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DA2GP_774_11.qxd 01/09/2006 11:11 PM Page 53 CHAPTER 11 Introduction to Geometry Content Summary Chapter 11 is a preview of geometry. Even so, in some places it uses linear equations. Students also focus on radical expressions and operations with radicals. Synthetic Geometry The original study of geometry is now called synthetic to distinguish it from the analytic geometry of coordinate systems, which developed much later. In the area of synthetic geometry, the book focuses on the Pythagorean Theorem and on similar figures—figures that are stretches or shrinks of each other by the same factor in every direction. Analytic Geometry Analytic geometry is the geometry of coordinate graphs, which use algebraic equations to represent geometric figures. Students have been working with analytic geometry all through this course. They’ve already seen how parallel lines have the same slope, and how that slope appears in the equations of the lines. In this chapter they see how the slopes and equations of perpendicular lines relate. They also see how to find the coordinates of midpoints of line segments. In considering how the Pythagorean Theorem translates into coordinate geometry, students learn how to work with square roots. 18 3 3 20 2 4 Trigonometry When a geometric figure is stretched or shrunk uniformly, all angles retain their measures and all side lengths are multiplied by the same amount; therefore, the ratio of one side length to another remains the same. This stretching and shrinking produces similar figures, which have equal corresponding angles and proportional corresponding side lengths. The study of relationships between sides and angles of similar right triangles is part of trigonometry. In particular, in similar right triangles, the ratio of, for example, the length of the side opposite a particular acute angle to the length of the hypotenuse is the same, no matter what the enlargement. Each acute angle of a right triangle has several of these ratios associated with it. They’re called trigonometric ratios. The book considers three such ratios: the sine, the cosine, and the tangent. Students use calculators to find values of these ratios for various angles and, in reverse, angles that correspond to various given ratios. They thus preview the deeper consideration of trigonometry found in Discovering Advanced Algebra. (continued) ©2007 Key Curriculum Press Discovering Algebra: A Guide for Parents 53 DA2GP_774_11.qxd 01/09/2006 11:11 PM Page 54 Chapter 11 • Introduction to Geometry (continued) Trigonometric Functions For acute angle A in a right angle, the trigonometric functions are length of opposite leg sine of angle A length of hypotenuse or o sin A h length of adjacent leg cosine of angle A length of hypotenuse or a cos A h length of opposite leg tangent of angle A length of adjacent leg or o tan A a h A o a Summary Problem You and your student might revisit this problem, adapted from Exercise 7 in Lesson 11.1, several times while working through the chapter. What can you say about the quadrilateral with vertices (5, 0), (1, 4), (6, 3), and (3, 3)? Questions you might ask in your role as student to your student include: ● ● ● ● ● ● ● Do any sides look parallel? Do any sides appear perpendicular? Can you confirm your conjectures? What are the equations of the four lines containing the sides of the quadrilateral? Do the diagonals meet at their midpoints? Can you find the lengths of the diagonals? Can you find the angle measures? Sample Answers Two of the sides are parallel y 1 23x and 3y 10 2x both have slope 23, and a third side is perpendicular to them 2y 15 3x with slope 32. The fourth side is contained in the line with equation 5y 21 x. The diagonals lie on the lines with equations y 74(x 1) 4 and y 131 (x 5). Graphing and tracing indicates that the diagonals meet at about (0.6, 1.2), which is not either of the midpoints, (1, 0.5) or (0.5, 1.5). The diagonals have lengths 65 and 130 . If you make a right triangle by dropping a perpendicular from the vertex at (1, 4) to the parallel side opposite, you can find that the sine of the angle at vertex (6, 3) 1 1 13 x to find that the angle is . Solve the equation sin1 2 2 26 is 45°. Therefore, the angle at (1, 4) will be 135°. 