{"id":83506,"date":"2021-04-05T16:30:34","date_gmt":"2021-04-05T11:00:34","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=83506"},"modified":"2022-04-01T10:26:05","modified_gmt":"2022-04-01T04:56:05","slug":"eureka-math-grade-8-module-7-lesson-15","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/eureka-math-grade-8-module-7-lesson-15\/","title":{"rendered":"Eureka Math Grade 8 Module 7 Lesson 15 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 15 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 7 Lesson 15 Classwork Answer Key<\/h3>\n

Classwork
\nProof of the Pythagorean Theorem
\n\"Engage
\nAnswer:
\nUsing one right triangle, we created 3 right triangles. Name those triangles.
\nThe three triangles are \u25b3ABC, \u25b3ACD, and \u25b3BCD.
\nWe can use our basic rigid motions to reorient the triangles so they are easier to compare, as shown below.<\/p>\n

\"Engage
\nAnswer:
\n\u2192 The next step is to show that these triangles are similar. Begin by showing that
\n\u25b3ADC ~ \u25b3ACB. Discuss in your group.
\n\u25b3ADC and \u25b3ACB are similar because they each have a right angle, and they each share \u2220A. Then, by the AA criterion for similarity, \u25b3ADC ~ \u25b3ACB.<\/p>\n

\u2192 Now, show that \u25b3ACB ~ \u25b3CDB. Discuss in your group.
\n\u25b3ACB ~ \u25b3CDB because they each have a right angle, and they each share \u2220B. Then, by the AA criterion for similarity,
\n\u25b3ACB ~ \u25b3CDB.<\/p>\n

\u2192 Are \u25b3ADC and \u25b3CDB similar? Discuss in your group.
\nWe know that similarity has the property of transitivity; therefore, since \u25b3ADC ~ \u25b3ACB, and
\n\u25b3ACB ~ \u25b3CDB, then \u25b3ADC ~ \u25b3CDB.<\/p>\n

\u2192 Let\u2019s identify the segments that comprise side c as follows: |AD| = x and |BD| = y. (Ensure that students note x and y in their student materials.) Using this notation, we see that side c is equal to the sum of the lengths x and y (i.e., x + y = c).
\n\"Engage
\n\u2192 If we consider \u25b3ADC and \u25b3ACB, we can write a statement about corresponding sides being equal in a ratio that helps us reach our goal of showing a2<\/sup> + b2<\/sup> = c2<\/sup>. Discuss in your group.
\nUsing \u25b3ADC and \u25b3ACB, we can write
\n\\(\\frac{x}{b}\\) = \\(\\frac{b}{c}\\).
\nNow solve the equation for x.
\nx = \\(\\frac{b^{2}}{c}\\)<\/p>\n

\u2192 Using \u25b3ACB and \u25b3CDB gives us another piece that we need. Discuss in your group.
\nUsing \u25b3ACB and \u25b3CDB, we can write
\n\\(\\frac{a}{y}\\) = \\(\\frac{c}{a}\\).
\nNow solve the equation for y.
\n\\(\\frac{a^{2}}{c}\\) = y<\/p>\n

\u2192 We know that x + y = c, and we just found expressions equal to x and y. Use this information to show that a2<\/sup> + b2<\/sup> = c2<\/sup>. Discuss in your group.
\nBy substituting \\(\\frac{b^{2}}{c}\\) for x and \\(\\frac{a^{2}}{c}\\) for y in c = x + y, we have
\n\\(\\frac{b^{2}}{c}\\) + \\(\\frac{a^{2}}{c}\\) = c.
\nMultiplying through by c we have
\nb2<\/sup> + a2<\/sup> = c2<\/sup>.
\nBy the commutative property of addition we can rewrite the left side as
\na2<\/sup> + b2<\/sup> = c2<\/sup>.<\/p>\n

