The Pythagorean Theorem Crown Jewel of Mathematics
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Gpindex: Generalized Price and Quantity Indexes
Package ‘gpindex’ August 4, 2021 Title Generalized Price and Quantity Indexes Version 0.3.4 Description A small package for calculating lots of different price indexes, and by extension quan- tity indexes. Provides tools to build and work with any type of bilateral generalized-mean in- dex (of which most price indexes are), along with a few important indexes that don't be- long to the generalized-mean family. Implements and extends many of the meth- ods in Balk (2008, ISBN:978-1-107-40496-0) and ILO, IMF, OECD, Euro- stat, UN, and World Bank (2020, ISBN:978-1-51354-298-0) for bilateral price indexes. Depends R (>= 3.5) Imports stats, utils License MIT + file LICENSE Encoding UTF-8 URL https://github.com/marberts/gpindex LazyData true Collate 'helper_functions.R' 'means.R' 'weights.R' 'price_indexes.R' 'operators.R' 'utilities.R' NeedsCompilation no Author Steve Martin [aut, cre, cph] Maintainer Steve Martin <[email protected]> Repository CRAN Date/Publication 2021-08-04 06:10:06 UTC R topics documented: gpindex-package . .2 contributions . .3 generalized_mean . .8 lehmer_mean . 12 logarithmic_means . 15 nested_mean . 18 offset_prices . 20 1 2 gpindex-package operators . 22 outliers . 23 price_data . 25 price_index . 26 transform_weights . 33 Index 37 gpindex-package Generalized Price and Quantity Indexes Description A small package for calculating lots of different price indexes, and by extension quantity indexes. Provides tools to build and work with any type of bilateral generalized-mean index (of which most price indexes are), along with a few important indexes that don’t belong to the generalized-mean family. Implements and extends many of the methods in Balk (2008, ISBN:978-1-107-40496-0) and ILO, IMF, OECD, Eurostat, UN, and World Bank (2020, ISBN:978-1-51354-298-0) for bilateral price indexes. -
Centroids by Composite Areas.Pptx
Centroids Method of Composite Areas A small boy swallowed some coins and was taken to a hospital. When his grandmother telephoned to ask how he was a nurse said 'No change yet'. Centroids ¢ Previously, we developed a general formulation for finding the centroid for a series of n areas n xA ∑ ii i=1 x = n A ∑ i i=1 2 Centroids by Composite Areas Monday, November 12, 2012 1 Centroids ¢ xi was the distance from the y-axis to the local centroid of the area Ai n xA ∑ ii i=1 x = n A ∑ i i=1 3 Centroids by Composite Areas Monday, November 12, 2012 Centroids ¢ If we can break up a shape into a series of smaller shapes that have predefined local centroid locations, we can use this formula to locate the centroid of the composite shape n xA ∑ ii i=1 x = n A ∑ i 4 Centroids by Composite Areas i=1 Monday, November 12, 2012 2 Centroid by Composite Bodies ¢ There is a table in the back cover of your book that gives you the location of local centroids for a select group of shapes ¢ The point labeled C is the location of the centroid of that shape. 5 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies ¢ Please note that these are local centroids, they are given in reference to the x and y axes as shown in the table. 6 Centroids by Composite Areas Monday, November 12, 2012 3 Centroid by Composite Bodies ¢ For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ¢ The x-centroid would be located at 0 and the y-centroid would be located -
AHM As a Measure of Central Tendency of Sex Ratio
Biometrics & Biostatistics International Journal Research Article Open Access AHM as a measure of central tendency of sex ratio Abstract Volume 10 Issue 2 - 2021 In some recent studies, four formulations of average namely Arithmetic-Geometric Mean Dhritikesh Chakrabarty (abbreviated as AGM), Arithmetic-Harmonic Mean (abbreviated as AHM), Geometric- Department of Statistics, Handique Girls’ College, Gauhati Harmonic Mean (abbreviated as GHM) and Arithmetic-Geometric-Harmonic Mean University, India (abbreviated as AGHM) have recently been derived from the three Pythagorean means namely Arithmetic Mean (AM), Geometric Mean (GM) and Harmonic Mean (HM). Each Correspondence: Dhritikesh Chakrabarty, Department of of these four formulations has been found to be a measure of central tendency of data. in Statistics, Handique Girls’ College, Gauhati University, India, addition to the existing measures of central tendency namely AM, GM & HM. This paper Email focuses on the suitability of AHM as a measure of central tendency of numerical data of ratio type along with the evaluation of central tendency of sex ratio namely male-female ratio and female-male ratio of the states in India. Received: May 03, 2021 | Published: May 25, 2021 Keywords: AHM, sex ratio, central tendency, measure Introduction however, involve huge computational tasks. Moreover, these methods may not be able to yield the appropriate value of the parameter if Several research had already been done on developing definitions/ observed data used are of relatively small size (and/or of moderately 1,2 formulations of average, a basic concept used in developing most large size too) In reality, of course, the appropriate value of the 3 of the measures used in analysis of data. -
