Geometr and Honors Geometr Summer Review Packet 04 This will not be graded. It is for our benefit onl. The problems in this packet are designed to help ou review topics from previous mathematics courses that are important to our success in Geometr or Honors Geometr. Please tr to do each problem and show the work that goes with that answer. Bring the packet with ou to our Geometr or Honors Geometr class on the first da of school. It is recommended that ou work with one or more people. Before ou leave school, write down the names, phone numbers, and/or email addresses for at least two people who are also taking Geometr or Honors Geometr in the fall. Name Phone Email Name Phone Email During the summer, answer kes will be posted on the Sherwood website. Enjo our summer. We are looking forward to seeing ou in the fall. (If ou have an questions, please contact the math office at (30)94-353)
Solve each equation. Then check our solution. Solving Linear Equations. 8 6 d 3 8. 6 0. 3 5 7 0 4 9. 7 3. 5r 55 0. 3k 5 7k 4. 45 v 5. 8s 9 7s 6 5. 80. 7( 3) 7 6. z 45 5 3. 8 4(3c 5) 7. 8 5p 3 4. 8 0 0 5. ( 7 5) ( 3) 45
Word Problems 6. ABC is an equilateral triangle. B Find. Find the length of side AB. 3 + + 7 A C 7. The perimeter of rectangle PQRS is 40 cm. Q Find. Find the area of PQRS. + 6 P - 8 R S 8. The two line segments below have the same length. Find the length of each segment. + 3 4-9. The perimeter of square JKLM is 64 units. Find the length of each side. J K a + 7 M L 0. ***The sides of a triangle are in the ratio :3:5. The perimeter of the triangle is 55 feet. Find the length of each side.. ***The sides of a rectangle are in the ratio of 3:4. If the area of the rectangle 300 square meters what is the length and width?. ***The base of a triangle is 3 less than twice the height. The area of the triangle is 7 cm. What is the length of the base and height?
Pthagorean Theorem 5 c a b c Eample: Step : Set-up Equation 5 c Step : Multipl 544 c Step 3: Add 69 c Step 4: Take square root of each side. 69 c Step 5: Simplif square roots. 3 c 3. 3 b 5 6. c 8 5 4. 8 0 7. a 5 9 5. 6 b 0 b Distributive Propert and FOIL Epand the following using distributive propert and/or FOIL. 8. 4( 5) 9. ( 5)( 3 8z 6)
c 30. 4 0 8 h 33. ( )( 3 4) 3. ( 3)( 5 ) 34. ( ) 7 3. ( 5)( 5) 35. ( 3 ) Evaluating Formulas/Solving for Missing Variables Eample : P l w; P = 48, l = 6, solve for w. Begin with given equation: Substitute given values for P and l: Solve for the remaining variable (w) Simplif: P l w 48 (6) w 48 w Subtract from both sides: 48 w 36 w Divide both sides b : 36 w Simplif both sides: 8 w ANSWER: w = 8
Eample : A 3.4r A = 34, solve for r Begin with given equation: Substitute given value for A: Solve for the remaining variable (r): Divide both sides b 3.4 ANSWER : r = 0 Simplif To get rid of the square, A 3.4r 34 3.4r 34 3.4r 3.4 3.4 00 r take the square root of both sides 00 r Simplif 0 r For each of the equations given below:. Substitute the given values for the variable(s) specified. Solve for the remaining variable (if necessar, round to nearest hundredth). 36. V lwh ; V = 60, l = 5, w = 4 4. V s h ; V = 0, h = 0 3 37. S 4( 34. ) r ; S = 34 4. S 6s ; S = 94 38. V 4 ( r 3 34. ) 3 ; r = 3 43. V s 3 ; s = 4 39. A ( b b ) h ; b 8, b 4 A 4, 44. S ( 34. ) rh ( 34. ) r ; r =, S = 00.4 40. V ( 34. ) r h ; V = 0, r = 6
GRAPHING Instruction: Plot each point on the graph below. Remember, coordinate pairs are labeled (, ). Label each point on the graph with the letter given. 45. A(3, 4) 46. B(4, 0) 47. C(-4, ) 48. D(-3, -) 49. E(0, 7) Eample: F(-6, ) + F F - + Determine the coordinates for each point below: 50. Eample. (, 3) 5. (, ) 5. (, )
53. (, ) 55. (, ) 57. (, ) 54. (, ) 56. (, ) 58. (, ) Slope and Midpoint slope midpo int, Before using the slope or midpoint formula, ou must label our,,, and. Eample: For the points (0, ) and (, 5), label our,,, and so that ou can use them in our slope or midpoint formula. Step : Label one point as point and the other as point. Point Point (0, ) and (, 5)
Step : Label the - and -coordinates of point as and, respectivel. Then, label the - and -coordinates of point as and. Point Point (0, ) and (, 5) Eample : Use the slope formula to find the slope of the line between (0, ) and (, 3). Step : Label,,, and. Point : (0, ) Point : (, 3) Step : Plug values into the slope formula. slope 3 slope 0 Step 3: Simplif. slope 59. Find the slope of the line between (, 3) and (5, 5). 60. Find the slope of the line between (, 3) and (9, 7).
