Find the Volume of Each Pyramid. 1. SOLUTION

Total Page:16

File Type:pdf, Size:1020Kb

Find the Volume of Each Pyramid. 1. SOLUTION 11-3 Volumes of Pyramids and Cones Find the volume of each pyramid. 1. 2. SOLUTION: SOLUTION: The volume of a pyramid is , where B is the The volume of a pyramid is , where B is the area of the base and h is the height of the pyramid. area of the base and h is the height of the pyramid. The base of this pyramid is a right triangle with legs The base of this pyramid is a regular pentagon with of 9 inches and 5 inches and the height of the sides of 4.4 centimeters and an apothem of 3 pyramid is 10 inches. centimeters. The height of the pyramid is 12 centimeters. ANSWER: ANSWER: 75 in3 132 cm3 3. a rectangular pyramid with a height of 5.2 meters and a base 8 meters by 4.5 meters SOLUTION: The volume of a pyramid is , where B is the area of the base and h is the height of the pyramid. The base of this pyramid is a rectangle with a length of 8 meters and a width of 4.5 meters. The height of the pyramid is 5.2 meters. ANSWER: 62.4 m3 eSolutions Manual - Powered by Cognero Page 1 11-3 Volumes of Pyramids and Cones 4. a square pyramid with a height of 14 meters and a base with 8-meter side lengths SOLUTION: 6. The volume of a pyramid is , where B is the SOLUTION: area of the base and h is the height of the pyramid. The base of this pyramid is a square with sides of 8 meters. The height of the pyramid is 14 meters. Use trigonometry to find the radius r. ANSWER: The volume of a circular cone is , or 298.7 m3 , where B is the area of the base, h is the Find the volume of each cone. Round to the height of the cone, and r is the radius of the base. nearest tenth. The height of the cone is 11.5 centimeters. 5. SOLUTION: ANSWER: The volume of a circular cone is , or 168.1 cm3 , where B is the area of the base, h is the height of the cone, and r is the radius of the base. Since the diameter of this cone is 7 inches, the radius is or 3.5 inches. The height of the cone is 4 inches. ANSWER: 51.3 in3 eSolutions Manual - Powered by Cognero Page 2 11-3 Volumes of Pyramids and Cones 7. an oblique cone with a height of 10.5 millimeters and 8. a cone with a slant height of 25 meters and a radius a radius of 1.6 millimeters of 15 meters SOLUTION: SOLUTION: The volume of a circular cone is , or , where B is the area of the base, h is the height of the cone, and r is the radius of the base. The radius of this cone is 1.6 millimeters and the Use the Pythagorean Theorem to find the height h of height is 10.5 millimeters. the cone. Then find its volume. ANSWER: So, the height of the cone is 20 meters. 3 28.1 mm ANSWER: 4712.4 m3 eSolutions Manual - Powered by Cognero Page 3 11-3 Volumes of Pyramids and Cones 9. HUTS The Caddo Indians, from the east part of STRUCTURE Find the volume of each pyramid. Texas, lived in tall cone-shaped grass huts made of Round to the nearest tenth. wooden pole frames with long prairie grasses dried and threaded through the poles in layers. These houses normally measured 40 feet tall. Suppose the diameter is also 40 feet. What is the volume inside the hut? 10. SOLUTION: The volume of a pyramid is , where B is the area of the base and h is the height of the pyramid. SOLUTION: The volume of a circular cone is , where B is the area of the base and h is the height of the cone. ANSWER: 605 in3 ANSWER: 11. about 16,755 ft3 SOLUTION: The volume of a pyramid is , where B is the area of the base and h is the height of the pyramid. ANSWER: 105.8 mm3 eSolutions Manual - Powered by Cognero Page 4 11-3 Volumes of Pyramids and Cones 12. SOLUTION: The volume of a pyramid is , where B is the area of the base and h is the height of the pyramid. The apothem is . ANSWER: 482.1 m3 ANSWER: 233.8 cm3 13. 