Download CHAPTER 2: MATH NOTES Angle Relationships Naming Parts of

 CHAPTER 2: MATH NOTES Angle Relationships Complementary angles­ two angles that add up to 90°. For example, in the diagram at right, ∠ABC and ∠CBD are complementary because together they form a right angle. Supplementary angles­ two angles that add up to 180°. For example, in the diagram at right, ∠EFG and ∠GFH are supplementary because together they form a straight angle. Naming Parts of Shapes Part of geometry is the study of parts of shapes, such as points, line segments, and angles. To avoid confusion, standard notation is used to name these parts. Point A, B, C a single capital letter Prime Notation ΔABC = ΔA′B′C′ used after a shape has been transformed Line Segment Line Angle a portion of a line ∠CDE, m∠HGI continues in either direction two rays that share a vertex Triangle Angle Sum Theorem The Triangle Angle Sum Theorem states that the measures of the angles in a triangle must add up to 180°. For example, in ΔABC m∠A + m∠B + m∠C = 180° Multiplying Binomials One method for multiplying binomials is to use a generic rectangle. Use each factor of the problem as a dimension of a rectangle and find its area. See the example ( 3x ­ 1 ) ( 2x + 5 ) = 6x2 + 13x ­ 5 More Angle Pair Relationships *These rules only apply when parallel lines are intersected by a transversal. Name Position Rule of angles Example Vertical angles two opposite angles (non­adjacent) examples are ∠c and ∠g or ∠h and ∠n Corresponding angles lie in the same position examples are ∠d and ∠m ∠c and ∠h alternate interior angles both are between the pair of lines (inside) one on the left and one on the right examples are ∠f and ∠m ∠g and ∠h same­side interior angles both are between the pair of lines (inside) one on the left and one on the right such as angles ∠g and ∠m ∠f and ∠h these angles are congruent (equal) these angles are congruent (equal) these angles are congruent (equal) these angles are supplemetary (add up to 180°) Conditional Statements A conditional statement is written in the form “If …, then ….” Here are some examples of conditional statements: If a shape is a rhombus, then it has four sides of equal length. If it is February 14th, then it is Valentine's Day. If a shape is a parallelogram, then its area is A = bh. Proof by Contradiction A proof by contradiction is a type of argument when you prove that something cannot be true and you provide a false example. This example as drawn is impossible as drawn, since the angles of a triangle must add up to 180. Instead you can prove that and must be parallel. To show that lines and must be parallel in the diagram below, you use a proof by contradiction. You argued since the Same­Side Interior angles add up to 180, the lines must be parallel.. Areas of a Triangle,Parallelogram,and Trapezoid Triangle A = bh Parallelogram A = bh Trapezoid A = (b1 + b2)h Right Triangle Vocabulary A Right Triangle is a triangle that contains a 90° angle. The two shortest sides of the right triangle (the sides that meet at the right angle) are called the legs of the triangle and the longest side (the side opposite the right angle) is called the hypotenuse of the triangle.
The Pythagorean Theorem The Pythagorean Theorem can be used to find a missing side length in a right triangle. The Theorem states that in a right triangle, (length of leg #1)2 + (length of leg #2)2 = (length of hypotenuse)2 a2 + b2 = c2 ‘c’ is always the hypotenuse. 52 + x2 = 82 In this example, is an example 2
25 + x = 64
of an exact answer, while 6.24 is an 2
x = 39 approximate answer. x = ≈ 6.24 Solving a System of Linear Equations To solve a system of linear equations, you must find the Point of intersection, the (x,y) value that will solve both equations. y = 2x
y = −3x + 5 Ways to solve: 1) Graph­ graph both lines if you can and find the exact point where they intersect (x,y). If the lines are parallel, there are no points of intersection If the lines are the same, then there are infinite number of intersections. Such lines are said to coincide. 2) Substitution Method is another way to find the point of intersection for the two lines. Since you can’t solve an equation with both an ‘x’ and a ‘y’ change the equations into one equation with one variable by substituting. Follow the example shown. Since ‘x’ equals ­3y+1, replace the ‘x’ in the second problem with (−3y + 1) to get 4(−3y + 1) − 3y = −11. This equation now has only one variable and can then be solved. Distribute, combine like­trms and solve for ‘y’. In this case, y = 1. Once you solve for ‘y’, plug it back into the original problem and solve for ‘x’. Remember the final answer is an ordered pair (x,y). This the point where the two lines will intersect.