Determine Whether the Solid Is a Polyhedron. Then Identify the Solid
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Arc Length. Surface Area
Calculus 2 Lia Vas Arc Length. Surface Area. Arc Length. Suppose that y = f(x) is a continuous function with a continuous derivative on [a; b]: The arc length L of f(x) for a ≤ x ≤ b can be obtained by integrating the length element ds from a to b: The length element ds on a sufficiently small interval can be approximated by the hypotenuse of a triangle with sides dx and dy: p Thus ds2 = dx2 + dy2 ) ds = dx2 + dy2 and so s s Z b Z b q Z b dy2 Z b dy2 L = ds = dx2 + dy2 = (1 + )dx2 = (1 + )dx: a a a dx2 a dx2 2 dy2 dy 0 2 Note that dx2 = dx = (y ) : So the formula for the arc length becomes Z b q L = 1 + (y0)2 dx: a Area of a surface of revolution. Suppose that y = f(x) is a continuous function with a continuous derivative on [a; b]: To compute the surface area Sx of the surface obtained by rotating f(x) about x-axis on [a; b]; we can integrate the surface area element dS which can be approximated as the product of the circumference 2πy of the circle with radius y and the height that is given by q the arc length element ds: Since ds is 1 + (y0)2dx; the formula that computes the surface area is Z b q 0 2 Sx = 2πy 1 + (y ) dx: a If y = f(x) is rotated about y-axis on [a; b]; then dS is the product of the circumference 2πx of the circle with radius x and the height that is given by the arc length element ds: Thus, the formula that computes the surface area is Z b q 0 2 Sy = 2πx 1 + (y ) dx: a Practice Problems. -
Volumes of Prisms and Cylinders 625
11-4 11-4 Volumes of Prisms and 11-4 Cylinders 1. Plan Objectives What You’ll Learn Check Skills You’ll Need GO for Help Lessons 1-9 and 10-1 1 To find the volume of a prism 2 To find the volume of • To find the volume of a Find the area of each figure. For answers that are not whole numbers, round to prism a cylinder the nearest tenth. • To find the volume of a 2 Examples cylinder 1. a square with side length 7 cm 49 cm 1 Finding Volume of a 2. a circle with diameter 15 in. 176.7 in.2 . And Why Rectangular Prism 3. a circle with radius 10 mm 314.2 mm2 2 Finding Volume of a To estimate the volume of a 4. a rectangle with length 3 ft and width 1 ft 3 ft2 Triangular Prism backpack, as in Example 4 2 3 Finding Volume of a Cylinder 5. a rectangle with base 14 in. and height 11 in. 154 in. 4 Finding Volume of a 6. a triangle with base 11 cm and height 5 cm 27.5 cm2 Composite Figure 7. an equilateral triangle that is 8 in. on each side 27.7 in.2 New Vocabulary • volume • composite space figure Math Background Integral calculus considers the area under a curve, which leads to computation of volumes of 1 Finding Volume of a Prism solids of revolution. Cavalieri’s Principle is a forerunner of ideas formalized by Newton and Leibniz in calculus. Hands-On Activity: Finding Volume Explore the volume of a prism with unit cubes. -
Build a Tetrahedral Kite
Aeronautics Research Mission Directorate Build a Tetrahedral Kite Suggested Grades: 8-12 Activity Overview Time: 90-120 minutes In this activity, you will build a tetrahedral kite from Materials household supplies. • 24 straws (8 inches or less) - NOTE: The straws need to be Steps straight and the same length. If only flexible straws are available, 1. Cut a length of yarn/string 4 feet long. then cut off the flexible portion. • Two or three large spools of 2. Take six straws and place them on a flat surface. cotton string or yarn (approximately 100 feet total) 3. Use your piece of string to join three straws • Scissors together in a triangular shape. On the side where • Hot glue gun and hot glue sticks the two strings are extending from it, one end • Ruler or dowel rod for kite bridle should be approximately 20 inches long, and the • Four pieces of tissue paper (24 x other should be approximately 4 inches long. 18 inches or larger) See Figure 1. • All-purpose glue stick Figure 1 4. Tie these two ends of the string tightly together. Make sure there is no room for the triangle to wiggle. 5. The three straws should form a tight triangle. 6. Cut another 4-inch piece of string. 7. Take one end of the 4-inch string, and tie that end to a corner of the triangle that does not have the string ends extending from it. Figure 2. 8. Add two more straws onto the longest piece of string. 9. Next, take the string that holds the two additional straws and tie it to the end of one of the 4-inch strings to make another tight triangle. -
THE GEOMETRY of PYRAMIDS One of the More Interesting Solid