14. a pentagonal pyramid with a base area of 590 square SOLUTION: feet and an altitude of 7 feet The volume of a pyramid is , where B is the SOLUTION: area of the base and h is the height of the pyramid. The volume of a pyramid is , where B is the The base is a hexagon, so we need to make a right area of the base and h is the height of the pyramid. triangle to determine the apothem. The interior angles of the hexagon are 120°. The radius bisects the angle, so the right triangle is a 30°-60°-90° triangle. ANSWER: 1376.7 ft3 eSolutions Manual - Powered by Cognero Page 5 11-3 Volumes of Pyramids and Cones 15. a triangular pyramid with a height of 4.8 centimeters Use the Pythagorean Theorem to find the other leg a and a right triangle base with a leg 5 centimeters and of the right triangle and then find the area of the hypotenuse 10.2 centimeters triangle. SOLUTION: Find the height of the right triangle. The length of the other leg of the right triangle is 6 cm. So, the area of the base B is 24 cm2. The volume of a pyramid is , where B is the area of the base and h is the height of the pyramid. Replace V with 144 and B with 24 in the formula for the volume of a pyramid and solve for the height h. ANSWER: 3 35.6 cm 16. A triangular pyramid with a right triangle base with a Therefore, the height of the triangular pyramid is 18 leg 8 centimeters and hypotenuse 10 centimeters has cm. a volume of 144 cubic centimeters. Find the height. ANSWER: SOLUTION: 18 cm The base of the pyramid is a right triangle with a leg of 8 centimeters and a hypotenuse of 10 centimeters. eSolutions Manual - Powered by Cognero Page 6 11-3 Volumes of Pyramids and Cones Find the volume of each cone. Round to the nearest tenth. 18. SOLUTION: 17. SOLUTION: The volume of a circular cone is , where r is the radius of the base and h is the height of the The volume of a circular cone is , cone. The radius of this cone is 4.2 centimeters and where r is the radius of the base and h is the height the height is 7.3 centimeters. of the cone. Since the diameter of this cone is 10 inches, the radius is or 5 inches. The height of the cone is 9 inches. Therefore, the volume of the cone is about 134.8 cm3. ANSWER: 134.8 cm3 Therefore, the volume of the cone is about 235.6 in3. ANSWER: 235.6 in3 eSolutions Manual - Powered by Cognero Page 7 11-3 Volumes of Pyramids and Cones 20. 19. SOLUTION: SOLUTION: Use trigonometric ratios to find the height h and the radius r of the cone. Use a trigonometric ratio to find the height h of the cone. The volume of a circular cone is , where The volume of a circular cone is , where r is the radius of the base and h is the height of the r is the radius of the base and h is the height of the cone. The radius of this cone is 8 centimeters. cone. Therefore, the volume of the cone is about 1473.1 cm3. Therefore, the volume of the cone is about 2.8 ft3. ANSWER: ANSWER: 1473.1 cm3 2.8 ft3 eSolutions Manual - Powered by Cognero Page 8 11-3 Volumes of Pyramids and Cones 21. an oblique cone with a diameter of 16 inches and an 22. a right cone with a slant height of 5.6 centimeters and altitude of 16 inches a radius of 1 centimeter SOLUTION: SOLUTION: The cone has a radius r of 1 centimeter and a slant The volume of a circular cone is , where height of 5.6 centimeters. Use the Pythagorean r is the radius of the base and h is the height of the Theorem to find the height h of the cone. cone. Since the diameter of this cone is 16 inches, the radius is or 8 inches. Therefore, the volume of the cone is about 1072.3 in 3. ANSWER: 1072.3 in3 Therefore, the volume of the cone is about 5.8 cm3. ANSWER: 5.8 cm3 eSolutions Manual - Powered by Cognero Page 9 11-3 Volumes of Pyramids and Cones 23. SNACKS Approximately how many cubic octagonal pyramid with a height of 5 feet. The base centimeters of roasted peanuts will completely fill a has side lengths of 2 feet. What is the volume of the paper cone that is 14 centimeters high and has a base greenhouse? diameter of 8 centimeters? Round to the nearest tenth. SOLUTION: The volume of a circular cone is , where r is the radius of the base and h is the height of the cone. Since the diameter of the cone is 8 centimeters, SOLUTION: the radius is or 4 centimeters. The height of the cone is 14 centimeters.
