{"id":83644,"date":"2022-06-01T16:30:39","date_gmt":"2022-06-01T11:00:39","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=83644"},"modified":"2022-05-10T14:19:42","modified_gmt":"2022-05-10T08:49:42","slug":"eureka-math-grade-8-module-7-lesson-16","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-7-lesson-16\/","title":{"rendered":"Eureka Math Grade 8 Module 7 Lesson 16 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 16 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 7 Lesson 16 Classwork Answer Key<\/h3>\n

Classwork
\nProof of the Converse of the Pythagorean Theorem
\n\"Engage
\nAnswer:
\n\u2192 What do we know or not know about each of these triangles?
\nIn the first triangle, ABC, we know that a2<\/sup> + b2<\/sup> = c2<\/sup>. We do not know if angle C is a right angle.
\nIn the second triangle, A’B’C’, we know that it is a right triangle.<\/p>\n

\u2192 What conclusions can we draw from this?
\nBy applying the Pythagorean theorem to \u25b3A’B’C’, we get |A’B’|2<\/sup> = a2<\/sup> + b2<\/sup>. Since we are given
\nc2<\/sup> = a2<\/sup> + b2<\/sup>, then by substitution, |A’B’|2<\/sup> = c2<\/sup>, and then |A’B’| = c. Since c is also |AB|, then |A’B’| = |AB|. That means that both triangles have sides a, b, and c that are the exact same lengths.<\/p>\n

\u2192 Recall that we would like to prove that \u2220ACB is a right angle, that it maps to \u2220A’ C’ B’. If we can translate
\n\u25b3ABC so that A goes to A’, B goes to B’, and C goes to C’, it follows that all three angles in the triangle will match. In particular, that \u2220ACB maps to the right angle \u2220A’ C’ B’, and so is a right angle, too.
\n\u2192 We can certainly perform a translation that takes B to B’ and C to C’ because segments BC and B’C’ are the same length. Must this translation take A to A’? What goes wrong mathematically if it misses and translates to a different point A” as shown below?<\/p>\n

In this picture, we\u2019ve drawn A” to the left of \\(\\overline{A{\\prime} C{\\prime}}\\). The reasoning that follows works just as well for a picture with A” to the right of \\(\\overline{A{\\prime} C{\\prime}}\\) instead.
\n\"Engage
\nProvide time for students to think of what may go wrong mathematically. If needed, prompt them to notice the two isosceles triangles in the diagram, \u25b3A”C’A’ and \u25b3A”B’A’ and the four angles w1<\/sub>,w2<\/sub>,w3<\/sub>,w4<\/sub> labeled as shown in the diagram below.
\n\"Engage
\n\u25b3A”C’A’ is isosceles and therefore has base angles that are equal in measure:
\nw1<\/sub> + w2<\/sub> = w3<\/sub>.
\n\u25b3A”B’A’ is isosceles and therefore has base angles that are equal in measure:
\nw2<\/sub> = w3<\/sub> + w4<\/sub>.
\nThese two equations give w1<\/sub> + w3<\/sub> + w4<\/sub> = w3<\/sub>, which is equal to w1<\/sub> + w4<\/sub> = 0, which is obviously not true.
\nTherefore, the translation must map A to A’, and since translations preserve the measures of angles, we can conclude that the measure of \u2220ACB is equal to the measure of \u2220A’C’B’, and \u2220ACB is a right angle.
\nFinally, if a triangle has side lengths of a,b and c, with c the longest length, that don\u2019t satisfy the equation a2<\/sup> + b2<\/sup> = c2<\/sup>, then the triangle cannot be a right triangle.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 16 Exercise Answer Key<\/h3>\n

Exercises 1\u20137<\/p>\n

Exercise 1.
\nIs the triangle with leg lengths of 3 mi. and 8 mi. and hypotenuse of length \\(\\sqrt{73}\\) mi. a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n32<\/sup> + 82<\/sup> = (\\(\\sqrt{73}\\))2<\/sup>
\n9 + 64 = 73
\n73 = 73
\nYes, the triangle with leg lengths of 3 mi. and 8 mi. and hypotenuse of length \\(\\sqrt{73}\\) mi. is a right triangle because it satisfies the Pythagorean theorem.<\/p>\n

Exercise 2.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
\n\"Engage
\nAnswer:
\nLet c in. represent the length of the hypotenuse of the triangle.
\n12<\/sup> + 42<\/sup> = c2<\/sup>
\n1 + 16 = c2<\/sup>
\n17 = c2<\/sup>
\n\\(\\sqrt{17}\\) = c
\n4.1\u2248c
\nThe length of the hypotenuse of the right triangle is exactly \\(\\sqrt{17}\\) inches and approximately 4.1 inches.<\/p>\n

