{"id":83891,"date":"2022-06-01T16:40:04","date_gmt":"2022-06-01T11:10:04","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=83891"},"modified":"2022-05-10T14:19:41","modified_gmt":"2022-05-10T08:49:41","slug":"eureka-math-grade-8-module-3-lesson-7","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-3-lesson-7\/","title":{"rendered":"Eureka Math Grade 8 Module 3 Lesson 7 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 3 Lesson 7 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 3 Lesson 7 Exercise Answer Key<\/h3>\n

Use the diagram below to prove the theorem: Dilations preserve the measures of angles.<\/p>\n

Let there be a dilation from center O with scale factor r. Given \u2220PQR, show that since P’=Dilation(P),
\nQ’=Dilation(Q), and R’=Dilation(R), then |\u2220PQR|=|\u2220P’Q’R’|. That is, show that the image of the angle after a dilation has the same measure, in degrees, as the original.
\n\"Eureka
\nAnswer:
\nUsing FTS, we know that the line containing \\(\\overrightarrow{Q^{\\prime} P^{\\prime}}\\) is parallel to the line containing \\(\\overrightarrow{Q P}\\) and that the line containing \\(\\overrightarrow{Q R}\\) is parallel to the line containing \\(\\overrightarrow{Q R}\\). We also know that there exists just one line through a given point, parallel to a given line. Therefore, we know that \"Eureka must intersect \"Eureka at a point. We know this because there is already a line that goes through point Q’ that is parallel to \"Eureka, and that line is \"Eureka. Since \"Eureka cannot be parallel to \"Eureka, it must intersect it. We let the intersection of \"Eureka and \"Eureka be named point B. Alternate interior angles of parallel lines cut by a transversal are equal in measure. Parallel lines QR and Q’R’ are cut by transversal \\(\\overline{\\boldsymbol{Q}^{\\prime} \\boldsymbol{B}}\\). Therefore, the alternate interior angles P’Q’R’ and Q’BQ are equal in measure. Parallel lines \"Eureka and \"Eureka are cut by transversal \\(\\overline{Q B} .\\) Therefore, the alternate interior angles PQR and Q’BQ are equal in measure. Since the measures of \u2220P’Q’R’ and \u2220PQR are equal to the measure of \u2220QB’Q, then the measure of \u2220PQR is equal to the measure of \u2220P’ Q’ R’.<\/p>\n

Eureka Math Grade 8 Module 3 Lesson 7 Problem Set Answer Key<\/h3>\n

Question 1.
\nA dilation from center O by scale factor r of a line maps to what? Verify your claim on the coordinate plane.
\nAnswer:
\nThe dilation of a line maps to a line.
\nSample student work is shown below.
\n\"Eureka<\/p>\n

Question 2.
\nA dilation from center O by scale factor r of a segment maps to what? Verify your claim on the coordinate plane.
\nAnswer:
\nThe dilation of a segment maps to a segment.
\nSample student work is shown below.
\n\"Eureka<\/p>\n

Question 3.
\nA dilation from center O by scale factor r of a ray maps to what? Verify your claim on the coordinate plane.
\nAnswer:
\nThe dilation of a ray maps to a ray.
\nSample student work is shown below.
\n\"Eureka<\/p>\n

Question 4.
\nChallenge Problem:
\nProve the theorem: A dilation maps lines to lines.
\nLet there be a dilation from center O with scale factor r so that P’=Dilation(P) and Q’=Dilation(Q). Show that line PQ maps to line P’Q’ (i.e., that dilations map lines to lines). Draw a diagram, and then write your informal proof of the theorem. (Hint: This proof is a lot like the proof for segments. This time, let U be a point on line PQ that is not between points P and Q.)
\nAnswer:
\nSample student drawing and response are below:
\n\"Eureka
\nLet U be a point on line PQ. By the definition of dilation, we also know that U’=Dilation(U). We need to show that U’ is a point on line P’Q’. If we can, then we have proven that a dilation maps lines to lines.
\nBy the definition of dilation and FTS, we know that \\(\\frac{\\left|O P^{\\prime}\\right|}{|O P|}\\) =\\(\\frac{\\left|O Q^{\\prime}\\right|}{|O Q|}\\) and that \\(\\overleftrightarrow{P Q}\\) is parallel to \"Eureka. Similarly, we know that \"Eureka and that line QU is parallel to line Q’U’. Since U is a point on line PQ, then we also know that line PQ is parallel to line Q’U’. But we already know that \"Eureka is parallel to \"Eureka. Since there can only be one line that passes through Q’ that is parallel to line PQ, then line P’Q’ and line Q’U’ must coincide. That places the dilation of point U, U’, on the line P’Q’, which proves that dilations map lines to lines.<\/p>\n

Eureka Math Grade 8 Module 3 Lesson 7 Exit Ticket Answer Key<\/h3>\n

Dilate \u2220ABC with center O and scale factor r=2. Label the dilated angle, \u2220A’B’C’.
\n\"Engage
\nAnswer:
\n\"Engage<\/p>\n

Question 1.
\nIf \u2220ABC=72\u00b0, then what is the measure of \u2220A’ B’ C’?
\nAnswer:
\nSince dilations preserve angles, then \u2220A’B’C’=72\u00b0.<\/p>\n

Question 2.
\nIf the length of segment AB is 2 cm, what is the length of segment A’B’?
\nAnswer:
\nThe length of segment A’B’ is 4 cm.<\/p>\n

Question 3.
\nWhich segments, if any, are parallel?
\nAnswer:
\nSince dilations map segments to parallel segments, then \\(\\overline{A B}\\) || \\(\\overline{A^{\\prime} B^{\\prime}}\\), and \\(\\overline{\\boldsymbol{B C}}\\) || \\(\\overline{\\boldsymbol{B}^{\\prime} \\boldsymbol{C}^{\\prime}}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 3 Lesson 7 Answer Key Eureka Math Grade 8 Module 3 Lesson 7 Exercise Answer Key Use the diagram below to prove the theorem: Dilations preserve the measures of angles. Let there be a dilation from center O with scale factor r. Given \u2220PQR, show that since P’=Dilation(P), … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83891"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=83891"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83891\/revisions"}],"predecessor-version":[{"id":137256,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83891\/revisions\/137256"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=83891"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=83891"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=83891"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}