Recommended publications
  • Pentagonal Pyramid
    Chapter 9 Surfaces and Solids Copyright © Cengage Learning. All rights reserved. Pyramids, Area, and 9.2 Volume Copyright © Cengage Learning. All rights reserved. Pyramids, Area, and Volume The solids (space figures) shown in Figure 9.14 below are pyramids. In Figure 9.14(a), point A is noncoplanar with square base BCDE. In Figure 9.14(b), F is noncoplanar with its base, GHJ. (a) (b) Figure 9.14 3 Pyramids, Area, and Volume In each space pyramid, the noncoplanar point is joined to each vertex as well as each point of the base. A solid pyramid results when the noncoplanar point is joined both to points on the polygon as well as to points in its interior. Point A is known as the vertex or apex of the square pyramid; likewise, point F is the vertex or apex of the triangular pyramid. The pyramid of Figure 9.14(b) has four triangular faces; for this reason, it is called a tetrahedron. 4 Pyramids, Area, and Volume The pyramid in Figure 9.15 is a pentagonal pyramid. It has vertex K, pentagon LMNPQ for its base, and lateral edges and Although K is called the vertex of the pyramid, there are actually six vertices: K, L, M, N, P, and Q. Figure 9.15 The sides of the base and are base edges. 5 Pyramids, Area, and Volume All lateral faces of a pyramid are triangles; KLM is one of the five lateral faces of the pentagonal pyramid. Including base LMNPQ, this pyramid has a total of six faces. The altitude of the pyramid, of length h, is the line segment from the vertex K perpendicular to the plane of the base.
    [Show full text]
  • [ENTRY POLYHEDRA] Authors: Oliver Knill: December 2000 Source: Translated Into This Format from Data Given In
    ENTRY POLYHEDRA [ENTRY POLYHEDRA] Authors: Oliver Knill: December 2000 Source: Translated into this format from data given in http://netlib.bell-labs.com/netlib tetrahedron The [tetrahedron] is a polyhedron with 4 vertices and 4 faces. The dual polyhedron is called tetrahedron. cube The [cube] is a polyhedron with 8 vertices and 6 faces. The dual polyhedron is called octahedron. hexahedron The [hexahedron] is a polyhedron with 8 vertices and 6 faces. The dual polyhedron is called octahedron. octahedron The [octahedron] is a polyhedron with 6 vertices and 8 faces. The dual polyhedron is called cube. dodecahedron The [dodecahedron] is a polyhedron with 20 vertices and 12 faces. The dual polyhedron is called icosahedron. icosahedron The [icosahedron] is a polyhedron with 12 vertices and 20 faces. The dual polyhedron is called dodecahedron. small stellated dodecahedron The [small stellated dodecahedron] is a polyhedron with 12 vertices and 12 faces. The dual polyhedron is called great dodecahedron. great dodecahedron The [great dodecahedron] is a polyhedron with 12 vertices and 12 faces. The dual polyhedron is called small stellated dodecahedron. great stellated dodecahedron The [great stellated dodecahedron] is a polyhedron with 20 vertices and 12 faces. The dual polyhedron is called great icosahedron. great icosahedron The [great icosahedron] is a polyhedron with 12 vertices and 20 faces. The dual polyhedron is called great stellated dodecahedron. truncated tetrahedron The [truncated tetrahedron] is a polyhedron with 12 vertices and 8 faces. The dual polyhedron is called triakis tetrahedron. cuboctahedron The [cuboctahedron] is a polyhedron with 12 vertices and 14 faces. The dual polyhedron is called rhombic dodecahedron.
