mass if the metal has a density of 9 g/cm 3 2. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by 2.5 m (1 litre = 1000 cm 3 ) 3. Determine how many cubic metres of con- crete are required for a 120 m long path, 150 mm wide and 80 mm deep 4.
eBook - PDFMathematical Literacy NQF4 SB
TVET FIRST
- K van Niekerk O Roberts(Author)
- 2017(Publication Date)
- Macmillan(Publisher)
We use volume to measure and compare how much space things take up. l The quantity of liquids such as water, milk, cool drink, and so on is always measured in units such as ℓ, ml, and so on. l We can also measure the volume of solid objects in cubic units. l We can convert between the units of liquid volume and solid units. You learnt about this in Topic 1. l Volume is calculated by multiplying three dimensions together. l Volume of a box = length × breadth × height. Nets of 3D objects To find the surface area of 3D objects, it helps to imagine or sketch their nets (when they are ‘opened up’ and laid flat). The total surface area is the sum of the areas of the flat shapes. Remember We calculate volume by multiplying three dimensions together, so the units are always cubic units. Did you know? The surface area of a sphere is exactly four times the area of a circle with the same radius. You can see this in the area formula, since the area of a circle is π r 2 and the surface area of a sphere is 4π r 2 .
- Frank R. Spellman, Nancy E. Whiting(Authors)
- 2013(Publication Date)
- CRC Press(Publisher)
Sphere— A container shaped like a ball. Square units— Measurements used to express area (e.g., square feet, square meters, acres). Volume— The capacity of a unit (how much it will hold), measured in cubic units (e.g., cubic feet, cubic meters) or in liquid volume units (e.g., gallons, liters, million gallons). Width— The distance from one side of the tank to the other, measured in linear units. 61 Basic Math Operations 2.11.2 R ELEVANT G EOMETRIC E QUATIONS Circumference C of a circle: C = π d = 2 π r Perimeter P of a square with side a : P = 4 a Perimeter P of a rectangle with sides a and b : P = 2 a + 2 b Perimeter P of a triangle with sides a , b , and c : P = a + b + c Area A of a circle with radius r ( d = 2 r ): A = π d 2 /4 = π r 2 Area A of duct in square feet when d is in inches: A = 0.005454 d 2 Area A of a triangle with base b and height h : A = 0.5 bh Area A of a square with sides a : A = a 2 Area A of a rectangle with sides a and b : A = ab Area A of an ellipse with major axis a and minor axis b : A = π ab Area A of a trapezoid with parallel sides a and b and height h : A = 0.5( a + b ) h Area A of a duct in square feet when d is in inches: A = π d 2 /576 = 0.005454 d 2 Volume V of a sphere with a radius r ( d = 2 r ): V = 1.33 π r 3 = 0.1667 π d 3 Volume V of a cube with sides a : V = a 3 Volume V of a rectangular solid (sides a and b and height c ): V = abc Volume V of a cylinder with a radius r and height H : V = π r 2 h = π d 2 h /4 Volume V of a pyramid: V = 0.33 2.11.3 G EOMETRICAL C ALCULATIONS 2.11.3.1 Perimeter and Circumference On occasion, it may be necessary to determine the distance around grounds or landscapes. To mea-sure the distance around property, buildings, and basin-like structures, it is necessary to determine either perimeter or circumference. The perimeter is the distance around an object; a border or outer boundary. Circumference is the distance around a circle or circular object, such as a clarifier.
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