Volume of Cone

10 Key excerpts on "Volume of Cone"

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  Mathematics for Elementary Teachers

  A Contemporary Approach

  • Gary L. Musser, Blake E. Peterson, William F. Burger(Authors)
  • 2013(Publication Date)
  • Wiley

    (Publisher)

  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. T H E O R E M 1 3 . 1 7 For a cone with a circular base of radius r , the volume of the cone is 1 3 2 p r h. Check for Understanding: Exercise/Problem Set A #8–9 ✔ r r 2r Figure 13.74 Volume of a Sphere The volume V of a sphere with radius r is V r = 4 3 3 p . r T H E O R E M 1 3 . 1 8 Section 13.4 Volume 703 Table 13.11 summarizes the volume and surface area formulas for right prisms, right circular cylinders, right regular pyramids, right circular cones, and spheres. The indicated dimensions are the area of the base, A; the height, h; the perimeter or cir- cumference of the base, P or C ; and the slant height, l . By observing similarities, one can minimize the amount of memorization. TABLE 13.11 GEOMETRIC SHAPE SURFACE AREA VOLUME Right prism S A Ph = + 2 V Ah = Right circular cylinder S A Ch = + 2 V Ah = Right regular pyramid S A Pl = + 1 2 V Ah = 1 3 Right circular cone S A Cl = + 1 2 V Ah = 1 3 Sphere S r = 4 2 π V r = 4 3 3 π The remainder of this section presents a more formal derivation of the volume and surface area of a sphere. First, to find the volume of a sphere, we use Cavalieri’s prin- ciple, which compares solids where cross-sections have equal areas. As an aid to recall and distinguish between the formulas for the surface area and volume of a sphere, observe that the r in 4 2 p r is squared, an area unit, whereas the r in 4 3 3 p r is cubed, a volume unit. An ice cream shop has sugar cones with a slant height of 13 cm and the diameter of the base is 10 cm (see Figure 13.75).

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  Mathematical Practices, Mathematics for Teachers

  Activities, Models, and Real-Life Examples

  • Ron Larson, Robyn Silbey(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Bases are congruent. h Volume of a Cone Words The volume V of a cone is one-third Height, h Area of base, B the product of the area of the base and the height of the cone. Algebra V = 1 — 3 Bh Area of base Height EXAMPLE 5 Finding the Volume of a Cone Find the volume of the cone. Round your answer 15 m 9 m to the nearest tenth. SOLUTION The diameter is 9 meters. So, the radius is 4.5 meters. V = 1 — 3 Bh = 1 — 3 πr 2 h = 1 — 3 π(4.5) 2 (15) = 101.25π Simplify. ≈ 317.925 Multiply. The volume is about 317.9 cubic meters. Write formula for volume of a cone. Substitute 4.5 for r and 15 for h. The formula for the volume of a cone applies to all cones, including oblique cones and cones with noncircular bases. Classroom Tip Copyright 2014 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. 482 Chapter 12 Circles and Circular Solids You sell 10 gallons of lemonade in the paper cup shown 8 cm 12 cm at the right. One gallon is about 3785 cubic centimeters. a. How many paper cups will you need if you fill each cup to 100% of its capacity? b. You decide to fill each cup to 80% of its capacity. How many more cups of lemonade can you sell? SOLUTION a. Begin by finding the volume of the cup. The diameter is 8 centimeters. So, the radius is 4 centimeters. V = 1 — 3 Bh = 1 — 3 πr 2 h Write formula for volume of a cone. = 1 — 3 π(4) 2 (12) Substitute 4 for r and 12 for h. = 64π Simplify. ≈ 201 Multiply. The volume is about 201 cm 3 . Ten gallons is about 10 × 3785 cm 3 = 37,850 cm 3 . So, you need 37,850 ÷ 201 ≈ 188 paper cups.

