Unit 5 Review Packet

Geometry/Trig Unit 5 Study Guide Section 5-1 Definitions: Parallelogram Theorems: 5-1: Opposite sides of a parallelogram are congruent. 5-2: Opposite angles in a parallelogram are congruent. 5-3: Diagonals of a parallelogram bisect one another. Know How To: set up algebraic equations by applying properties of parallelograms. Suggested Exercises: p. 168 #1-14 p. 169 #1-10 p. 170 #19-28 Section 5-3 Theorems: 5-8: If two lines are parallel, then all points on one line are equidistant from the other lines. 5-9: If three parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal, 5-10: A line that contains the midpoint of one side of a triangle and is parallel to another side passes through the midpoint of the third side. 5-11: The segment that joins the midpoints of two sides of a triangle 1. is parallel to the third side. 2. is half the length of the third side. Know How To: apply these theorem by setting up and solving algebraic equations. Suggested Exercises: p. 179 #2-6 p. 180-1 #1-17 Name: Date: Section 5-2 Five ways to prove that a quadrilateral is a parallelogram (You will have to list these on your test.) 1. If both pairs of opposite sides are parallel, then the quadrilateral is a parallelogram (definition of a parallelogram). 2. If both pairs of opposite sides are congruent, then the quadrilateral is a parallelogram. 3. If one pair of opposite sides are both congruent and parallel, then the quadrilateral is a parallelogram. 4. If both pairs of opposite angles are congruent, then the quadrilateral is a parallelogram. 5. If the diagonals bisect one another, then the quadrilateral is a parallelogram. Know How To: determine and prove (informally) that a quadrilateral is a parallelogram. Suggested Exercises: p. 173 #1-11 p. 174-5 #1-10, 19-22 Section 5-4 Definitions: Rectangle, Rhombus, Square Properties of Special Parallelograms: Refer to property chart from 5.4 Exploration note pages. Know How To: Classify a give quadrilateral, apply properties of special quadrilaterals by setting up and solving algebraic equations. Suggested Exercises: p. 187 #1-16 Geometry/Trig Name: Unit 5 Study Guide - page 2 Date: Section 5-5 Definitions: Trapezoid, Base, Leg, Isosceles Trapezoid, Median of a Trapezoid Theorems: 5-18: Base angles of an isosceles triangle are congruent. 5-19: The median of a trapezoid 1. is parallel to the bases. 2. has the length equal to the average of the base lengths. Know How To: Determine whether a given quadrilateral is a trapezoid, determine whether a trapezoid is isosceles; calculate the length of the median of a trapezoid; apply theorem 5-19 by setting up and solving algebraic equations. Suggested Exercises: p. 192 C.E #1-11 p. 192-3 #1-11, 14-18 Other Suggested Exercises: p. 197-8 Chapter Review #1-8, 13-16, 19-22 p. 199 Chapter Test (not 10-11) Additionally: There will be problems on the test that involve systems of equations and factoring. Geometry/Trig Unit 5 Review Name: Answer Key Date: For each problem (1-4) a and b are segment lengths, x and y are angle measures, 1. Figure is a Parallelogram 2. Figure is a Square 21 *104* 16 41 yo 40 3. Figure is a Rectangle a = 21 y 32 6 b 4. Figure is a Rhombus 14 25 25 65 6 65 130 3.2 25 25 32 32 32 b = 3.2 a = 3.2 6) TA = RP 4x + 9 = 9x + b x = 3/5 3 = 5x 5). 8. TR = 1/2 * I T e 2x = 1/2 * (5x - 4) 4x = 5x - 4 - x = - 9 2y - 7 = y + 18 y = 25 6 P 5. If TR = 2x and RP = 5x - 9 then x 9 2(x + 7) = 5x - 2 6. If TA = 4x + 9 and RP = 9x + 6 then x= 3/5 7. IfTA = 16 then TP 24 x = ib / 3 2x + 14 = 5x - 2 x= 16/3 y=\ 25 a = 21 x = q \mathfrak{a} = 14 y = 50 x = 104 a = 14 x = 65 b = 16 y = 40 b = 6 b = 3.2/(9D) gamma + 18 2y - 7 b = 6 5x - 2 x + 7 3 * k deg y = 32 32/(y deg) 104 deg / x ^ n * 1 65 deg a = (3sqrt(2))/b x= 45 90 36 deg Geometry/Trig Unit 5 Review - page 2 1 A Name: Date: 2. 3. A B A B B 15y - 35 80 D C D 80 C ABCD is a square m angle DAB = 4x + 4y = 90 m angle ADC = Bx - 4y = 90 4E 4(15) + 4y = 90 4y = 30 y = 15/2 12x = 190 x = 15 D ABCD is a rectangle. C ABCD is a rhombus BD = x + 72 x ^ 2 = x + 72 x ^ 2 - x - 70 = 0 ( x-4) (x + 6) = 0 7-9-0 x=9 x + 8 = 0 x = - 8 triangle c = 60 x = (- G)/(x = 9) AC = x ^ 2 m angle DCB = 20x 20X = 100 15y - 35 = 100 15y = 135 m angle DAB = 15y - 35 m angle ADC = 80 4. 5. A 6. B A 8x - 2 в 14 A B 79 101 E 79/ 5x - 7 = 3x + 1 5X - 7 = r/2 * (6X + 2) x = 4 2X = 8 F 13 E F D C D C ABCD is a parallelogram 3x + 6 12 ABCD is a trapezoid; overline EF ; the median D C AE = 5x - 7 y ^ 2 + y - 6 = 0 y + z = c y+3) (y - 2) = 0 AD = 6x + 2 y ^ 2 = - y + w y - 2 = 0 x = 4 \\ underline -3 y = 2 FC = - y + 6 BF = gamma ^ 2 y = - 3 ABCD is an isosceles trapezoid overline EF the median EF = 3x + 7 5X=10 11x + 4 = 6x + 14 X = 2 (8x - 2 + 3x + b)/2 = 3x + 7 AB = 8x - 2 DC = 3x + b 5X + 9 + 7x + 3 = 180 12X + 12 = 180 x = 14 12x = 168' 5(14)+9= 79 7(14)+3=101 m angle A = 5x + 9 m angle B = 7x + 3 x = 15/16 \\ y= underline 16 2 aligned x = 5/9 3x + 7 y = |0| 100/(20x) x = 14/101