Geometry Course Outline
GEOMETRY COURSE OUTLINE Content Area Formative Assessment # of Lessons Days G0 INTRO AND CONSTRUCTION 12 G-CO Congruence 12, 13 G1 BASIC DEFINITIONS AND RIGID MOTION Representing and 20 G-CO Congruence 1, 2, 3, 4, 5, 6, 7, 8 Combining Transformations Analyzing Congruency Proofs G2 GEOMETRIC RELATIONSHIPS AND PROPERTIES Evaluating Statements 15 G-CO Congruence 9, 10, 11 About Length and Area G-C Circles 3 Inscribing and Circumscribing Right Triangles G3 SIMILARITY Geometry Problems: 20 G-SRT Similarity, Right Triangles, and Trigonometry 1, 2, 3, Circles and Triangles 4, 5 Proofs of the Pythagorean Theorem M1 GEOMETRIC MODELING 1 Solving Geometry 7 G-MG Modeling with Geometry 1, 2, 3 Problems: Floodlights G4 COORDINATE GEOMETRY Finding Equations of 15 G-GPE Expressing Geometric Properties with Equations 4, 5, Parallel and 6, 7 Perpendicular Lines G5 CIRCLES AND CONICS Equations of Circles 1 15 G-C Circles 1, 2, 5 Equations of Circles 2 G-GPE Expressing Geometric Properties with Equations 1, 2 Sectors of Circles G6 GEOMETRIC MEASUREMENTS AND DIMENSIONS Evaluating Statements 15 G-GMD 1, 3, 4 About Enlargements (2D & 3D) 2D Representations of 3D Objects G7 TRIONOMETRIC RATIOS Calculating Volumes of 15 G-SRT Similarity, Right Triangles, and Trigonometry 6, 7, 8 Compound Objects M2 GEOMETRIC MODELING 2 Modeling: Rolling Cups 10 G-MG Modeling with Geometry 1, 2, 3 TOTAL: 144 HIGH SCHOOL OVERVIEW Algebra 1 Geometry Algebra 2 A0 Introduction G0 Introduction and A0 Introduction Construction A1 Modeling With Functions G1 Basic Definitions and Rigid -
Right Triangles and the Pythagorean Theorem Related?
Activity Assess 9-6 EXPLORE & REASON Right Triangles and Consider △ ABC with altitude CD‾ as shown. the Pythagorean B Theorem D PearsonRealize.com A 45 C 5√2 I CAN… prove the Pythagorean Theorem using A. What is the area of △ ABC? Of △ACD? Explain your answers. similarity and establish the relationships in special right B. Find the lengths of AD‾ and AB‾ . triangles. C. Look for Relationships Divide the length of the hypotenuse of △ ABC VOCABULARY by the length of one of its sides. Divide the length of the hypotenuse of △ACD by the length of one of its sides. Make a conjecture that explains • Pythagorean triple the results. ESSENTIAL QUESTION How are similarity in right triangles and the Pythagorean Theorem related? Remember that the Pythagorean Theorem and its converse describe how the side lengths of right triangles are related. THEOREM 9-8 Pythagorean Theorem If a triangle is a right triangle, If... △ABC is a right triangle. then the sum of the squares of the B lengths of the legs is equal to the square of the length of the hypotenuse. c a A C b 2 2 2 PROOF: SEE EXAMPLE 1. Then... a + b = c THEOREM 9-9 Converse of the Pythagorean Theorem 2 2 2 If the sum of the squares of the If... a + b = c lengths of two sides of a triangle is B equal to the square of the length of the third side, then the triangle is a right triangle. c a A C b PROOF: SEE EXERCISE 17. Then... △ABC is a right triangle. -
Archimedes and the Arbelos1 Bobby Hanson October 17, 2007
Archimedes and the Arbelos1 Bobby Hanson October 17, 2007 The mathematician’s patterns, like the painter’s or the poet’s must be beautiful; the ideas like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics. — G.H. Hardy, A Mathematician’s Apology ACBr 1 − r Figure 1. The Arbelos Problem 1. We will warm up on an easy problem: Show that traveling from A to B along the big semicircle is the same distance as traveling from A to B by way of C along the two smaller semicircles. Proof. The arc from A to C has length πr/2. The arc from C to B has length π(1 − r)/2. The arc from A to B has length π/2. ˜ 1My notes are shamelessly stolen from notes by Tom Rike, of the Berkeley Math Circle available at http://mathcircle.berkeley.edu/BMC6/ps0506/ArbelosBMC.pdf . 1 2 If we draw the line tangent to the two smaller semicircles, it must be perpendicular to AB. (Why?) We will let D be the point where this line intersects the largest of the semicircles; X and Y will indicate the points of intersection with the line segments AD and BD with the two smaller semicircles respectively (see Figure 2). Finally, let P be the point where XY and CD intersect. D X P Y ACBr 1 − r Figure 2 Problem 2. Now show that XY and CD are the same length, and that they bisect each other. -