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  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2019(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Then the volume of the right circular cone is V 5 1 3 Bh. Because the area of the base of the cone is B 5 pr 2 , an alternative formula for the volume of the cone is V 5 1 3 pr 2 h We state this result as a theorem. Theorem 9.3.7 The volume V of a right circular cone with base area B and altitude of length h is given by V 5 1 3 Bh. Alternative Form: Where r is the length of the radius of the base, the formula for the volume of the cone is usually written as V 5 1 3 pr 2 h. Discover Complete this analogy: Prism is to Cylinder as Pyramid is to ___. ANSWER Cone Figure 9.34 Using a kit that contains “hollow” models of a right circular cylinder and a right circular cone of the same dimensions (altitude, radius of base), compare their volumes. ANSWER Geometry in the Real World r r h V cone 5 1 3 V cylinder 9.3 ■ Cylinders and Cones 425 Table 9.3 should help us to recall and compare the area and volume formulas found in Sections 9.2 and 9.3. TABLE 9.3 Lateral Area Total Area Volume Slant Height Pyramid L 5 1 2 /P T 5 B 1 L V 5 1 3 Bh / 2 5 a 2 1 h 2 Cone L 5 1 2 /C T 5 B 1 L V 5 1 3 Bh / 2 5 r 2 1 h 2 NOTE: The formulas that contain the slant height / are used only with a regular pyramid and a right circular cone. SSG EXS. 12, 13 SOLIDS OF REVOLUTION Suppose that part of the boundary for a plane region is a line segment. When the plane region is revolved about this line segment, the locus of points generated in space is called a solid of revolution. The complete 360° rotation moves the region about the edge until the region returns to its original position. The side (edge) used is called the axis of the result- ing solid of revolution. Consider Example 5. EXAMPLE 5 Describe the solid of revolution that results when a) a rectangular region with dimensions 2 ft by 5 ft is revolved about the 5-ft side, as shown in Figure 9.35(a). b) a semicircular region with radius of length 3 cm is revolved about the diameter, shown in Figure 9.35(b).

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  Geometry

  A Self-Teaching Guide

  • Steve Slavin, Ginny Crisonino(Authors)
  • 2004(Publication Date)
  • Wiley

    (Publisher)

  Volume = πr 2 h = (3.14)(4.5) 2 (12) = (3.14)(20.25)(12) = 763.02 cubic yards c. Volume = πr 2 h = (3.14)(8) 2 (23) = (3.14)(64)(23) = 4,622.08 cubic inches 2. a. SA = 2πrh = 2(3.14)(12)(9) = (6.28)(108) = 678.24 square feet b. SA = 2πrh = 2(3.14)(4.5)(12) = (6.28)(54) = 339.12 square yards c. SA = 2πrh = 2(3.14)(8)(23) = (6.28)(184) = 1,155.52 cubic inches 9 2 d 2 182 GEOMETRY Cones You know cones. We eat ice cream cones and snow cones, and we see orange traf- fic cones whenever there is highway construction. And we’ve all seen conical headgear ranging from party hats to coneheads and dunce caps. A cone is the three-dimensional shape formed by a straight line when one end is moved around a simple closed curve, while the other end of the line is kept fixed at a point that is not in the plane of the curve. The following figure is a cone with a radius of 3 and a height of 6. Volume of a cone formula V = 1 3 πr 2 h See if you can find the volume of this cone. Solution: Volume = πr 2 h = (.33)(3.14)(3) 2 (6) = (1.0362)(54) = 55.95 Example 26: Find the volume of a cone with a diameter of 10 inches and a height of 27 inches. Solution: Volume = πr 2 h = (.33)(3.14)(5) 2 (27) = (1.0362)(675) = 699.435 cubic inches SELF-TEST 5 1. Find the volume of a cone with a radius of 12 inches and a height of 25 inches. 2. Find the volume of a cone with a radius of 3.5 feet and a height of 6 feet. 3. Find the volume of a cone with a diameter of 8 yards and a height of 13 yards. 1 3 1 3 6 3 Volume and Surface Area of Three-dimensional Polygons 183 ANSWERS 1. Volume = πr 2 h = (.33)(3.14)(12) 2 (25) = (1.0362)(144)(25) = 3,730.32 cubic inches 2. Volume = πr 2 h = (.33)(3.14)(3.5) 2 (6) = (1.0362)(12.25)(6) = 76.1607 cubic feet 3. Volume = πr 2 h = (.33)(3.14)(4) 2 (13) = (1.0362)(16)(13) = 215.53 cubic yards Complex Geometric Solids So far we’ve found the volume and surface area of cubes, rectangular solids, pyra- mids, spheres, cylinders, and cones.

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  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein, , , Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  35. The volume V of a washer that has an inside radius of length r , an outside radius of length R , and an altitude of measure h is given by . 36. For a right circular cone, the slant height has a measure equal to twice that of the radius length of the base. If the total area of the cone is , what are the dimensions of the cone? 37. For a right circular cone, the ratio of the slant height to the length of the radius is 5:3. If the volume of the cone is , find the lateral area of the cone. 38. If the length of the radius and the height of a right circular cylinder are both doubled to form a larger cylinder, what is the ratio of the volume of the larger cylinder to the volume of the smaller cylinder? (Note: The two cylinders are said to be “similar.”) 39. For the two similar cylinders in Exercise 38, what is the ratio of the lateral area of the larger cylinder to that of the smaller cylinder? 40. For a right circular cone, the dimensions are and . If the length of the radius is doubled while the height is made half as large in forming a new cone, will the volumes of the two cones be equal? 41. A cylindrical storage tank has a depth of 5 ft and a radius measuring 2 ft. If each cubic foot can hold 7.5 gal of gasoline, what is the total storage capacity of the tank measured in gallons? 42. If the tank in Exercise 41 needs to be painted and 1 pt of paint covers 50 ft , how many pints are needed to paint the exterior of the storage tank? 43. A frustrum of a cone is the portion of the cone bounded between the circular base and a plane parallel to the base. With dimensions as indicated, show that the volume of the frustrum of the cone is In Exercises 44 and 45, use the formula from Exercise 43. Similar triangles were used to find h and H. 44. A margarine tub has the shape of the frustrum of a cone. With the lower base having diameter length 11 cm and the upper base having diameter length 14 cm, the volume of such a container tall can be determined by using , and .