Exercise 3.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
\n\"Engage
\nAnswer:
\nLet c mm represent the length of the hypotenuse of the triangle.
\n22<\/sup> + 62<\/sup> = c2<\/sup>
\n4 + 36 = c2<\/sup>
\n40 = c2<\/sup>
\n\\(\\sqrt{40}\\) = c
\n\\(\\sqrt{2^{3}}\\)\u00d7\\(\\sqrt{5}\\) = c
\n\\(\\sqrt{2^{2}}\\)\u00d7\\(\\sqrt{2}\\)\u00d7\\(\\sqrt{5}\\) = c
\n2\\(\\sqrt{10}\\) = c
\nThe length of the hypotenuse of the right triangle is exactly 2\\(\\sqrt{10}\\) mm and approximately 6.3 mm.<\/p>\n

Exercise 4.
\nIs the triangle with leg lengths of 9 in. and 9 in. and hypotenuse of length \\(\\sqrt{175}\\) in. a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n92<\/sup> + 92<\/sup> = (\\(\\sqrt{175}\\))2<\/sup>
\n81 + 81 = 175
\n162 \u2260 175
\nNo, the triangle with leg lengths of 9 in. and 9 in. and hypotenuse of length \\(\\sqrt{175}\\) in. is not a right triangle because the lengths do not satisfy the Pythagorean theorem.<\/p>\n

Exercise 5.
\nIs the triangle with leg lengths of \u221a(28 ) cm and 6 cm and hypotenuse of length 8 cm a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n(\\(\\sqrt{28}\\))2<\/sup> + 62<\/sup> = 82<\/sup>
\n28 + 36 = 64
\n64 = 64
\nYes, the triangle with leg lengths of \\(\\sqrt{28}\\) cm and 6 cm and hypotenuse of length 8 cm is a right triangle because the lengths satisfy the Pythagorean theorem.<\/p>\n

Exercise 6.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence.
\n\"Engage
\nAnswer:
\nLet c ft. represent the length of the hypotenuse of the triangle.
\n32<\/sup> + (\\(\\sqrt{27}\\))2<\/sup> = c2<\/sup>
\n9 + 27 = c2<\/sup>
\n36 = c2<\/sup>
\n\\(\\sqrt{36}\\) = \\(\\sqrt{c^{2}}\\)
\n6 = c
\nThe length of the hypotenuse of the right triangle is 6 ft.<\/p>\n

Exercise 7.
\nThe triangle shown below is an isosceles right triangle. Determine the length of the legs of the triangle. Show your work, and answer in a complete sentence.
\n\"Engage
\nAnswer:
\nLet x cm represent the length of each of the legs of the isosceles triangle.
\nx2<\/sup> + x2<\/sup> = (\\(\\sqrt{18}\\))2<\/sup>
\n2x2<\/sup> = 18
\n\\(\\frac{2 x^{2}}{2}\\) = \\(\\frac{18}{2}\\)
\nx2<\/sup> = 9
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{9}\\)
\nx = 3
\nThe leg lengths of the isosceles triangle are 3 cm.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 16 Problem Set Answer Key<\/h3>\n

Question 1.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
\n\"Eureka
\nAnswer:
\nLet c cm represent the length of the hypotenuse of the triangle.
\n12<\/sup> + 12<\/sup> = c2<\/sup>
\n1 + 1 = c2<\/sup>
\n2 = c2<\/sup>
\n\\(\\sqrt{2}\\) = \\(\\sqrt{c^{2}}\\)
\n1.4\u2248c
\nThe length of the hypotenuse is exactly \\(\\sqrt{2}\\) cm and approximately 1.4 cm.<\/p>\n

Question 2.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
\n\"Eureka
\nAnswer:
\nLet x ft. represent the unknown length of the triangle.
\n72<\/sup> + x2<\/sup> = 112<\/sup>
\n49 + x2<\/sup> = 121
\n49 – 49 + x2<\/sup> = 121 – 49
\nx2<\/sup> = 72
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{72}\\)
\nx = \\(\\sqrt{2^{2}}\\) \u22c5 \\(\\sqrt{2}\\) \u22c5 \\(\\sqrt{3^{2}}\\)
\nx = 6\\(\\sqrt{2}\\)
\nx\u22488.5
\nThe length of the unknown side of the triangle is exactly 6\\(\\sqrt{2}\\) ft. and approximately 8.5 ft.<\/p>\n

Question 3.
\nIs the triangle with leg lengths of \\(\\sqrt{3}\\) cm and 9 cm and hypotenuse of length \\(\\sqrt{84}\\) cm a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n(\\(\\sqrt{3}\\))2<\/sup> + 92<\/sup> = (\\(\\sqrt{84}\\))2<\/sup>
\n3 + 81 = 84
\n84 = 84
\nYes, the triangle with leg lengths of \\(\\sqrt{3}\\) cm and 9 cm and hypotenuse of length \\(\\sqrt{84}\\) cm is a right triangle because the lengths satisfy the Pythagorean theorem.<\/p>\n

