GH Chapter 2 Test Review-includes Constructions
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1 Name: Class: Date: Show All Work. Test will include 2 proofs from the proof practice worksheet assigned week of 9/8. GH Chapter 2 Test Review-includes Constructions ID: A 1. What is the value of x? State the postulate or theorem that justifies your work. 4. A gardener has 26 feet of fencing for a garden. To find the width of the rectangular garden, the gardener uses the formula P=2l+2w, where P is the perimeter, l is the length, and w is the width of the rectangle. The gardener wants to fence a garden that is 8 feet long. How wide is the garden? Solve the equation for w, and justify each step using an algebraic proof. 5. Write a two-column proof to prove that x= 12, if.m JKL = 100, m JKM = 6x+8, and m MKL=2x Two angles with measures (2x 2 + 3x 5) and (x x 7) are supplementary. Find the value of x and the measure of each angle. 3. Find the values of x and y. State the postulate(s) or theorem(s) that justifies your work. m JKL= Two lines intersect to form two pairs of vertical angles. 1 with measure (20x+ 7) and 3 with measure (5x+ 7y+ 49) are vertical angles. 2 with measure (3x 2y+ 30) and 4 are vertical angles. Find the values x and y and the measures of all four angles. 1
2 Name: ID: A 7. Write a justification for each step, proving that EF=GH, given that EG= FH. Statement Reason 1. EG= FH 1. Given information 2 EG=EF+FG 2. FH= FG+ GH 3 EF+FG= FG+GH EF=GH Use the given plan to write a two-column proof. Given: m 1 + m 2 = 90, m 3 + m 4 = 90, m 2 = m 3 9. Which is not a possible value for y in the figure below? Prove: m 1 = m 4 Plan: Since both pairs of angle measures add to 90, use substitution to show that the sums of both pairs are equal. Since m 2 = m 3, use substitution again to show that sums of the other pairs are equal. Use the Subtraction Property of Equality to conclude that m 1 = m 4. a. 70 b. 115 c. 55 d Solve for x. State the postulate or theorem that justifies your work. 2
3 Name: ID: A Write a two-column proof. 11. FG, HI, JK, LM, and NO are all parallel to each other. HI AB and LM AB. Prove 2 5. Choose the phrase that completes the following statement as stated by the Point, Line, and Plane Postulates: 14. A line contains at least two points. a. always b. never c. sometimes d. Point, Line, and Plane Postulates do not address this topic directly. 15. Two distinct points are connected by two distinct lines. 12. Given: AB CD; FG HJ. Prove: 3 8 a. never b. always c. sometimes d. Point, Line, and Plane Postulates do not address this topic directly. 16. A line passing through two distinct points in one plane lies completely in that plane. a. always b. never c. sometimes d. Point, Line, and Plane Postulates do not address this topic directly. 17. If two distinct lines intersect, their intersection is one point. 13. If the ratio of m 1 to m 2 is 5 to 4, find m 2. a. always b. never c. sometimes d. Point, Line, and Plane Postulates do not address this topic directly. 18. Decide which one of the following statements is false. a. Three noncollinear points determine a plane. b. A line contains at least two points. c. Through any two distinct points there exists exactly one line. d. Any three points lie on a distinct line. 3
4 Name: ID: A State the postulate indicated by the diagram. 26. Construct the bisector of the segment. 19. Identify the property that makes the statement true. 20. If XY = MN, then MN = XY. 21. If m P = m R and m R = m T, then m P = m T. 27. Construct B so that B A. 22. If MP = PQ and PQ = QR, then MP = QR. 23. If m 1 + m 2 = 25 and m 1 = 9, then 9 o + m 2= 25 o. 24. Construct JK so that JK FG Construct CJ, the bisector of C. 4