THE GEOMETRY OF PYRAMIDS One of the more interesting solid structures which has fascinated individuals for thousands of years going all the way back to the ancient Egyptians is the pyramid. It is a structure in which one takes a closed curve in the x-y plane and connects straight lines between every point on this curve and a fixed point P above the centroid of the curve. Classical pyramids such as the structures at Giza have square bases and lateral sides close in form to equilateral triangles. When the closed curve becomes a circle one obtains a cone and this cone becomes a cylindrical rod when point P is moved to infinity. It is our purpose here to discuss the properties of all N sided pyramids including their volume and surface area using only elementary calculus and geometry. Our starting point will be the following sketch- The base represents a regular N sided polygon with side length ‘a’ . The angle between neighboring radial lines r (shown in red) connecting the polygon vertices with its centroid is θ=2π/N. From this it follows, by the law of cosines, that the length r=a/sqrt[2(1- cos(θ))] . The area of the iscosolis triangle of sides r-a-r is- a a 2 a 2 1 cos( ) A r 2 T 2 4 4 (1 cos( ) From this we have that the area of the N sided polygon and hence the pyramid base will be- 2 2 1 cos( ) Na A N base 2 4 1 cos( ) N 2 It readily follows from this result that a square base N=4 has area Abase=a and a hexagon 2 base N=6 yields Abase= 3sqrt(3)a /2. -
Cones, Pyramids and Spheres
The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 12 CONES, PYRAMIDS AND SPHERES A guide for teachers - Years 9–10 June 2011 YEARS 910 Cones, Pyramids and Spheres (Measurement and Geometry : Module 12) For teachers of Primary and Secondary Mathematics 510 Cover design, Layout design and Typesetting by Claire Ho The Improving Mathematics Education in Schools (TIMES) Project 2009‑2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations. The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations. © The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE‑EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution‑NonCommercial‑NoDerivs 3.0 Unported License. 2011. http://creativecommons.org/licenses/by‑nc‑nd/3.0/ The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 12 CONES, PYRAMIDS AND SPHERES A guide for teachers - Years 9–10 June 2011 Peter Brown Michael Evans David Hunt Janine McIntosh Bill Pender Jacqui Ramagge YEARS 910 {4} A guide for teachers CONES, PYRAMIDS AND SPHERES ASSUMED KNOWLEDGE • Familiarity with calculating the areas of the standard plane figures including circles. • Familiarity with calculating the volume of a prism and a cylinder. • Familiarity with calculating the surface area of a prism. • Facility with visualizing and sketching simple three‑dimensional shapes. • Facility with using Pythagoras’ theorem. • Facility with rounding numbers to a given number of decimal places or significant figures. -
Lateral and Surface Area of Right Prisms 1 Jorge Is Trying to Wrap a Present That Is in a Box Shaped As a Right Prism
CHAPTER 11 You will need Lateral and Surface • a ruler A • a calculator Area of Right Prisms c GOAL Calculate lateral area and surface area of right prisms. Learn about the Math A prism is a polyhedron (solid whose faces are polygons) whose bases are congruent and parallel. When trying to identify a right prism, ask yourself if this solid could have right prism been created by placing many congruent sheets of paper on prism that has bases aligned one above the top of each other. If so, this is a right prism. Some examples other and has lateral of right prisms are shown below. faces that are rectangles Triangular prism Rectangular prism The surface area of a right prism can be calculated using the following formula: SA 5 2B 1 hP, where B is the area of the base, h is the height of the prism, and P is the perimeter of the base. The lateral area of a figure is the area of the non-base faces lateral area only. When a prism has its bases facing up and down, the area of the non-base faces of a figure lateral area is the area of the vertical faces. (For a rectangular prism, any pair of opposite faces can be bases.) The lateral area of a right prism can be calculated by multiplying the perimeter of the base by the height of the prism. This is summarized by the formula: LA 5 hP. Copyright © 2009 by Nelson Education Ltd. Reproduction permitted for classrooms 11A Lateral and Surface Area of Right Prisms 1 Jorge is trying to wrap a present that is in a box shaped as a right prism. -
Name Date Period 1
