Pythagorean Theorem
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Circle Theorems
Circle theorems A LEVEL LINKS Scheme of work: 2b. Circles – equation of a circle, geometric problems on a grid Key points • A chord is a straight line joining two points on the circumference of a circle. So AB is a chord. • A tangent is a straight line that touches the circumference of a circle at only one point. The angle between a tangent and the radius is 90°. • Two tangents on a circle that meet at a point outside the circle are equal in length. So AC = BC. • The angle in a semicircle is a right angle. So angle ABC = 90°. • When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference. So angle AOB = 2 × angle ACB. • Angles subtended by the same arc at the circumference are equal. This means that angles in the same segment are equal. So angle ACB = angle ADB and angle CAD = angle CBD. • A cyclic quadrilateral is a quadrilateral with all four vertices on the circumference of a circle. Opposite angles in a cyclic quadrilateral total 180°. So x + y = 180° and p + q = 180°. • The angle between a tangent and chord is equal to the angle in the alternate segment, this is known as the alternate segment theorem. So angle BAT = angle ACB. Examples Example 1 Work out the size of each angle marked with a letter. Give reasons for your answers. Angle a = 360° − 92° 1 The angles in a full turn total 360°. = 268° as the angles in a full turn total 360°. -
Angles ANGLE Topics • Coterminal Angles • Defintion of an Angle
Angles ANGLE Topics • Coterminal Angles • Defintion of an angle • Decimal degrees to degrees, minutes, seconds by hand using the TI-82 or TI-83 Plus • Degrees, seconds, minutes changed to decimal degree by hand using the TI-82 or TI-83 Plus • Standard position of an angle • Positive and Negative angles ___________________________________________________________________________ Definition: Angle An angle is created when a half-ray (the initial side of the angle) is drawn out of a single point (the vertex of the angle) and the ray is rotated around the point to another location (becoming the terminal side of the angle). An angle is created when a half-ray (initial side of angle) A: vertex point of angle is drawn out of a single point (vertex) AB: Initial side of angle. and the ray is rotated around the point to AC: Terminal side of angle another location (becoming the terminal side of the angle). Hence angle A is created (also called angle BAC) STANDARD POSITION An angle is in "standard position" when the vertex is at the origin and the initial side of the angle is along the positive x-axis. Recall: polynomials in algebra have a standard form (all the terms have to be listed with the term having the highest exponent first). In trigonometry, there is a standard position for angles. In this way, we are all talking about the same thing and are not trying to guess if your math solution and my math solution are the same. Not standard position. Not standard position. This IS standard position. Initial side not along Initial side along negative Initial side IS along the positive x-axis. -
Calculus Terminology
AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential -
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. -
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. -
Copyrighted Material
INDEX Abel, Niels Henrik, 234 algebraic manipulation, of integrand, finding by numerical methods, 367 abscissa, Web-H1 412 and improper integrals, 473 absolute convergence, 556, 557 alternating current, 324 as a line integral, 981 ratio test for, 558, 559 alternating harmonic series, 554 parametric curves, 369, 610 absolute error, 46 alternating series, 553–556 polar curve, 632 Euler’s Method, 501 alternating series test, 553, 560, 585 from vector viewpoint, 760, 761 absolute extrema amplitude arc length parametrization, 761, 762 Extreme-Value Theorem, 205 alternating current, 324 finding, 763, 764 finding on closed and bounded sets, simple harmonic motion, 124, properties, 765, 766 876–879 Web-P5 arccosine, 57 finding on finite closed intervals, 205, sin x and cos x, A21, Web-D7 Archimedean spiral, 626, 629 206 analytic geometry, 213, Web-F3 Archimedes, 253, 255 finding on infinite intervals, 206, 207 Anderson, Paul, 376 palimpsest, 255 functions with one relative extrema, angle(s), A1–A2, Web-A1–Web-A2 arcsecant 57 208, 209 finding from trigonometric functions, arcsine, 57 on open intervals, 207, 208 Web-A10 arctangent, 57 absolute extremum, 204, 872 of inclination, A7, Web-A10 area absolute maximum, 204, 872 between planes, 719, 720 antiderivative method, 256, 257 absolute minimum, 204, 872 polar, 618 calculated as double integral, 906, 907 absolute minimum values, 872 rectangular coordinate system, computing exact value of, 279 absolute value, Web-G1 A3–A4, Web-A3–Web-A4 definition, 277 and square roots, Web-G1, Web-B2 standard position, -
Classify Each Triangle by Its Side Lengths and Angle Measurements