    [Show full text]
  • Supporting Information
    Supporting Information Triple-ringed Luminescent Heptanuclear Zn(II) Cluster for Efficient Ag(I) Ion Sensing Materials Qian-Jun Deng,*,1 Min Chen,1 Dong-Chu Chen,1 Zhong-Hong Zhu,2,* Hua-Hong Zou,2,* 1 School of Material Science and Energy Engineering, Foshan University, Foshan 528000, Guangdong, People’s Republic of China 2 State Key Laboratory for Chemistry and Molecular Engineering of Medicinal Resources, School of Chemistry & Pharmacy of Guangxi Normal University, Guilin 541004, P. R. China Supporting Tables Table S1 Selected bond lengths (Å) and angles (°) of 1. Table S2 SHAPE analysis in complex 1. Supporting Figures 1 2 Figure S1 The scheme with the structures of H2L and H2L . Figure S2 The coordination pattern diagram of Zn(II) ions in compound 1. Thermogravimetry of the compounds at a heating rate of 5 °C/min Figure S3 under N2 atmosphere for 1. The UV-visible absorption spectrum of the ligand HL1 and compound Figure S4 1 dissolved in DMF, respectively. Fluorescence spectra of ligand HL1 (a) and compound 1 (b) dissolved Figure S5 in DMF, respectively. The complex 1 and different metal ions were dissolved in an Figure S6 ultraviolet-visible absorption test in DMF. Crystals 2019, 9, 374; doi: 10.3390/cryst9070374 www.mdpi.com/journal/crystals Crystals 2019, 9, 374 2 of 7 Table S1. Selected bond lengths (Å) and angles (°) of 1. 1 Zn5—O15 1.914 (5) Zn2—N4 1.961 (6) Zn6—N12 2.001 (6) Zn5—O7 1.946 (6) Zn2—O4 2.047 (5) Zn6—N11 2.316 (7) Zn5—O10 2.190 (6) Zn2—N3 2.420 (6) Zn1—O3 1.924 (6) Zn5—N9 2.400 (7) Zn4—O8 2.196 (6) Zn1—O8 1.938
    [Show full text]
  • Supporting Information
    Electronic Supplementary Material (ESI) for CrystEngComm. This journal is © The Royal Society of Chemistry 2020 Supporting Information Table S1. Analysis results of Continuous Shape Measure (CSM) for Ln1 in cpdc-Ln CSM parameters * Ln1 JSD-8 BTPR-8 JBTPR-8 JETBPY- JGBF-8 TDD-8 SAPR- CU-8 HBPY- geometry 8 8 8 Eu1 9.453 6.003 6.644 27.526 20.710 5.891 5.282 13.610 20.003 Tb1 9.481 5.931 6.679 27.383 20.647 5.837 5.256 13.577 20.001 SAPR-8 Dy1 9.321 5.975 6.534 26.808 20.412 5.841 5.179 13.514 20.023 * HBPY-8→ Hexagonal bipyramid (D6h symmetry) CU-8→ Cube (Oh symmetry) SAPR-8→ Square antiprism (D4d symmetry) TDD-8→ Triangular dodecahedron (D2d symmetry) JGBF-8→ Johnson - Gyrobifastigium (J26) (D2d symmetry) JETBPY-8→ Johnson - Elongated triangular bipyramid (J14) (D3h symmetry) JBTP-8→ Johnson - Biaugmented trigonal prism (J50) (C2v symmetry) BTPR-8→ Biaugmented trigonal prism (C2v symmetry) JSD-8→ Snub disphenoid (J84) (D2d symmetry) Table S2. Analysis results of Continuous Shape Measure (CSM) for Ln2 in cpdc- Ln CSM parameters * Ln2 JTDIC- TCTPR- JTCTPR- CSAPR- JCSAPR- CCU-9 JCCU-9 JTC-9 HBPY- geometry 9 9 9 9 9 9 Eu2 17.653 18.362 17.250 17.382 15.938 10.071 8.314 18.552 18.091 Tb2 18.055 18.685 17.314 17.693 16.135 10.357 8.469 18.937 18.196 JCCU-9 Dy2 18.602 18.713 17.193 17.730 16.131 10.403 8.465 19.327 18.014 * HBPY-9→ Heptagonal bipyramid (D7h symmetry) JTC-9→ Triangular cupola (J3) = trivacant cuboctahedron (C3v symmetry) JCCU-9→ Capped cube (Elongated square pyramid, J8) (C4v symmetry) CCU-9→ Capped cube (C4v symmetry) JCSAPR-9→ Capped sq.