Lesson 20: Composite Area Problems
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 20 7•3 Lesson 20: Composite Area Problems Student Outcomes . Students find the area of regions in the coordinate plane with polygonal boundaries by decomposing the plane into triangles and quadrilaterals, including regions with polygonal holes. Students find composite areas of regions in the coordinate plane by decomposing the plane into familiar figures (triangles, quadrilaterals, circles, semicircles, and quarter circles). Lesson Notes In Lessons 17 through 20, students learned to find the areas of various regions, including quadrilaterals, triangles, circles, semicircles, and those plotted on coordinate planes. In this lesson, students use prior knowledge to use the sum and/or difference of the areas to find unknown composite areas. Classwork Example 1 (5 minutes) Scaffolding: For struggling students, display Example 1 posters around the room displaying the visuals and the Find the composite area of the shaded region. Use ퟑ. ퟏퟒ for 흅. formulas of the area of a circle, a triangle, and a quadrilateral for reference. Allow students to look at the problem and find the area independently before solving as a class. What information can we take from the image? Two circles are on the coordinate plane. The diameter of the larger circle is 6 units, and the diameter of MP.1 the smaller circle is 4 units. How do we know what the diameters of the circles are? We can count the units along the diameter of the circles, or we can subtract the coordinate points to find the length of the diameter. Lesson 20: Composite Area Problems 283 This work is derived from Eureka Math ™ and licensed by Great Minds. -
Chapter 6 the Arbelos
Chapter 6 The arbelos 6.1 Archimedes’ theorems on the arbelos Theorem 6.1 (Archimedes 1). The two circles touching CP on different sides and AC CB each touching two of the semicircles have equal diameters AB· . P W1 W2 A O1 O C O2 B A O1 O C O2 B Theorem 6.2 (Archimedes 2). The diameter of the circle tangent to all three semi- circles is AC CB AB · · . AC2 + AC CB + CB2 · We shall consider Theorem ?? in ?? later, and for now examine Archimedes’ wonderful proofs of the more remarkable§ Theorems 6.1 and 6.2. By synthetic reasoning, he computed the radii of these circles. 1Book of Lemmas, Proposition 5. 2Book of Lemmas, Proposition 6. 602 The arbelos 6.1.1 Archimedes’ proof of the twin circles theorem In the beginning of the Book of Lemmas, Archimedes has established a basic proposition 3 on parallel diameters of two tangent circles. If two circles are tangent to each other (internally or externally) at a point P , and if AB, XY are two parallel diameters of the circles, then the lines AX and BY intersect at P . D F I W1 E H W2 G A O C B Figure 6.1: Consider the circle tangent to CP at E, and to the semicircle on AC at G, to that on AB at F . If EH is the diameter through E, then AH and BE intersect at F . Also, AE and CH intersect at G. Let I be the intersection of AE with the outer semicircle. Extend AF and BI to intersect at D. -
Pythagorean Theorem Word Problems Ws #1 Name ______
Pythagorean Theorem word problems ws #1 Name __________________________ Solve each of the following. Please draw a picture and use the Pythagorean Theorem to solve. Be sure to label all answers and leave answers in exact simplified form. 1. The bottom of a ladder must be placed 3 feet from a wall. The ladder is 12 feet long. How far above the ground does the ladder touch the wall? 2. A soccer field is a rectangle 90 meters wide and 120 meters long. The coach asks players to run from one corner to the corner diagonally across the field. How far do the players run? 3. How far from the base of the house do you need to place a 15’ ladder so that it exactly reaches the top of a 12’ wall? 4. What is the length of the diagonal of a 10 cm by 15 cm rectangle? 5. The diagonal of a rectangle is 25 in. The width is 15 in. What is the area of the rectangle? 6. Two sides of a right triangle are 8” and 12”. A. Find the the area of the triangle if 8 and 12 are legs. B. Find the area of the triangle if 8 and 12 are a leg and hypotenuse. 7. The area of a square is 81 cm2. Find the perimeter of the square. 8. An isosceles triangle has congruent sides of 20 cm. The base is 10 cm. What is the area of the triangle? 9. A baseball diamond is a square that is 90’ on each side. -