5 GH Chapter 2 Test Review-includes Constructions Answer Section x = 6; 85 ; 95 Step 1 Create an equation The angles are supplements and their sum equals 180. (2x 2 + 3x 5)+(x x 7)=180 Step 2 Solve the equation 3x x 12= 180 3x x 192= 0 (3x+ 32)(x 6)=0 x= 32 3 or 6. When x= 32 3, the measurement of the second angle is x x 7= Angles cannot have negative measurements, so x= 6. Step 3 Solve for the required values The measurement of the first angle is 2x 2 + 3x 5 = 2(6) 2 + 3(6) 5 = 85. The measurement of the second angle is x x 7 = (6) (6) 7= x = 15, y = [1] Substitution Property of Equality [2] Division Property of Equality The garden is 5 ft wide. P=2l+2w Given equation 26= 2(8)+2w [1] Substitution Property of Equality 26= 16+2w Simplify. 16= 16 10= 2w Subtraction Property of Equality Simplify = 2w 2 [2] Division Property of Equality 5= w Simplify. w=5 Symmetric Property of Equality 1
6 5. [1] Angle Addition Postulate [2] Division Property of Equality m JKL = m JKM + m MKL [1] Angle Addition Postulate 100 = (6x + 8) + (2x 4) Substitution Property of Equality 100 = 8x + 4 Simplify. 96= 8x Subtraction Property of Equality 12= x [2] Division Property of Equality x= 12 Symmetric Property of Equality 6. x= 7; y= 9; 147 ; 147 ; 33 ; 33 Step 1 Create a system of equations. m 1 = m 3 20x+ 7=5x+ 7y x 7y= 42 The sum of the measures of supplementary angles equals 180. m = x+ 7+ 3x 2y+ 30= x 2y= 143 Create a system of equations. 15x 7y= 42 23x 2y= 143 Step 2 Solve the system of equations. 15x 7y= 42 23x 2y= x+ 14y= x 14y= 1001 Multiply the first equation by 2. Multiply the second equation by x= 917 Add the two equations together. x= 7 Divide both sides by 131. Solve for y. Substitute x= 7 into 15x 7y= (7) 7y= 42 y= 9 The values are x= 7 and y= 9. Step 3 Solve for the four angles. Angle 1: (20(7) + 7) = 147 Angle 2: (3(7) 2(9)+30) = 33 Angle 3: (5(7)+7(9)+ 49) =147 Angle 4 and angle 2 are vertical and thus have equal measures. The measurement of angle 4 is 33. The measures of all four angles are 147, 147, 33, and 33. 2
7 7. [1] Segment Addition Postulate [2] Substitution Property of Equality EG = FH Given information EG = EF + FG Segment Addition Postulate FH = FG + GH Segment Addition Postulate EF + FG = FG + GH Substitution Property of Equality EF=GH Subtraction Property of Equality 8. [1] m 3 + m 4 = 90 [2] Substitution Property [3] Subtraction Property of Equality Proof: Statements Reasons 1. m 1 + m 2 = Given 2. m 3 + m 4 = Given 3. m 1 + m 2 = m 3 + m 4 3. Substitution Property 4. m 2 = m 3 4. Given 5. m 1 + m 2 = m 2 + m 4 5. Substitution Property 6. m 1 = m 4 6. Subtraction Property of Equality 9. C 10. x = Sample: Given: FG Ä HI Ä JK Ä LM Ä NO Prove: 2 5 Proof: Statements Reasons 1. FG Ä HI Ä JK Ä LM Ä NO 1. Given 2. HI AB, LM AB 2. Given 3. 2, 5 are right angles. 3. Definition of perpendicular 4. m 2=90, m 5=90 4. Definition of right angle 5. m 2= m 5 5. Substitution Property 6. m 2 m 5 6. Definition of congruent angles 12. Sample: Given: AB CD, FG HJ Prove: 3 8 Proof: Statements Reasons 1. AB CD, FG HJ 1. Given 2. 3 and 8 are right angles. 2. Definition of perpendicular 3. m 3=90, m 8=90 3. Definition of right angles 4. m 3= m 8 4. Substitution Property 5. m 3 m 8 5. Definition of congruent angles 3
8 A 15. A 16. A 17. A 18. D 19. Through any three noncollinear points there exists exactly one plane. 20. Symmetric Property of Equality 21. Transitive Property of Equality 22. Transitive Property of Equality 23. Substitution Property of Equality
9 27. 5
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