{"id":70279,"date":"2021-03-24T19:31:23","date_gmt":"2021-03-24T14:01:23","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=70279"},"modified":"2022-04-01T10:27:41","modified_gmt":"2022-04-01T04:57:41","slug":"eureka-math-grade-7-module-6-lesson-4","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/eureka-math-grade-7-module-6-lesson-4\/","title":{"rendered":"Eureka Math Grade 7 Module 6 Lesson 4 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 6 Lesson 4 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 6 Lesson 4 Example Answer Key<\/h3>\n

Example 1.
\nFind the measurements of \u2220FAE and \u2220CAD.
\n\"Engage
\nAnswer:
\n2x + 6x + 90 = 170 Vert. \u2220s
\n8x + 90 = 170
\n8x + 90 – 90 = 170 – 90
\n8x = 80
\n(\\(\\frac{1}{8}\\))8x = (\\(\\frac{1}{8}\\))80
\nx = 10
\nThe measurement of \u2220FAE: 2(10)\u00b0 = 20\u00b0
\nThe measurement of \u2220CAD: 6(10)\u00b0 = 60\u00b0
\nTwo lines meet at a point. List the relevant angle relationship in the diagram. Set up and solve an equation to find the value of x. Find the measurement of one of the vertical angles.
\n\"Engage
\nStudents use information in the figure and a protractor to solve for x.
\ni) Students measure a 64\u00b0 angle as shown; the remaining portion of the angle must be x\u00b0 (\u2220s add).
\nii) Students can use their protractors to find the measurement of x\u00b0 and use this measurement to partition the other angle in the vertical pair.
\nAs a check, students should substitute the measured x value into each expression and evaluate; each angle of the vertical pair should be equal to the other. Students can also use their protractor to measure each angle of the vertical angle pair.
\nWith a modified figure, students can write an algebraic equation that they have the skills to solve.
\nv
\n2x = 64 Vert.\u2220s
\n(\\(\\frac{1}{2}\\))2x = (\\(\\frac{1}{2}\\))64
\nx = 32<\/p>\n

Extension:
\n3x = x + 64 vert.\u2220s
\n3x – x = x – x + 64
\n2x = 64
\n(\\(\\frac{1}{2}\\))2x = (\\(\\frac{1}{2}\\))64
\nx = 32
\nMeasurement of each angle in the vertical pair: 3(32)\u00b0 = 96\u00b0<\/p>\n

Example 2.
\nThree lines meet at a point. List the relevant angle relationships in the diagram. Set up and solve an equation to find the value of b.
\n\"Engage
\nAnswer:
\nLet b = c.
\nc + 37 + 43 = 180 \u2220s on a line
\nc + 80 = 180
\nc + 80 – 80 = 180 – 80
\nc = 100
\nSince b = c, b = 100.<\/p>\n

Example 3.
\nThe measurement of an angle is \\(\\frac{2}{3}\\) the measurement of its supplement. Find the measurements of the angle and its supplement.
\nAnswer:
\nangle = \\(\\frac{2}{3}\\)(supplement)
\nangle = \\(\\frac{2}{3}\\)(180 – angle)
\nUsing a tape diagram:
\n\"Engage
\nThe measurements of the two supplementary angles that satisfy these criteria are 72\u00b0 and 108\u00b0.
\nThe tape diagram model is an ideal strategy for this question. If students are not familiar with the tape diagram model, use a Guess and Check table with them. Here is an example of such a table with two entries for guesses that did not result in a correct answer.
\n\"Engage<\/p>\n

Example 4.
\nThree lines meet at a point that is also the endpoint of a ray. List the relevant angle relationships in the diagram. Set up and solve an equation to find the value of z.
\n\"Engage
\nAnswer:
\nLet x\u00b0 be the measurement of the indicated angle.
\nx + 90 + 29 = 180 \u2220s on a line
\nx + 119 = 180
\nx + 119 – 119 = 180 – 119
\nx = 61<\/p>\n

z = x + 90 \u2220s add
\nz = 61 + 90
\nz = 151<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 4 Exercise Answer Key<\/h3>\n

Exercise 1.
\nSet up and solve an equation to find the value of x. List the relevant angle relationship in the diagram. Find the measurement of one of the vertical angles.
\n\"Engage
\nAnswer:
\nStudents use information in the figure and a protractor to solve for x.
\ni) Measure a 132\u00b0 angle as shown; the remaining portion of the original angle must be x\u00b0 (\u2220s add).
\nii) Partition the other angle in the vertical pair into equal angles of x\u00b0.
\nStudents should perform a check (as in Example 1) before solving an equation that matches the modified figure.
\n\"Engage
\nExtension:
\n4x = 132 vert.\u2220s
\n(\\(\\frac{1}{4}\\))4x = (\\(\\frac{1}{4}\\))132
\nx = 33
\nMeasurement of each vertical angle: 5(33)\u00b0 = 165\u00b0
\nNote: Students can check their answers for any question by measuring each unknown angle with a protractor, as all diagrams are drawn to scale.<\/p>\n

