Standards - Greenville Public School District
... Theorem; complete the square to find the center and radius of a circle given by an equation. Use coordinates to prove simple geometric theorems algebraically Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in ...
... Theorem; complete the square to find the center and radius of a circle given by an equation. Use coordinates to prove simple geometric theorems algebraically Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in ...
Fill in the blank in each sentence with the vocabulary term that best
... The polygon has 3 sides. So, it is a triangle. No line containing any of the sides will pass through the interior of the triangle, so it is convex. All of the sides are congruent, so it is equilateral. All of the angles are congruent, so it is equiangular. Since the polygon is convex, equilateral, a ...
... The polygon has 3 sides. So, it is a triangle. No line containing any of the sides will pass through the interior of the triangle, so it is convex. All of the sides are congruent, so it is equilateral. All of the angles are congruent, so it is equiangular. Since the polygon is convex, equilateral, a ...
here
... Expressing Geometric Properties with Equations (G-GPE) Translate between the geometric description and the equation for a conic section G-GPE.1 ...
... Expressing Geometric Properties with Equations (G-GPE) Translate between the geometric description and the equation for a conic section G-GPE.1 ...
Topic 1 - Dr Frost Maths
... Question: A square sheet of paper ABCD is folded along FG, as shown, so that the corner B is folded onto the midpoint M of CD. Prove that the sides of triangle GCM have lengths of ratio 3 : 4 : 5. Starting point: How might I label the sides? Ensure you use information in the question! The paper is f ...
... Question: A square sheet of paper ABCD is folded along FG, as shown, so that the corner B is folded onto the midpoint M of CD. Prove that the sides of triangle GCM have lengths of ratio 3 : 4 : 5. Starting point: How might I label the sides? Ensure you use information in the question! The paper is f ...
Geometry Definitions
... sphere - Set of all points in space that are a given distance from a given point called the center. square - A quadrilateral with four right angles and four congruent sides. square unit - Square region having sides that measure one unit in length. supplementary angles - Two angles with measure whose ...
... sphere - Set of all points in space that are a given distance from a given point called the center. square - A quadrilateral with four right angles and four congruent sides. square unit - Square region having sides that measure one unit in length. supplementary angles - Two angles with measure whose ...
Contents
... Albert and Betty are playing the following game. There are 100 blue balls in a red bowl and 100 red balls in a blue bowl. In each turn a player must make one of the following ...
... Albert and Betty are playing the following game. There are 100 blue balls in a red bowl and 100 red balls in a blue bowl. In each turn a player must make one of the following ...
texts and listening tasks
... line of reflection symmetry and it has rotational symmetry around the centre for every angle. Its symmetry group is the orthogonal group O(2,R). The group of rotations alone is the circle group. ...
... line of reflection symmetry and it has rotational symmetry around the centre for every angle. Its symmetry group is the orthogonal group O(2,R). The group of rotations alone is the circle group. ...
- wced curriculum development
... Dividing a line into equal parts Given the line to be divided Draw a light construction line at any convenient angle from one end of the given line. With dividers or scale, set off from the intersections of the lines as many equal divisions as needed (in this example, three). Connect the last divisi ...
... Dividing a line into equal parts Given the line to be divided Draw a light construction line at any convenient angle from one end of the given line. With dividers or scale, set off from the intersections of the lines as many equal divisions as needed (in this example, three). Connect the last divisi ...
Inscribed Angles
... to solve problems. (2) The student will be able to use properties of inscribed polygons. Toolbox: Summary: Inscribed Angle – an angle whose vertex is on a circle and whose sides contain chords of the circle. Intercepted Arc – arc that lies in the interior of an inscribed angle and has endpoints on t ...
... to solve problems. (2) The student will be able to use properties of inscribed polygons. Toolbox: Summary: Inscribed Angle – an angle whose vertex is on a circle and whose sides contain chords of the circle. Intercepted Arc – arc that lies in the interior of an inscribed angle and has endpoints on t ...
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.