Discussion
\n\"Engage
\nLet\u2019s draw three squares on the right triangle. Notice that we can use the altitude to divide the large square, of area c2<\/sup>, into two rectangles as shown. Call them rectangle I and rectangle II.
\n\"Engage
\n\u2192 What would it mean, geometrically, for a2<\/sup> + b2<\/sup> to equal c2<\/sup>?
\nIt means that the sum of the areas of a2<\/sup> and b2<\/sup> is equal to the area c2<\/sup>.
\nThere are two possible ways to continue; one way is by examining special cases on grid paper, as mentioned in the scaffolding box on the previous page, and showing the relationship between the squares physically. The other way is by using the algebraic proof of the general case that continues below.
\n\u2192 What is the area of rectangle I?
\nThe area of rectangle I is xc.
\nThis is where the proof using similar triangles just completed is helpful. We said that x = \\(\\frac{b^{2}}{c}\\). Therefore, the area of rectangle I is
\nxc = \\(\\frac{b^{2}}{c}\\)\u22c5c = b2<\/sup>.<\/p>\n

\u2192 Now use similar reasoning to determine the area of rectangle II.
\nThe area of rectangle II is yc. When we substitute \\(\\frac{a^{2}}{c}\\) for y we get yc = \\(\\frac{a^{2}}{c}\\)\u22c5c = a2<\/sup>.
\n\u2192 Explain how the work thus far shows that the Pythagorean theorem is true.
\nThe Pythagorean theorem states that given a right triangle with lengths a, b, c, then a2<\/sup> + b2<\/sup> = c2<\/sup>. The diagram shows that the area of the rectangles drawn off of side c have a sum of a2<\/sup> + b2<\/sup>. The square constructed off of side c clearly has an area of c2<\/sup>. Thus, the diagram shows that the areas
\na2<\/sup> + b2<\/sup> are equal to the area of c2<\/sup>, which is exactly what the theorem states.<\/p>\n

To solidify student understanding of the proof, consider showing students the six-minute video located at http:\/\/www.youtube.com\/watch?v = QCyvxYLFSfU. If multiple computers or tablets are available, have small groups of students watch the video together so they can pause and replay parts of the proof as necessary.
\n\"Engage<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 15 Problem Set Answer Key<\/h3>\n

Question 1.
\nFor the right triangle shown below, identify and use similar triangles to illustrate the Pythagorean theorem.
\n\"Eureka
\nAnswer:
\nFirst, I must draw a segment that is perpendicular to side AB that goes through point C. The point of intersection of that segment and side AB will be marked as point D.
\n\"Eureka
\nThen, I have three similar triangles, \u25b3ABC, \u25b3CBD, and \u25b3ACD, as shown below.
\n\"Eureka
\n\u25b3ABC and \u25b3CBD are similar because each one has a right angle, and they both share \u2220B. By AA criterion,
\n\u25b3ABC ~ \u25b3CBD. \u25b3ABC and \u25b3ACD are similar because each one has a right angle, and they both share \u2220A.
\nBy AA criterion, \u25b3ABC ~ \u25b3ACD. By the transitive property, we also know that \u25b3ACD ~ \u25b3CBD.
\nSince the triangles are similar, they have corresponding sides that are equal in ratio. For \u25b3ABC and \u25b3CBD,
\n\\(\\frac{9}{15}\\) = \\(\\frac{|B D|}{9}\\),
\nwhich is the same as 92<\/sup> = 15(|BD|).
\nFor \u25b3ABC and \u25b3ACD,
\n\\(\\frac{12}{15}\\) = \\(\\frac{|A D|}{12}\\),
\nwhich is the same as 122<\/sup> = 15(|AD|).
\nAdding these two equations together I get
\n92<\/sup> + 122<\/sup> = 15(|BD|) + 15(|AD|).
\nBy the distributive property,
\n92<\/sup> + 122<\/sup> = 15(|BD| + |AD|);
\nhowever, |BD| + |AD| = |AC| = 15. Therefore,
\n92<\/sup> + 122<\/sup> = 15(15)
\n92<\/sup> + 122<\/sup> = 152<\/sup>.<\/p>\n

Question 2.
\nFor the right triangle shown below, identify and use squares formed by the sides of the triangle to illustrate the Pythagorean theorem.
\n\"Eureka
\nAnswer:
\nThe sum of the areas of the smallest squares is 152<\/sup> cm2<\/sup> + 202<\/sup> cm2<\/sup> = 625 cm2<\/sup>. The area of the largest square is
\n252<\/sup> cm2<\/sup> = 625 cm2<\/sup>. The sum of the areas of the squares off of the legs is equal to the area of the square off of the hypotenuse; therefore, a2<\/sup> + b2<\/sup> = c2<\/sup>.<\/p>\n