Exercise 2.
\nTwo lines meet at a point that is also the endpoint of two rays. List the relevant angle relationships in the diagram. Set up and solve an equation to find the value of b.
\n\"Engage
\nAnswer:
\nb + 95 = 90 + 80 Vert. \u2220s
\nb + 95 = 170
\nb + 95 – 95 = 170 – 95
\nb = 75<\/p>\n

Exercise 3.
\nThe measurement of an angle is \\(\\frac{1}{4}\\) the measurement of its complement. Find the measurements of the two complementary angles.
\nAnswer:
\nangle = \\(\\frac{1}{4}\\)(complement)
\nangle = \\(\\frac{1}{4}\\)(90 – angle)
\nUsing a tape diagram:
\n\"Engage
\n5 units = 90
\n1 unit = 18
\n4 units = 72
\nThe measurements of the two complementary angles that satisfy these criteria are 18\u00b0 and 72\u00b0.<\/p>\n

Exercise 4.
\nTwo lines meet at a point that is also the vertex of an angle. Set up and solve an equation to find the value of x. Find the measurements of \u2220GAF and \u2220BAC.
\n\"Engage
\nAnswer:
\nLet y\u00b0 be the measurement of the indicated angle.
\ny = 180 – (90 + 36) \u2220s on a line
\ny = 54<\/p>\n

4x + y + 5x = 180 \u2220s on a line
\n4x + 54 + 5x = 180
\n9x + 54 = 180
\n9x + 54 – 54 = 180 – 54
\n9x = 126
\n(\\(\\frac{1}{9}\\))9x = (\\(\\frac{1}{9}\\))126
\nx = 14
\nThe measurement of \u2220GAF: 4(14)\u00b0 = 56\u00b0
\nThe measurement of \u2220BAC: 5(14)\u00b0 = 70\u00b0<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 4 Problem Set Answer Key<\/h3>\n

Question 1.
\nFour rays have a common endpoint on a line. Set up and solve an equation to find the value of c.
\n\"Eureka
\nAnswer:
\n59 + d = 90 Complementary \u2220s
\n59 – 59 + d = 90 – 59
\nd = 31<\/p>\n

d + c + 140 = 180 \u2220s on a line
\n31 + c + 140 = 180
\nc + 171 = 180
\nc + 171 – 171 = 180 – 171
\nc = 9<\/p>\n

Question 2.
\nLines BC and EF meet at A. Set up and solve an equation to find the value of x. Find the measurements of \u2220EAH and \u2220HAC.
\n\"Eureka
\nAnswer:
\n\u2220BAE + 57 = 90 Complementary \u2220s
\n\u2220BAE + 57 – 57 = 90 – 57
\n\u2220BAE = 33<\/p>\n

\u2220BAE + 3x + 4x = 180 \u2220s on a line
\n33 + 3x + 4x = 180
\n33 + 7x = 180
\n33 – 33 + 7x = 180 – 33
\n7x = 147
\n(\\(\\frac{1}{7}\\))7x = (\\(\\frac{1}{7}\\))147
\nx = 21
\nThe measurement of \u2220EAH: 3(21)\u00b0 = 63\u00b0
\nThe measurement of \u2220HAC: 4(21)\u00b0 = 84\u00b0<\/p>\n

Question 3.
\nFive rays share a common endpoint. Set up and solve an equation to find the value of x. Find the measurements of \u2220DAG and \u2220GAH.
\n\"Eureka
\nAnswer:
\n2x + 36.5 + 36.5 + 2x + 3x = 360 \u2220s at a point
\n7x + 73 = 360
\n7x + 73 – 73 = 360 – 73
\n7x = 287
\n(\\(\\frac{1}{7}\\))7x = (\\(\\frac{1}{7}\\))287
\nx = 41
\nThe measurement of \u2220EAF: 2(41)\u00b0 = 82\u00b0
\nThe measurement of \u2220GAH: 3(41)\u00b0 = 123\u00b0<\/p>\n

Question 4.
\nFour lines meet at a point which is also the endpoint of three rays. Set up and solve an equation to find the values of x and y.
\n\"Eureka
\nAnswer:
\n2y + 12 + 15 + 90 = 180 \u2220s on a line
\n2y + 117 = 180
\n2y + 117 – 117 = 180 – 117
\n2y = 63
\n(\\(\\frac{1}{2}\\))2y = (\\(\\frac{1}{2}\\))63
\ny = 31.5<\/p>\n

3x = 2y Vert. \u2220s
\n3x = 2(31.5)
\n3x = 63
\n(\\(\\frac{1}{3}\\))3x = (\\(\\frac{1}{3}\\))63
\nx = 21<\/p>\n