    [Show full text]
  • Surface Areas of Pyramids
    8.3 Surface Areas of Pyramids How can you use a net to fi nd the surface area of a pyramid? 1 ACTIVITY: Identifying Pyramids Work with a partner. Label one of the faces as a “base” and the other as a “lateral face.” Use the shape of the base to identify the pyramid. a. b. Pyramid Pyramid 2 ACTIVITY: Using a Net Work with a partner. a. Copy the net shown below onto grid paper. b. Cut out the net and fold it to form a pyramid. What type of rectangle is the base? Use this shape to name the pyramid. c. Find the surface area of the pyramid.pyramid. COMMON CORE Geometry In this lesson, you will ● use nets to represent pyramids. ● fi nd the surface area of pyramids. ● solve real-life problems. Learning Standard 6.G.4 368 Chapter 8 Surface Area and Volume 3 ACTIVITY: Estimating the Surface Area of a Triangular Pyramid Work with a partner. Label each face in the net of the triangular pyramid as a “base” or a “lateral face.” Then estimate the surface area of the pyramid. Math Practice Analyze Givens What information is given in the diagram? How does this help you estimate the surface area of the pyramid? 6 6 4 ACTIVITY: Finding the Surface Area of a Square Pyramid Work with a partner. Draw a net for each square pyramid. Use the net to fi nd the surface area of the pyramid. a. b. 3 5 3 5 5. IN YOUR OWN WORDS How can you use a net to fi nd the surface area of a pyramid? 6.
    [Show full text]
  • The Octahedron from Spheres
    CPS Geometry Part 2 – Platonic Solids 8. Introduction - Platonic Solids And CPS 9. The Tetrahedron in CPS 10. The Octahedron in CPS 11. The Cube in CPS 12. The Icosahedron in CPS 13. The Dodecahedron in CPS Nick Trif Ottawa, Ontario, Canada – 2018 www.platonicstructures.com CPS Geometry Part 2 – Platonic Solids – 10: The Octahedron from Spheres YouTube: https://youtu.be/s6hFZzlIADc The octahedron pattern is again relatively easy to spot in the Close Packing of Sphere arrangement. The octahedron pattern is again relatively easy to spot in the Close Packing of Sphere arrangement. Starting from a square pattern, a square pyramid can be constructed in a very direct way - just stuck smaller and smaller square planes on top of each other, until you get to the top. 2 Layers Starting from a square pattern, a square pyramid can be constructed in a very direct way - just stuck smaller and smaller square planes on top of each other, until you get to the top. 3 Layers Starting from a square pattern, a square pyramid can be constructed in a very direct way - just stuck smaller and smaller square planes on top of each other, until you get to the top. 5 Layers Starting from a square pattern, a square pyramid can be constructed in a very direct way - just stuck smaller and smaller square planes on top of each other, until you get to the top. 6 Layers Starting from a square pattern, a square pyramid can be constructed in a very direct way - just stuck smaller and smaller square planes on top of each other, until you get to the top.
    [Show full text]
  • Lateral and Surface Area of Right Pyramids You Will Need C GOAL • a Ruler Calculate Lateral Area and Surface Area of Right Pyramids
    CHAPTER 11 B Lateral and Surface Area of Right Pyramids You will need c GOAL • a ruler Calculate lateral area and surface area of right pyramids. • a calculator Learn about the Math A pyramid is a polyhedron (a solid whose faces are polygons) with one polygonal base. A right pyramid is a pyramid with right pyramid a base that is a regular polygon and whose apex is directly pyramid with a base that is a regular above the centre of the base. polygon and whose apex is directly above the centre of the base apex the vertex of a pyramid at which the triangular faces meet The surface area of a right pyramid can be calculated, using 1 the following formula: SA 5 B 1 Ps , where B is the area Q2 R of the base, P is the perimeter of the base, and s is the slant height. The slant height is the distance from the centre slant height of a side of the base to the apex. the distance from the centre of a side of the The lateral area of a right pyramid can be calculated by base of a right pyramid to the apex multiplying half of the perimeter of the base by the slant 1 height. This is summarized by the formula: LA Ps. 5 2 We can relate this formula to the square pyramid below and its net. The side length of the base of the pyramid is b, and the slant height is s. s b s s bbs b s b Copyright © 2009 by Nelson Education Ltd.