Question 3.
\nReese claimed that any figure can be drawn off the sides of a right triangle and that as long as they are similar figures, then the sum of the areas off of the legs will equal the area off of the hypotenuse. She drew the diagram by constructing rectangles off of each side of a known right triangle. Is Reese\u2019s claim correct for this example? In order to prove or disprove Reese\u2019s claim, you must first show that the rectangles are similar. If they are, then you can use computations to show that the sum of the areas of the figures off of the sides a and b equals the area of the figure off of side c.
\n\"Eureka
\nAnswer:
\nThe rectangles are similar because their corresponding side lengths are equal in scale factor. That is, if we compare the longest side of the rectangle to the side with the same length as the right triangle sides, we get the ratios
\n\\(\\frac{4.8}{3}\\) = \\(\\frac{6.4}{4}\\) = \\(\\frac{8}{5}\\) = 1.6.
\nSince the corresponding sides were all equal to the same constant, then we know we have similar rectangles. The areas of the smaller rectangles are 14.4 cm2<\/sup> and 25.6 cm2<\/sup>, and the area of the largest rectangle is 40 cm2<\/sup>. The sum of the smaller areas is equal to the larger area:
\n14.4 + 25.6 = 40
\n40 = 40.
\nTherefore, we have shown that the sum of the areas of the two smaller rectangles is equal to the area of the larger rectangle, and Reese\u2019s claim is correct.<\/p>\n

Question 4.
\nAfter learning the proof of the Pythagorean theorem using areas of squares, Joseph got really excited and tried explaining it to his younger brother. He realized during his explanation that he had done something wrong. Help Joseph find his error. Explain what he did wrong.
\n\"Eureka
\nAnswer:
\nBased on the proof shown in class, we would expect the sum of the two smaller areas to be equal to the sum of the larger area (i.e., 16 + 25 should equal 49). However, 16 + 25 = 41. Joseph correctly calculated the areas of each square, so that was not his mistake. His mistake was claiming that a triangle with side lengths of 4, 5, and 7 was a right triangle. We know that the Pythagorean theorem only works for right triangles. Considering the converse of the Pythagorean theorem, when we use the given side lengths, we do not get a true statement.
\n42<\/sup> + 52<\/sup> = 72<\/sup>
\n16 + 25 = 49
\n41 \u2260 49
\nTherefore, the triangle Joseph began with is not a right triangle, so it makes sense that the areas of the squares were not adding up like we expected.<\/p>\n

Question 5.
\nDraw a right triangle with squares constructed off of each side that Joseph can use the next time he wants to show his younger brother the proof of the Pythagorean theorem.
\nAnswer:
\nAnswers will vary. Verify that students begin, in fact, with a right triangle and do not make the same mistake that Joseph did. Consider having students share their drawings and explanations of the proof in future class meetings.<\/p>\n

Question 6.
\nExplain the meaning of the Pythagorean theorem in your own words.
\nAnswer:
\nIf a triangle is a right triangle, then the sum of the squares of the legs will be equal to the square of the hypotenuse. Specifically, if the leg lengths are a and b, and the hypotenuse is length c, then for right triangles a2<\/sup> + b2<\/sup> = c2<\/sup>.<\/p>\n

Question 7.
\nDraw a diagram that shows an example illustrating the Pythagorean theorem.
\nAnswer:
\nDiagrams will vary. Verify that students draw a right triangle with side lengths that satisfy the Pythagorean theorem.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 15 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nExplain a proof of the Pythagorean theorem in your own words. Use diagrams and concrete examples, as necessary, to support your explanation.
\nAnswer:
\nProofs will vary. The critical parts of the proof that demonstrate proficiency include an explanation of the similar triangles \u25b3ADC, \u25b3ACB, and \u25b3CDB, including a statement about the ratio of their corresponding sides being equal, leading to the conclusion of the proof.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 15 Answer Key Eureka Math Grade 8 Module 7 Lesson 15 Classwork Answer Key Classwork Proof of the Pythagorean Theorem Answer: Using one right triangle, we created 3 right triangles. Name those triangles. The three triangles are \u25b3ABC, \u25b3ACD, and \u25b3BCD. We can use our basic … 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\/83506"}],"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=83506"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/83506\/revisions"}],"predecessor-version":[{"id":91510,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/83506\/revisions\/91510"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=83506"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=83506"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=83506"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}