Question 5.
\nTwo lines meet at a point that is also the vertex of a right angle. Set up and solve an equation to find the value of x. Find the measurements of \u2220CAE and \u2220BAG.
\n\"Eureka
\nAnswer:
\n\u2220DAB = 4x vert. \u2220s
\n\u2220DAG = 90 + 15 = 105 \u2220s add
\n4x + 3x = 105
\n7x = 105 \u2220s add
\n(\\(\\frac{1}{7}\\))7x = (\\(\\frac{1}{7}\\))105
\nx = 15
\nThe measurement of \u2220CAE: 4(15)\u00b0 = 60\u00b0
\nThe measurement of \u2220BAG: 3(15)\u00b0 = 45\u00b0<\/p>\n

Question 6.
\n6. Five angles are at a point. The measurement of each angle is one of five consecutive, positive whole numbers.
\na. Determine the measurements of all five angles.
\nb. Compare the expressions you used for the five angles and their combined expression. Explain how they are equivalent and how they reveal different information about this situation.
\nAnswer:
\nx + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 360
\n5x + 10 = 360
\n5x + 10 – 10 = 360 – 10
\n5x = 350
\n(\\(\\frac{1}{5}\\))5x = (\\(\\frac{1}{5}\\))350
\nx = 70
\nAngle measures are 70\u00b0, 71\u00b0, 72\u00b0, 73\u00b0, and 74\u00b0.<\/p>\n

b. By the commutative and associative laws, x + (x + 1) + (x + 2) + (x + 3) + (x + 4) is equal to (x + x + x + x + x) + (1 + 2 + 3 + 4), which is equal to 5x + 10. The first expression, x + (x + 1) + (x + 2) + (x + 3) + (x + 4), shows the sum of five unknown numbers where the second is 1 degree more than the first, the third is 1 degree more than the second, and so on. The expression 5x + 10 shows the sum of five times an unknown number with 10.<\/p>\n

Question 7.
\nLet x\u00b0 be the measurement of an angle. The ratio of the measurement of the complement of the angle to the measurement of the supplement of the angle is 1:3. The measurement of the complement of the angle and the measurement of the supplement of the angle have a sum of 180\u00b0. Use a tape diagram to find the measurement of this angle.
\nAnswer:
\n(90 – x):(180 – x) = 1:3
\n\"Eureka
\nThe measurement of the angle that satisfies these criteria is 45\u00b0.<\/p>\n

Question 8.
\nTwo lines meet at a point. Set up and solve an equation to find the value of x. Find the measurement of one of the vertical angles.
\n\"Eureka
\nAnswer:
\nA solution can include a modified diagram (as shown) and the supporting algebra work:
\n3x = 117 vert. \u2220s
\n(\\(\\frac{1}{3}\\))3x = (\\(\\frac{1}{3}\\))117
\nx = 39
\n\"Eureka
\nEach vertical angle: 4(39)\u00b0 = 156\u00b0
\nSolutions may also include the full equation and solution:
\n4x = x + 117
\n4x – x = x – x + 117
\n3x = 117
\n(\\(\\frac{1}{3}\\))3x = (\\(\\frac{1}{3}\\))117
\nx = 39<\/p>\n

Question 9.
\nThe difference between three times the measurement of the complement of an angle and the measurement of the supplement of that angle is 20\u00b0. What is the measurement of the angle?
\nAnswer:
\n3(90 – x) – (180 – x) = 20
\n270 – 3x – 180 + x = 20
\n90 – 2x = 20
\n90 – 90 – 2x = 20 – 90
\n– 2x = – 70
\n( – \\(\\frac{1}{2}\\))( – 2x) = ( – \\(\\frac{1}{2}\\))( – 70)
\nx = 35
\nThe measurement of the angle is 35\u00b0.<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 4 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nLines BC and EF meet at A. Rays AG and AD form a right angle. Set up and solve an equation to find the values of x and w.
\n\"Eureka
\nAnswer:
\n\u2220BAE = 60 Vert. \u2220s
\n\u2220BAG = 10\u00a0 \u00a0 \u00a0\u2220s add<\/p>\n

x + \u2220BAG = 90\u00a0 \u00a0 \u00a0 \u00a0 \u00a0Complementary \u2220s
\nx + 10 = 90
\nx + 10 – 10 = 90 – 10
\nx = 80<\/p>\n

x + w + 60 = 180\u00a0 \u00a0 \u00a0\u2220s on a line
\n80 + w + 60 = 180
\n140 + w = 180
\n140 + w – 140 = 180 – 140
\nw = 40<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 6 Lesson 4 Answer Key Eureka Math Grade 7 Module 6 Lesson 4 Example Answer Key Example 1. Find the measurements of \u2220FAE and \u2220CAD. Answer: 2x + 6x + 90 = 170 Vert. \u2220s 8x + 90 = 170 8x + 90 – 90 = 170 – … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/70279"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=70279"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/70279\/revisions"}],"predecessor-version":[{"id":92205,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/70279\/revisions\/92205"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=70279"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=70279"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=70279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}