6. Find the slope of the line between (, 6) and (-, 3). 6. Find the slope of the line between (-3, 9) and (-7, 6). 63. Find the slope of the line between (4.5, -) and (5.3, ). Eample : Use the midpoint formula to find the midpoint of the segment below. Step : Find the coordinates of the endpoints (-4, ) and (4, 3) Step : Label,,, and. (-4, ) and (4, 3) Step 3: Plug into midpoint formula. and simplif. midpo int, midpo int ( 4) 4 3, 0 4 midpo int, = (0, )
64. Find the midpoint of the segment with endpoints (0, 0) and (4, ). 65. Find the midpoint of the segment with endpoints (-3, -) and (3, 3). 66. Find the midpoint of the segment with endpoints (-5, ) and (, -3). 67. Find the midpoint of the segment below. 5 0-5 0 5 68. Find the midpoint of the segment below. -5 5 0-5 0 5-5
Factoring (Honors Onl Section) 69. 3 74. 3 40 70. 9 75. 0 7. 6 5 76. 6 7. 7 0 77. 3 0 5 73. 9 8 78. 4 8 49 Solve for (Honors Onl Section) 79. 6 8. 6 80. m 0m 8. 3 6
Sstem of Equations (Honors Onl Section) Substitution Method Eample solve: + = 4 3 + = 48. Solve for a variable in one equation. + = 4 We solved for in the st equation. = 4. Substitute this value for in the other 3 + = 48 equation and solve for. 3 + (4 ) = 48 3 +8 = 48 + 8 = 48 = 0 3. Substitute this value for in the equation found in Step and find. 4. Go back and check answers in both equations. + = 4 3 + = 48 0 + ( 6) = 4 3(0) + ( 6) = 48 4 = 4 60 + ( ) = 48 48 = 48 = 4 = 4 0 = -6 Solve b substitution... 3 5 3 3. 5a b 7 4. 3b 4a 6 3 4 4
Solve b linear combination. Eample solve: 3 = 8 + = 7. Since the coefficients of are opposites 3 = 8 add the two equations together. + = 7 5 = 5. Solve the equation. = 3 3. Substitute answer into one of the 3 = 8 original equations and solve. 3(3) = 8 - = 8 9 = 4. Check solution in both equations. 3 = 8 + = 7 Solution (3,) 3(3) = 8 (3) + = 7 8 = 8 7 = 7 Eample solve: 3 = 6 3 + 4 =-5. The coefficients of either or are not opposite. However, the variable could be eliminated b adding the equations together if we first multipl the first equation b 3 and the second b -. The variable could be eliminated b adding the equations together if we first multipl the first equation b 4 and the second b 3. We will do the latter. 4( 3 = 6) 8 = 4 3(3 + 4 = -5) 9 + = -75. Add two equations 7 = - 5 3. Solve = -3 4. Substitute into either of the original equations to find. 3 = 6 (-3)- 3 = 6-6 - 3 = 6 5. Check answer in both equations -3 = Solution ( -3, -4) = -4 Solve b linear combination.. 3 7. 5 30 3 7 6 3. 5 4. 3 7 4 5 47 3 4 0