NAME DATE PERIOD Unit 1, Lesson 13: Polyhedra Let’s investigate polyhedra. 13.1: What are Polyhedra? Here are pictures that represent polyhedra: Here are pictures that do not represent polyhedra: 1. Your teacher will give you some figures or objects. Sort them into polyhedra and non- polyhedra. 2. What features helped you distinguish the polyhedra from the other figures? 13.2: Prisms and Pyramids 1. Here are some polyhedra called prisms. 1 NAME DATE PERIOD Here are some polyhedra called pyramids. a. Look at the prisms. What are their characteristics or features? b. Look at the pyramids. What are their characteristics or features? 2 NAME DATE PERIOD 2. Which of the following nets can be folded into Pyramid P? Select all that apply. 3. Your teacher will give your group a set of polygons and assign a polyhedron. a. Decide which polygons are needed to compose your assigned polyhedron. List the polygons and how many of each are needed. b. Arrange the cut-outs into a net that, if taped and folded, can be assembled into the polyhedron. Sketch the net. If possible, find more than one way to arrange the polygons (show a different net for the same polyhedron). Are you ready for more? What is the smallest number of faces a polyhedron can possibly have? Explain how you know. 13.3: Assembling Polyhedra 1. Your teacher will give you the net of a polyhedron. Cut out the net, and fold it along the edges to assemble a polyhedron. Tape or glue the flaps so that there are no unjoined edges. -
Systematics of Atomic Orbital Hybridization of Coordination Polyhedra: Role of F Orbitals
molecules Article Systematics of Atomic Orbital Hybridization of Coordination Polyhedra: Role of f Orbitals R. Bruce King Department of Chemistry, University of Georgia, Athens, GA 30602, USA; [email protected] Academic Editor: Vito Lippolis Received: 4 June 2020; Accepted: 29 June 2020; Published: 8 July 2020 Abstract: The combination of atomic orbitals to form hybrid orbitals of special symmetries can be related to the individual orbital polynomials. Using this approach, 8-orbital cubic hybridization can be shown to be sp3d3f requiring an f orbital, and 12-orbital hexagonal prismatic hybridization can be shown to be sp3d5f2g requiring a g orbital. The twists to convert a cube to a square antiprism and a hexagonal prism to a hexagonal antiprism eliminate the need for the highest nodality orbitals in the resulting hybrids. A trigonal twist of an Oh octahedron into a D3h trigonal prism can involve a gradual change of the pair of d orbitals in the corresponding sp3d2 hybrids. A similar trigonal twist of an Oh cuboctahedron into a D3h anticuboctahedron can likewise involve a gradual change in the three f orbitals in the corresponding sp3d5f3 hybrids. Keywords: coordination polyhedra; hybridization; atomic orbitals; f-block elements 1. Introduction In a series of papers in the 1990s, the author focused on the most favorable coordination polyhedra for sp3dn hybrids, such as those found in transition metal complexes. Such studies included an investigation of distortions from ideal symmetries in relatively symmetrical systems with molecular orbital degeneracies [1] In the ensuing quarter century, interest in actinide chemistry has generated an increasing interest in the involvement of f orbitals in coordination chemistry [2–7]. -
Math 366 Lecture Notes Section 11.4 – Geometry in Three Dimensions
Section 11-4 Math 366 Lecture Notes Section 11.4 – Geometry in Three Dimensions Simple Closed Surfaces A simple closed surface has exactly one interior, no holes, and is hollow. A sphere is the set of all points at a given distance from a given point, the center . A sphere is a simple closed surface. A solid is a simple closed surface with all interior points. (see p. 726) A polyhedron is a simple closed surface made up of polygonal regions, or faces . The vertices of the polygonal regions are the vertices of the polyhedron, and the sides of each polygonal region are the edges of the polyhedron. (see p. 726-727) A prism is a polyhedron in which two congruent faces lie in parallel planes and the other faces are bounded by parallelograms. The parallel faces of a prism are the bases of the prism. A prism is usually names after its bases. The faces other than the bases are the lateral faces of a prism. A right prism is one in which the lateral faces are all bounded by rectangles. An oblique prism is one in which some of the lateral faces are not bounded by rectangles. To draw a prism: 1) Draw one of the bases. 2) Draw vertical segments of equal length from each vertex. 3) Connect the bottom endpoints to form the second base. Use dashed segments for edges that cannot be seen. A pyramid is a polyhedron determined by a polygon and a point not in the plane of the polygon. The pyramid consists of the triangular regions determined by the point and each pair of consecutive vertices of the polygon and the polygonal region determined by the polygon. -