GRADE 4 | MODULE 4 | TOPIC D | LESSONS 12–16 KEY CONCEPT OVERVIEW In Lessons 12 through 16, students explore lines of symmetry and characteristics of triangles and quadrilaterals. You can expect to see homework that asks your child to do the following: ▪ Find and draw lines of symmetry. ▪ Given half of a figure and the line of symmetry, draw the other half of the figure. ▪ Classify triangles by side lengths (e.g., equilateral, isosceles, scalene) and by angle measurements (e.g., acute, right, obtuse). ▪ Draw triangles that fit different classifications (e.g., acute and scalene). ▪ Name quadrilaterals, identify attributes (i.e., characteristics) that define them, and construct them based on given attributes. SAMPLE PROBLEM (From Lesson 13) Classify each triangle by its side lengths and angle measurements. Circle the correct names. Additional sample problems with detailed answer steps are found in the Eureka Math Homework Helpers books. Learn more at GreatMinds.org. For more resources, visit » Eureka.support GRADE 4 | MODULE 4 | TOPIC D | LESSONS 12–16 HOW YOU CAN HELP AT HOME ▪ Ask your child to look around the house for objects that have lines of symmetry. Examples include the headboard of a bed, dressers, chairs, couches, and place mats. Ask him to show where the line of symmetry would be and what makes it a line of symmetry. Be careful of objects such as doors and windows. They could have a line of symmetry, but if there’s a knob or crank on just one side, then they are not symmetrical. ▪ Ask your child to name and draw all the quadrilaterals she can think of (e.g., square, rectangle, parallelogram, trapezoid, and rhombus). -
The Pythagorean Theorem and Area: Postulates Into Theorems Paul A
Humanistic Mathematics Network Journal Issue 25 Article 13 8-1-2001 The Pythagorean Theorem and Area: Postulates into Theorems Paul A. Kennedy Texas State University Kenneth Evans Texas State University Follow this and additional works at: http://scholarship.claremont.edu/hmnj Part of the Mathematics Commons, Science and Mathematics Education Commons, and the Secondary Education and Teaching Commons Recommended Citation Kennedy, Paul A. and Evans, Kenneth (2001) "The Pythagorean Theorem and Area: Postulates into Theorems," Humanistic Mathematics Network Journal: Iss. 25, Article 13. Available at: http://scholarship.claremont.edu/hmnj/vol1/iss25/13 This Article is brought to you for free and open access by the Journals at Claremont at Scholarship @ Claremont. It has been accepted for inclusion in Humanistic Mathematics Network Journal by an authorized administrator of Scholarship @ Claremont. For more information, please contact [email protected]. The Pythagorean Theorem and Area: Postulates into Theorems Paul A. Kennedy Contributing Author: Department of Mathematics Kenneth Evans Southwest Texas State University Department of Mathematics, Retired San Marcos, TX 78666-4616 Southwest Texas State University [email protected] San Marcos, TX 78666-4616 Considerable time is spent in high school geometry equal to the area of the square on the hypotenuse. building an axiomatic system that allows students to understand and prove interesting theorems. In tradi- tional geometry classrooms, the theorems were treated in isolation with some of -
The Pythagorean Theorem
6.2 The Pythagorean Theorem How are the lengths of the sides of a right STATES triangle related? STANDARDS MA.8.G.2.4 MA.8.A.6.4 Pythagoras was a Greek mathematician and philosopher who discovered one of the most famous rules in mathematics. In mathematics, a rule is called a theorem. So, the rule that Pythagoras discovered is called the Pythagorean Theorem. Pythagoras (c. 570 B.C.–c. 490 B.C.) 1 ACTIVITY: Discovering the Pythagorean Theorem Work with a partner. a. On grid paper, draw any right triangle. Label the lengths of the two shorter sides (the legs) a and b. c2 c a a2 b. Label the length of the longest side (the hypotenuse) c. b b2 c . Draw squares along each of the three sides. Label the areas of the three squares a 2, b 2, and c 2. d. Cut out the three squares. Make eight copies of the right triangle and cut them out. a2 Arrange the fi gures to form 2 two identical larger squares. c b2 e. What does this tell you about the relationship among a 2, b 2, and c 2? 236 Chapter 6 Square Roots and the Pythagorean Theorem 2 ACTIVITY: Finding the Length of the Hypotenuse Work with a partner. Use the result of Activity 1 to fi nd the length of the hypotenuse of each right triangle. a. b. c 10 c 3 24 4 c. d. c 0.6 2 c 3 0.8 1 2 3 ACTIVITY: Finding the Length of a Leg Work with a partner. -
Measuring Angles and Angular Resolution
Angles Angle θ is the ratio of two lengths: R: physical distance between observer and objects [km] Measuring Angles S: physical distance along the arc between 2 objects Lengths are measured in same “units” (e.g., kilometers) and Angular θ is “dimensionless” (no units), and measured in “radians” or “degrees” Resolution R S θ R Trigonometry “Angular Size” and “Resolution” 22 Astronomers usually measure sizes in terms R +Y of angles instead of lengths R because the distances are seldom well known S Y θ θ S R R S = physical length of the arc, measured in m Y = physical length of the vertical side [m] Trigonometric Definitions Angles: units of measure R22+Y 2π (≈ 6.28) radians in a circle R 1 radian = 360˚ ÷ 2π≈57 ˚ S Y ⇒≈206,265 seconds of arc per radian θ Angular degree (˚) is too large to be a useful R S angular measure of astronomical objects θ ≡ R 1º = 60 arc minutes opposite side Y 1 arc minute = 60 arc seconds [arcsec] tan[]θ ≡= adjacent side R 1º = 3600 arcsec -1 -6 opposite sideY 1 1 arcsec ≈ (206,265) ≈ 5 × 10 radians = 5 µradians sin[]θ ≡== hypotenuse RY22+ R 2 1+ Y 2 1 Number of Degrees per Radian Trigonometry in Astronomy Y 2π radians per circle θ S R 360° Usually R >> S (particularly in astronomy), so Y ≈ S 1 radian = ≈ 57.296° 2π SY Y 1 θ ≡≈≈ ≈ ≈ 57° 17'45" RR RY22+ R 2 1+ Y 2 θ ≈≈tan[θθ] sin[ ] Relationship of Trigonometric sin[θ ] ≈ tan[θ ] ≈ θ for θ ≈ 0 Functions for Small Angles 1 sin(πx) Check it! tan(πx) 0.5 18˚ = 18˚ × (2π radians per circle) ÷ (360˚ per πx circle) = 0.1π radians ≈ 0.314 radians 0 Calculated Results