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  Spellman's Standard Handbook for Wastewater Operators

  Volume I, Fundamental Level, Second Edition

  • Frank R. Spellman(Author)
  • 2010(Publication Date)
  • CRC Press

    (Publisher)

  Pi (π)—A number used in calculations involving circles, spheres, or cones (π = 3.14). Radius— The distance, measured in linear units, from the center of a circle to the edge. 3 Basic Mathematics 45 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 from one side of the tank to the other, measured in linear units. 3.10.2 Relevant Geometric Equations Circumference C of a circle: C = π d = 2π 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 a square with sides a : A = a 2 Area A of a rectangle with sides a and b : A = ab Area A of a triangle with base b and height h : A = 0.5 bh 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 A = 0.5( a + b ) h and height h : Area A of a duct (in ft 2 ) 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 = 1.33π r 3 = 0.1667π 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 with base area B and height h : V = 0.33 Bh 46 Spellman’s Standard Handbook for Wastewater Operators: Volume I, Fundamental Level 3.10.3 Geometrical Calculations 3.10.3.1 Perimeter and Circumference On occasion, it may be necessary to determine the distance around grounds or landscapes.

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  Engineering Mathematics

  • John Bird(Author)
  • 2017(Publication Date)
  • Routledge

    (Publisher)

  The volume of a frustum of a pyramid or cone is given by the volume of the whole pyramid or cone minus the volume of the small pyramid or cone cut off. The surface area of the sides of a frustum of a pyra- mid or cone is given by the surface area of the whole pyramid or cone minus the surface area of the small 190 Engineering Mathematics pyramid or cone cut off. This gives the lateral surface area of the frustum. If the total surface area of the frus- tum is required then the surface area of the two parallel ends are added to the lateral surface area. There is an alternative method for finding the volume and surface area of a frustum of a cone. With reference to Fig. 20.12: h R r I Figure 20.12 Volume = 1 3 πh(R 2 + Rr + r 2 ) Curved surface area = π l (R + r) Total surface area = π l (R + r) + π r 2 + π R 2 Problem 16. Determine the volume of a frustum of a cone if the diameter of the ends are 6.0 cm and 4.0 cm and its perpendicular height is 3.6 cm Method 1 A section through the vertex of a complete cone is shown in Fig. 20.13 Using similar triangles AP DP = DR BR Hence AP 2.0 = 3.6 1.0 from which AP = (2.0)(3.6) 1.0 = 7.2cm The height of the large cone = 3.6 + 7.2 = 10.8 cm. Volume of frustum of cone = volume of large cone − volume of small cone cut off = 1 3 π(3.0) 2 (10.8) − 1 3 π(2.0) 2 (7.2) = 101.79 − 30.16 = 71.6 cm 3 4.0cm P E A D R B Q C 2.0cm 3.6cm 1.0cm 3.0cm 6.0cm Figure 20.13 Method 2 From above, volume of the frustum of a cone = 1 3 πh(R 2 + Rr + r 2 ) where R = 3.0cm, r = 2.0cm and h = 3.6cm Hence volume of frustum = 1 3 π(3.6)[(3.0) 2 + (3.0)(2.0) + (2.0) 2 ] = 1 3 π(3.6)(19.0) = 71.6 cm 3 Problem 17. Find the total surface area of the frustum of the cone in Problem 16 Method 1 Curved surface area of frustum = curved surface area of large cone—curved surface area of small cone cut off.

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  eBook - ePub

  Science and Mathematics for Engineering

  • John Bird(Author)
  • 2019(Publication Date)
  • Routledge

    (Publisher)

  Chapter 14 Volumes of common solids

  Why it is important to understand: Volumes of common solids

  There are many practical applications where volumes and surface areas of common solids are required. Examples include determining capacities of oil, water, petrol and fish tanks, ventilation shafts and cooling towers, determining volumes of blocks of metal, ball-bearings, boilers and buoys, and calculating the cubic metres of concrete needed for a path. Finding the surface areas of loudspeaker diaphragms and lampshades provide further practical examples. Understanding these calculations is essential for the many practical applications in engineering, construction, architecture and science.