VOLUME of POLYHEDRA USING a TETRAHEDRON BREAKUP We
VOLUME OF POLYHEDRA USING A TETRAHEDRON BREAKUP We have shown in an earlier note that any two dimensional polygon of N sides may be broken up into N-2 triangles T by drawing N-3 lines L connecting every second vertex. Thus the irregular pentagon shown has N=5,T=3, and L=2- With this information, one is at once led to the question-“ How can the volume of any polyhedron in 3D be determined using a set of smaller 3D volume elements”. These smaller 3D eelements are likely to be tetrahedra . This leads one to the conjecture that – A polyhedron with more four faces can have its volume represented by the sum of a certain number of sub-tetrahedra. The volume of any tetrahedron is given by the scalar triple product |V1xV2∙V3|/6, where the three Vs are vector representations of the three edges of the tetrahedron emanating from the same vertex. Here is a picture of one of these tetrahedra- Note that the base area of such a tetrahedron is given by |V1xV2]/2. When this area is multiplied by 1/3 of the height related to the third vector one finds the volume of any tetrahedron given by- x1 y1 z1 (V1xV2 ) V3 Abs Vol = x y z 6 6 2 2 2 x3 y3 z3 , where x,y, and z are the vector components. The next question which arises is how many tetrahedra are required to completely fill a polyhedron? We can arrive at an answer by looking at several different examples. Starting with one of the simplest examples consider the double-tetrahedron shown- It is clear that the entire volume can be generated by two equal volume tetrahedra whose vertexes are placed at [0,0,sqrt(2/3)] and [0,0,-sqrt(2/3)]. -
10-4 Surface and Lateral Area of Prism and Cylinders.Pdf
Aim: How do we evaluate and solve problems involving surface area of prisms an cylinders? Objective: Learn and apply the formula for the surface area of a prism. Learn and apply the formula for the surface area of a cylinder. Vocabulary: lateral face, lateral edge, right prism, oblique prism, altitude, surface area, lateral surface, axis of a cylinder, right cylinder, oblique cylinder. Prisms and cylinders have 2 congruent parallel bases. A lateral face is not a base. The edges of the base are called base edges. A lateral edge is not an edge of a base. The lateral faces of a right prism are all rectangles. An oblique prism has at least one nonrectangular lateral face. An altitude of a prism or cylinder is a perpendicular segment joining the planes of the bases. The height of a three-dimensional figure is the length of an altitude. Surface area is the total area of all faces and curved surfaces of a three-dimensional figure. The lateral area of a prism is the sum of the areas of the lateral faces. Holt Geometry The net of a right prism can be drawn so that the lateral faces form a rectangle with the same height as the prism. The base of the rectangle is equal to the perimeter of the base of the prism. The surface area of a right rectangular prism with length ℓ, width w, and height h can be written as S = 2ℓw + 2wh + 2ℓh. The surface area formula is only true for right prisms. To find the surface area of an oblique prism, add the areas of the faces. -
Year 6 – Wednesday 24Th June 2020 – Maths
1 Year 6 – Wednesday 24th June 2020 – Maths Can I identify 3D shapes that have pairs of parallel or perpendicular edges? Parallel – edges that have the same distance continuously between them – parallel edges never meet. Perpendicular – when two edges or faces meet and create a 90o angle. 1 4. 7. 10. 2 5. 8. 11. 3. 6. 9. 12. C1 – Using the shapes above: 1. Name and sort the shapes into: a. Pyramids b. Prisms 2. Draw a table to identify how many faces, edges and vertices each shape has. 3. Write your own geometric definition: a. Prism b. pyramid 4. Which shape is the odd one out? Explain why. C2 – Using the shapes above; 1. Which of the shapes have pairs of parallel edges in: a. All their faces? b. More than one half of the faces? c. One face only? 2. The following shapes have pairs of perpendicular edges. Identify the faces they are in: 2 a. A cube b. A square based pyramid c. A triangular prism d. A cuboid 3. Which shape with straight edges has no perpendicular edges? 4. Which shape has perpendicular edges in the shape but not in any face? C3 – Using the above shapes: 1. How many faces have pairs of parallel edges in: a. A hexagonal pyramid? b. A decagonal (10-sided) based prism? c. A heptagonal based prism? d. Which shape has no face with parallel edges but has parallel edges in the shape? 2. How many faces have perpendicular edges in: a. A pentagonal pyramid b. A hexagonal pyramid c.