    [Show full text]
  • Linear Trigonal Planar Tetrahedral Trigonal Bipyramid Octahedral Bent
    There are only FIVE possible electronic geometries which you establish by counting the number of electron regions surrounding the central atom electron electron electron electron electron 2regions 3regions 4regions 5regions 6regions AX2 AX3 AX4 AX5 AX6 3 sp sp2 sp sp3d sp3d2 linear trigonal planar tetrahedral trigonal bipyramid octahedral note that AX E AX4E the lone 5 AX E .. .. pairs all AX2E .. 3 go in the equatorial position occupied positions .. 1 by a lone pair bent trigonal pyramid see-saw square pyramid Molecular Geometries AX E can be any of the shapes on .. 4 2 AX2E2 .. .. .. the whole page. The .. AX E electronic geometries are only positions occupied 3 2 .. those in the box (and orbital 2 by a lone pair hybridizations). The molecular bent T-shaped square planar geometry will be different from the electronic when there is at least one or more lone pairs AX3E3 on the central atom. Look at the top of the table and go .. DOWN a column. As you change from bonding electrons .. AX E .. positions occupied .. 2 3 .. to lone pair electrons, the molecular shape is now different by a lone pair from the electronic because some of the positions are missing 3 .. atoms. The new shape is then renamed based on the shape of the atoms. linear T-shaped Remember, once you have estabilished the correct electronic geometry, the molecular .... AX2E4 geometry MUST be either the same as the electronic or one of the shapes listed directly positions occupied .. under the electronic geometry. In other words, each shape in a given column here has the 4 by a lone pair .
    [Show full text]
  • Square Pyramid
    THE TWO DOZEN RIGHT TRIANGLES OF A SQUARE PYRAMID NEAL BERNTSON Empire State College . CLASSROOM DAY TOPIC OBJECTIVE ITEMS 1 Division of a Manipulating Card Stock Square into 8 a square base Right Triangles of a pyramid Ruler yields several * * * * * triangles if a Scissors half-division Hypotenuse x 2 technique is Color Markers = Diagonal performed. 2 Division of a Comprehend Scotch Tape Pyramid's 4 bisecting the sides into 8 sides of any Card Stock Right Triangles pyramid will produce two Ruler * * * * * times the number of Scissors "Slant Height" triangles and Altitudes all of them Color Markers will be right. 3 Design of a Understand Transparencies Vertical Right how a see- Triangle inside through Card cut in Pyramid fitting pyramid will Right Triangle into the Corner. demonstrate how a new Scotch Tape * * * * * interior right triangle is to Ruler "Lateral Edge" be called the Hypotenuse "CORNER - Scissors SUPPORT" 4 Design of a Understand Transparencies Vertical Right how a see- Triangle inside through Card cut in Pyramid fitting pyramid will Right Triangle onto the Side. demonstrate how a new Scotch Tape * * * * * interior right triangle is to Ruler "Slant Height" be called the Hypotenuse "SIDE - Scissors SUPPORT" CLASSROOM DAY TOPIC OBJECTIVE ITEMS 5 Finding the Demonstrate Calculator Mystery most efficient Segment use of the Pyramid Pythagorean Constructions * * * * * Theorem to find length of Ruler for Most Direct any of the Estimations Strategy pyramid line Identification segments. White-Board First Lesson: "DIVIDING A SQUARE"
    [Show full text]
  • Chapter 17 Mensuration of Pyramid
    344 Applied Math Mensuration of Pyramid Chapter 17 Mensuration of Pyramid 17.1 Pyramid: A pyramid is a solid whose base is a plane polygon and sides are triangles that meet in a common vertex. The triangular sides are called lateral faces. The common vertex is also called Apex. A pyramid is named according to the shape of its base. If the base is a triangle, square, hexagon etc. the pyramid is called as a triangular pyramid, a square pyramid, a hexagonal pyramid etc. respectively. Altitude (or height): The altitude of a pyramid is the perpendicular distance from the vertex to the base. Axis: The axis of a pyramid is the distance from the vertex to the centre of the base. 17.2 Right or Regular Pyramid: A pyramid whose base is a regular polygon and congruent isosceles triangles as lateral faces. In a regular pyramid the axis is perpendicular to the base. Thus in a regular pyramid th axis and the altitude are identical. Slant Height: The slant height of a regular pyramid is the length of the median through the apex of any lateral face. In the Fig 17.1 OG is the slant height. It is denoted by . Fig. 17.1 Fig. 17.1 Lateral edge: It is the common side where the two, faces meet. In the OA is the lateral edge. 17.3 The surface area and Volume of a Regular Pyramid: If a is the side of the polygon base, h is the height and is the slant height of a regular pyramid, then (i) Lateral surface area = Sum of the triangular sides forming the pyramid, which are all equal in areas = n (area of one triangle of base a and slant height ) 345 Applied Math Mensuration of Pyramid 1 = n a 2 1 = (n a) x 2 1 = Perimeter of the base x slant height 2 (ii) Total surface area = Lateral surface area + area of the base (iii) Volume: The volume of a pyramid may be easily derived from the volume of a cube.