54 Discovering Algebra: A Guide for Parents ©2007 Key Curriculum Press DA2GP_774_11.qxd 01/09/2006 11:11 PM Page 55 Chapter 11 • Review Exercises Name Period Date 1. (Lessons 11.1–11.3, 11.6) Plot the points A(0, 2), B(3, 1), and C(3, 4). . a. Find the equation of the line through B that is parallel to AC . b. Find the midpoint of AB c. Find the equation of the line containing the median of the triangle . through C, and show that it is the perpendicular bisector of AB d. Find the length of each side. e. What kind of triangle is ABC? f. Find the area of triangle ABC. 2. (Lesson 11.5) Rewrite each expression with as few square root symbols as possible, and no parentheses. Your final result should not have any perfect-square factors under a radical. a. 26 32 8 21 2 3 c. b. 53 212 3. (Lesson 11.7) The two triangles shown below are similar. Write a proportion and solve it to find w. w 5 4 3 4. (Lessons 11.4, 11.8) Use the Pythagorean Theorem to find x. Then find the following ratios: A 5 C 13 x a. tan A ©2007 Key Curriculum Press B b. sin B c. cos B Discovering Algebra: A Guide for Parents 55 DA2GP_774_11.qxd 01/11/2006 05:48 PM Page 56 SOLUTIONS TO CHAPTER 11 REVIEW EXERCISES 1. 5 2. a. 2 6 3 2 2 3 6 2 C B 5 2 3 6•2 5 A 612 643 5 has slope 2, so write the equation a. The segment AC of the line with slope 2 that passes through the point (3, 1). In point-slope form, this is y 1 2(x 3), or y 2(x 3) 1. is b. The x-coordinate of the midpoint of AB 0 3 1.5, and the y-coordinate is 2 2 1 0.5. The midpoint is (1.5, 0.5). 2 c. Find the equation of the line through the points 0.5 4 (3, 4) and (1.5, 0.5). The slope is 1.5 3 4.5 1, so the equation is y 4 1(x 3), or 4.5 is 1 and the slope of y x 1. The slope of AB the median is 1, so the product of the slopes is 1. Therefore the two lines are perpendicular. The line y x 1 is the perpendicular bisector of AB and it passes because it is perpendicular to AB . through the midpoint of AB d. (Lesson 11.6) Use the distance formula d x 12 y 12 to calculate each x 2 y 2 length. For example: AB (3 0)2 [1 (2)] 2 2 2 AB (3) (3) AB 18 , or 3 2 For help in changing 18 to 3 2 , look ahead to Exercise 2. Compute the other two lengths in a similar fashion. The lengths are AC 45 , or 3 5 ; and BC 45 , or 3 5. e. Triangle ABC is a isosceles because AC BC. f. Draw a rectangle around triangle ABC as shown on the graph for 1a. The rectangle has area 36. The and AC have right triangles with hypotenuses BC areas 0.5(3)(6), or 9 square units. The smaller right triangle has area 0.5(3)(3), or 4.5 square units. Subtract the areas of the triangles from the area of the rectangle: 36 (9 + 9 + 4.5) 13.5. The area of triangle ABC is 13.5 square units. 56 Discovering Algebra: A Guide for Parents 6 23 123 b. 5 3 2 12 5 3 2 4 3 Multiplication of radical expressions. Multiply. 4 is a perfectsquare factor of 12. Take the square root. Multiply. 4 is a perfectsquare factor of 12. 53 43 Take the square root and multiply. (5 4)3 Addition of radical expressions. 3 2 18 292 c. 3 2 32 2 • 3 2 3 2 2 Commutative property of multiplication. Subtract. 9 is a perfectsquare factor of 18. Take the square root. Reduce. 3. Proportions may vary. w 53 Proportion to be solved. 4 3 w 8 4 4 4 3 Multiply both sides by 4. 32 2 w 3, or 103 Multiply. 4. 52 x 2 132 Equation to be solved. 25 x 2 169 Square each term. x 2 144 Subtract 25 from both sides. x 12 Take the square root of 144. Only the positive square root is a solution, because x is the length of a segment. opposite 12 a. tan A adjacent 5 opposite 5 b. sin B hypotenuse 13 adjacent 12 c. cos B hypotenuse 13 ©2007 Key Curriculum Press