5-7 the Pythagorean Theorem 5-7 the Pythagorean Theorem
55-7-7 TheThe Pythagorean Pythagorean Theorem Theorem Warm Up Lesson Presentation Lesson Quiz HoltHolt McDougal Geometry Geometry 5-7 The Pythagorean Theorem Warm Up Classify each triangle by its angle measures. 1. 2. acute right 3. Simplify 12 4. If a = 6, b = 7, and c = 12, find a2 + b2 2 and find c . Which value is greater? 2 85; 144; c Holt McDougal Geometry 5-7 The Pythagorean Theorem Objectives Use the Pythagorean Theorem and its converse to solve problems. Use Pythagorean inequalities to classify triangles. Holt McDougal Geometry 5-7 The Pythagorean Theorem Vocabulary Pythagorean triple Holt McDougal Geometry 5-7 The Pythagorean Theorem The Pythagorean Theorem is probably the most famous mathematical relationship. As you learned in Lesson 1-6, it states that in a right triangle, the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. a2 + b2 = c2 Holt McDougal Geometry 5-7 The Pythagorean Theorem Example 1A: Using the Pythagorean Theorem Find the value of x. Give your answer in simplest radical form. a2 + b2 = c2 Pythagorean Theorem 22 + 62 = x2 Substitute 2 for a, 6 for b, and x for c. 40 = x2 Simplify. Find the positive square root. Simplify the radical. Holt McDougal Geometry 5-7 The Pythagorean Theorem Example 1B: Using the Pythagorean Theorem Find the value of x. Give your answer in simplest radical form. a2 + b2 = c2 Pythagorean Theorem (x – 2)2 + 42 = x2 Substitute x – 2 for a, 4 for b, and x for c. x2 – 4x + 4 + 16 = x2 Multiply. -
An Elementary Proof of the Mean Inequalities
Advances in Pure Mathematics, 2013, 3, 331-334 http://dx.doi.org/10.4236/apm.2013.33047 Published Online May 2013 (http://www.scirp.org/journal/apm) An Elementary Proof of the Mean Inequalities Ilhan M. Izmirli Department of Statistics, George Mason University, Fairfax, USA Email: [email protected] Received November 24, 2012; revised December 30, 2012; accepted February 3, 2013 Copyright © 2013 Ilhan M. Izmirli. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. ABSTRACT In this paper we will extend the well-known chain of inequalities involving the Pythagorean means, namely the har- monic, geometric, and arithmetic means to the more refined chain of inequalities by including the logarithmic and iden- tric means using nothing more than basic calculus. Of course, these results are all well-known and several proofs of them and their generalizations have been given. See [1-6] for more information. Our goal here is to present a unified approach and give the proofs as corollaries of one basic theorem. Keywords: Pythagorean Means; Arithmetic Mean; Geometric Mean; Harmonic Mean; Identric Mean; Logarithmic Mean 1. Pythagorean Means 11 11 x HM HM x For a sequence of numbers x xx12,,, xn we will 12 let The geometric mean of two numbers x1 and x2 can n be visualized as the solution of the equation x j AM x,,, x x AM x j1 12 n x GM n 1 n GM x2 1 n GM x12,,, x xnj GM x x j1 1) GM AM HM and 11 2) HM x , n 1 x1 1 HM x12,,, x xn HM x AM x , n 1 1 x1 x j1 j 11 1 3) x xx n2 to denote the well known arithmetic, geometric, and 12 n xx12 xn harmonic means, also called the Pythagorean means This follows because PM . -
Pappus of Alexandria: Book 4 of the Collection
Pappus of Alexandria: Book 4 of the Collection For other titles published in this series, go to http://www.springer.com/series/4142 Sources and Studies in the History of Mathematics and Physical Sciences Managing Editor J.Z. Buchwald Associate Editors J.L. Berggren and J. Lützen Advisory Board C. Fraser, T. Sauer, A. Shapiro Pappus of Alexandria: Book 4 of the Collection Edited With Translation and Commentary by Heike Sefrin-Weis Heike Sefrin-Weis Department of Philosophy University of South Carolina Columbia SC USA [email protected] Sources Managing Editor: Jed Z. Buchwald California Institute of Technology Division of the Humanities and Social Sciences MC 101–40 Pasadena, CA 91125 USA Associate Editors: J.L. Berggren Jesper Lützen Simon Fraser University University of Copenhagen Department of Mathematics Institute of Mathematics University Drive 8888 Universitetsparken 5 V5A 1S6 Burnaby, BC 2100 Koebenhaven Canada Denmark ISBN 978-1-84996-004-5 e-ISBN 978-1-84996-005-2 DOI 10.1007/978-1-84996-005-2 Springer London Dordrecht Heidelberg New York British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Control Number: 2009942260 Mathematics Classification Number (2010) 00A05, 00A30, 03A05, 01A05, 01A20, 01A85, 03-03, 51-03 and 97-03 © Springer-Verlag London Limited 2010 Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licenses issued by the Copyright Licensing Agency.