    [Show full text]
  • A Square Pyramid 50 Mm Base Side and Axis 90 Mm Long Is Resting on HP at Its Base with a Side of Base Parallel to VP
    TYPES OF SECTION PLANES V T Y X Y X No HT for this CP SECTION PLANE / CUTTING PLANE Parallel to HP & Perpendicular to VP TYPES OF SECTION PLANES No VT for this CP Y X Y X H T SECTION PLANE / CUTTING PLANE Perpendicular to HP & Parallel to VP TYPES OF SECTION PLANES V θ(cp) Y X ┐ Y T φ (cp) = 900 θ X H SECTION PLANE / CUTTING PLANE Inclined to HP & Perpendicular to VP TYPES OF SECTION PLANES V φ ┐θ (cp) = 900 Y X Y Cutting plane T φ(cp) X H SECTION PLANE / CUTTING PLANE Perpendicular to HP & Inclined to VP V X Y T Y X H SECTION PLANE / CUTTING PLANE Perpendicular to HP & Perpendicular to VP φ Cutting plane.Y Y Inclined to VP X X Y Y Y X X X Sections of solids – Q1 A square pyramid 50 mm base side and axis 90 mm long is resting on HP at its base with a side of base parallel to VP. The pyramid is cut by a section plane parallel to HP and perpendicular to VP, bisecting the axis. Draw the sectional views and the true shape of the section. Sections of solids – Q1 Sections of solids – Q1 o’ FV of the Section V 3’,4’ 1’, 2’ T CP Ʇ to VP → Sectional FV is a straight line. X c’,d’ a’,b’ Y CP ‖ to HP → c b Sectional TV is a closed figure & 3 2 is the True shape of the section o 4 1 TV of the Section & True shape of the d a section Sections of solids – Q2 A square pyramid 50 mm base side and axis 90 mm long is resting on HP at its base with a side of base parallel to VP.
    [Show full text]
  • Convex Polyhedra with Regular Faces
    CONVEX POLYHEDRA WITH REGULAR FACES NORMAN W. JOHNSON 1. Introduction. An interesting set of geometric figures is composed of the convex polyhedra in Euclidean 3-space whose faces are regular polygons (not necessarily all of the same kind). A polyhedron with regular faces is uniform if it has symmetry operations taking a given vertex into each of the other vertices in turn (5, p. 402). If in addition all the faces are alike, the polyhedron is regular. That there are just five convex regular polyhedra—the so-called Platonic solids—was proved by Euclid in the thirteenth book of the Elements (10, pp. 467-509). Archimedes is supposed to have described thirteen other uniform, "semi-regular" polyhedra, but his work on the subject has been lost. Kepler (12, pp. 114-127) showed that the convex uniform polyhedra consist of the Platonic and Archimedean solids together with two infinite families—the regular prisms and antiprisms. It was Kepler also who gave the Archimedean polyhedra their generally accepted names. It is fairly easy to show that there are only a finite number of non-uniform regular-faced polyhedra (11; 13), but it is no simple matter to establish the exact number. However, it appears that there are just ninety-two such solids. Some special cases were discussed by Freudenthal and van der Waerden (8), and a more general treatment was attempted by Zalgaller (13). Subsequently, Zalgaller et al. (14) determined all the regular-faced polyhedra having one or more trivalent vertices and all those having only pentavalent vertices. Griin- baum and Johnson (9) proved that the only kinds of faces that a regular-faced solid, other than a prism or an antiprism, may have are triangles, squares, pentagons, hexagons, octagons, and decagons.
    [Show full text]