G_PP_11-2_ChordsArcs2015
... 1. Draw a large circle 2. Add a chord to the circle 3. Construct a perpendicular bisector of the chord – use the compass and the ruler 4. Repeat steps 2 and 3 with a congruent chord 5. What do you notice about the two bisector lines? ...
... 1. Draw a large circle 2. Add a chord to the circle 3. Construct a perpendicular bisector of the chord – use the compass and the ruler 4. Repeat steps 2 and 3 with a congruent chord 5. What do you notice about the two bisector lines? ...
Unit # 3 Name of unit Circles and Spheres
... A segment between two points on a circle, which passes through the center of the circle. (The diameter is the longest chord of a circle). ...
... A segment between two points on a circle, which passes through the center of the circle. (The diameter is the longest chord of a circle). ...
Answer Key 1 5.1 Copies of Line Segments and Angles
... 2. Create two circles with the same radius centered at each endpoint. The line connecting the intersection points of the circles is the perpendicular bisector. 3. The midpoint is the point where the line segment and perpendicular bisector intersect. 4. A bisector cuts a line segment in half while a ...
... 2. Create two circles with the same radius centered at each endpoint. The line connecting the intersection points of the circles is the perpendicular bisector. 3. The midpoint is the point where the line segment and perpendicular bisector intersect. 4. A bisector cuts a line segment in half while a ...
Solutions 13-14 - Durham University
... and making angle ∠A1 with A1 B1 . There exists exactly one such a ray in each half-plane with respect to the line A1 B1 (this is especially clear if A1 is the centre of the Poincaré disc model). Similarly, C1 and C2 lie on a (hyperbolic) ray starting from B1 and making angle ∠B1 with A1 B1 . As tw ...
... and making angle ∠A1 with A1 B1 . There exists exactly one such a ray in each half-plane with respect to the line A1 B1 (this is especially clear if A1 is the centre of the Poincaré disc model). Similarly, C1 and C2 lie on a (hyperbolic) ray starting from B1 and making angle ∠B1 with A1 B1 . As tw ...
Learning Area
... Join points A and E with a straight line. i) Does BAE have elements in common with ABC ? ii) Can BAE and BAC be called supplementary angles? Explain ...
... Join points A and E with a straight line. i) Does BAE have elements in common with ABC ? ii) Can BAE and BAC be called supplementary angles? Explain ...
“Straight” and “Angle” on Non-Planar Surfaces
... 1. Any two great circles intersect, but parallel straight lines on the plane do not intersect. 2. Great circles are finite and come back on themselves, but straight lines go on forever on the plane. 3. Even though great circles are finite you can walk along the great circle forever in either directi ...
... 1. Any two great circles intersect, but parallel straight lines on the plane do not intersect. 2. Great circles are finite and come back on themselves, but straight lines go on forever on the plane. 3. Even though great circles are finite you can walk along the great circle forever in either directi ...
Polygons, Circles, and Angles - mcs6
... but have no interior points in common – supplementary or complementary ...
... but have no interior points in common – supplementary or complementary ...
Welcome Course: Geometry Teacher: A. Ferraro Contact: AFerraro
... 1. Determine the negation of a statement and establish its truth value 2. Know and apply the conditions under which a compound statement is true A. Conjunction B. Disjunction C. Conditional D. Biconditional 3. Identify and write the inverse, converse, and contrapositive of a given conditional statem ...
... 1. Determine the negation of a statement and establish its truth value 2. Know and apply the conditions under which a compound statement is true A. Conjunction B. Disjunction C. Conditional D. Biconditional 3. Identify and write the inverse, converse, and contrapositive of a given conditional statem ...
Problem of Apollonius
In Euclidean plane geometry, Apollonius's problem is to construct circles that are tangent to three given circles in a plane (Figure 1). Apollonius of Perga (ca. 262 BC – ca. 190 BC) posed and solved this famous problem in his work Ἐπαφαί (Epaphaí, ""Tangencies""); this work has been lost, but a 4th-century report of his results by Pappus of Alexandria has survived. Three given circles generically have eight different circles that are tangent to them (Figure 2) and each solution circle encloses or excludes the three given circles in a different way: in each solution, a different subset of the three circles is enclosed (its complement is excluded) and there are 8 subsets of a set whose cardinality is 3, since 8 = 23.In the 16th century, Adriaan van Roomen solved the problem using intersecting hyperbolas, but this solution does not use only straightedge and compass constructions. François Viète found such a solution by exploiting limiting cases: any of the three given circles can be shrunk to zero radius (a point) or expanded to infinite radius (a line). Viète's approach, which uses simpler limiting cases to solve more complicated ones, is considered a plausible reconstruction of Apollonius' method. The method of van Roomen was simplified by Isaac Newton, who showed that Apollonius' problem is equivalent to finding a position from the differences of its distances to three known points. This has applications in navigation and positioning systems such as LORAN.Later mathematicians introduced algebraic methods, which transform a geometric problem into algebraic equations. These methods were simplified by exploiting symmetries inherent in the problem of Apollonius: for instance solution circles generically occur in pairs, with one solution enclosing the given circles that the other excludes (Figure 2). Joseph Diaz Gergonne used this symmetry to provide an elegant straightedge and compass solution, while other mathematicians used geometrical transformations such as reflection in a circle to simplify the configuration of the given circles. These developments provide a geometrical setting for algebraic methods (using Lie sphere geometry) and a classification of solutions according to 33 essentially different configurations of the given circles.Apollonius' problem has stimulated much further work. Generalizations to three dimensions—constructing a sphere tangent to four given spheres—and beyond have been studied. The configuration of three mutually tangent circles has received particular attention. René Descartes gave a formula relating the radii of the solution circles and the given circles, now known as Descartes' theorem. Solving Apollonius' problem iteratively in this case leads to the Apollonian gasket, which is one of the earliest fractals to be described in print, and is important in number theory via Ford circles and the Hardy–Littlewood circle method.