  At the end of this chapter, you should be able to:

  • state the SI unit of volume
  • calculate the volumes and surface areas of cuboids, cylinders, prisms, pyramids, cones and spheres
  • appreciate that volumes of similar bodies are proportional to the cubes of the corresponding linear dimensions

  14.1   Introduction

  The volume of any solid is a measure of the space occupied by the solid. Volume is measured in cubic units such as mm3 , cm3 and m3 .

  This chapter deals with finding volumes of common solids; in engineering it is often important to be able to calculate volume or capacity, to estimate, say, the amount of liquid, such as water, oil or petrol, in differently shaped containers.

  A prism is a solid with a constant cross-section and with two ends parallel. The shape of the end is used to describe the prism. For example, there are rectangular prisms (called cuboids), triangular prisms and circular prisms (called cylinders).

  On completing this chapter you will be able to calculate the volumes and surface areas of rectangular and other prisms, cylinders, pyramids, cones and spheres. Volumes of similar shapes are also considered.

  14.2   Calculating volumes and surface areas of common solids

  14.2.1   Cuboid or rectangular prism

  Science and Mathematics for Engineering. 978-0-367-20475-4, © John Bird. Published by Taylor & Francis. All rights reserved.

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  Believe me, you really want to choose this axis so that the cross-sections are as simple as possible. It helps if you can ensure that the cross-sections are in fact similar to each other, that is, different-sized copies of each other. This isn’t always possible, though. Let’s use the above technique to find the volume of a “generalized” cone. What this means is that we have some shape in a plane of area A square units, and an apex point P which hovers some distance above the plane: Section 29.2: Volumes of General Solids • 633 P h Now we draw a line segment from each point on the edge of the shape up to P . This gives us a surface whose base is the shape we started with. The solid we’re interested in is the filled-in version of the surface, or if you prefer, the interior of the surface. Here’s sort of what the surface looks like, in skeleton form: P h For example, if the base is a circle and the point P sits directly above the center of the circle, then we’ll get an ordinary cone. If the base is a square and P is directly above the center of the square (that is, where the diagonals of the square intersect), then we’ll get a square pyramid. You should think about what choice of base and point P gives you (a) a regular pyramid, or (b) a skew-cone (which looks like a weird hat—sort of like a witch’s hat but it doesn’t go straight up). It turns out that the only quantities which are relevant to finding the volume of the solid are the area of the base, A square units, and the perpendicular distance from P to the plane. We’ll call this last quantity h units (it’s labeled in the above figures). So, how do we find the volume? We first have to choose an axis. P seems to be a special point, so the line we choose should probably pass through P . Where else should it go? You could try all sorts of things, but the only thing that works is to make the line perpendicular to the plane which contains the base.

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  Introductory Technical Mathematics

  • John Peterson, Robert Smith(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  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. 736 Section V GEOMETRIC FIGURES: AREAS AND VOLUMES 3. A regular pyramid with a base area of 54.6 square feet contains 210.5 cubic feet of material. Find the height of the pyramid. 4. Compute the base area of a right circular cone that is 15.8 centimeters high and has a volume of 1070 cubic centimeters. 5. A regular pyramid has a base perimeter of 56.3 inches and a slant height of 14.9 inches. Find the lateral area of the pyramid. 6. A right circular cone has a slant height of 3.76 feet. The base circumference is 17.58 feet, and the base area is 24.62 square feet. a. Compute the lateral area of the cone. b. Compute the total surface area of the cone. 7. The frustum of a right circular cone has a larger base area of 40.0 square centimeters and a smaller base area of 19.0 square centimeters. The height is 22.0 centimeters. Find the volume. 8. The frustum of a regular pyramid has a smaller base perimeter of 18.0 meters and a larger base perimeter of 26.0 meters. The slant height is 5.60 meters. Find the lateral area. 9. A building has a roof in the shape of a regular pyramid. Each of the 4 outside walls of the building is 38 feet 0 inches long, and the roof is 16 feet 0 inches high. Compute the number of cubic feet of attic space in the building. 10. A solid brass casting in the shape of a right circular cone has a base diameter of 4.36 inches and a height of 3.94 inches. Find the weight of the casting. Brass weighs 0.302 pound per cubic inch. 11. A vessel in the shape of a right circular cone has a capacity of 0.690 liter. The base diameter is 12.3 centimeters. What is the height of the vessel? 12. A plywood form is constructed in the shape of a right pyramid with a square base.

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