Question 4.
\nIs the triangle with leg lengths of \\(\\sqrt{7}\\) km and 5 km and hypotenuse of length \\(\\sqrt{48}\\) km a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n(\\(\\sqrt{7}\\))2<\/sup> + 52<\/sup> = (\\(\\sqrt{48}\\))2<\/sup>
\n7 + 25 = 48
\n32 \u2260 48
\nNo, the triangle with leg lengths of \\(\\sqrt{7}\\) km and 5 km and hypotenuse of length \\(\\sqrt{48}\\) km is not a right triangle because the lengths do not satisfy the Pythagorean theorem.<\/p>\n

Question 5.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
\n\"Eureka
\nAnswer:
\nLet c mm represent the length of the hypotenuse of the triangle.
\n52<\/sup> + 102<\/sup> = c2<\/sup>
\n25 + 100 = c2<\/sup>
\n125 = c2<\/sup>
\n\\(\\sqrt{125}\\) = \\(\\sqrt{c^{2}}\\)
\n\\(\\sqrt{5^{3}}\\) = c
\n\\(\\sqrt{5^{2}}\\)\u00d7\\(\\sqrt{5}\\) = c
\n5\\(\\sqrt{5}\\) = c
\n11.2\u2248c
\nThe length of the hypotenuse is exactly 5\\(\\sqrt{5}\\) mm and approximately 11.2 mm.<\/p>\n

Question 6.
\nIs the triangle with leg lengths of 3 and 6 and hypotenuse of length \\(\\sqrt{45}\\) a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n32<\/sup> + 62<\/sup> = (\\(\\sqrt{45}\\))2<\/sup>
\n9 + 36 = 45
\n45 = 45
\nYes, the triangle with leg lengths of 3 and 6 and hypotenuse of length \\(\\sqrt{45}\\) is a right triangle because the lengths satisfy the Pythagorean theorem.<\/p>\n

Question 7.
\nWhat is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
\n\"Eureka
\nAnswer:
\nLet x in. represent the unknown side length of the triangle.
\n22<\/sup> + x2<\/sup> = 82<\/sup>
\n4 + x2<\/sup> = 64
\n4 – 4 + x2<\/sup> = 64 – 4
\nx2<\/sup> = 60
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{60}\\)
\nx = \\(\\sqrt{2^{2}}\\) \u22c5 \\(\\sqrt{3}\\) \u22c5 \\(\\sqrt{5}\\)
\nx = 2\\(\\sqrt{15}\\)
\nx\u22487.7
\nThe length of the unknown side of the triangle is exactly 2\\(\\sqrt{15}\\) inches and approximately 7.7 inches.<\/p>\n

Question 8.
\nIs the triangle with leg lengths of 1 and \\(\\sqrt{3}\\) and hypotenuse of length 2 a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n12<\/sup> + (\\(\\sqrt{3}\\))2<\/sup> = 22<\/sup>
\n1 + 3 = 4
\n4 = 4
\nYes, the triangle with leg lengths of 1 and \\(\\sqrt{3}\\) and hypotenuse of length 2 is a right triangle because the lengths satisfy the Pythagorean theorem.<\/p>\n

Question 9.
\nCorey found the hypotenuse of a right triangle with leg lengths of 2 and 3 to be \\(\\sqrt{13}\\). Corey claims that since \\(\\sqrt{13}\\) = 3.61 when estimating to two decimal digits, that a triangle with leg lengths of 2 and 3 and a hypotenuse of 3.61 is a right triangle. Is he correct? Explain.
\nAnswer:
\nNo, Corey is not correct.
\n22<\/sup> + 32<\/sup> = (3.61)2<\/sup>
\n4 + 9 = 13.0321
\n13 \u2260 13.0321
\nNo, the triangle with leg lengths of 2 and 3 and hypotenuse of length 3.61 is not a right triangle because the lengths do not satisfy the Pythagorean theorem.<\/p>\n

Question 10.
\nExplain a proof of the Pythagorean theorem.
\nAnswer:
\nConsider having students share their proof with a partner while their partner critiques their reasoning. Accept any of the three proofs that students have seen.<\/p>\n

Question 11.
\nExplain a proof of the converse of the Pythagorean theorem.
\nAnswer:
\nConsider having students share their proof with a partner while their partner critiques their reasoning. Accept either of the proofs that students have seen.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 16 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nIs the triangle with leg lengths of 7 mm and 7 mm and a hypotenuse of length 10 mm a right triangle? Show your work, and answer in a complete sentence.
\nAnswer:
\n72<\/sup> + 72<\/sup> = 102<\/sup>
\n49 + 49 = 100
\n98\u2260100
\nNo, the triangle with leg lengths of 7 mm and 7 mm and hypotenuse of length 10 mm is not a right triangle because the lengths do not satisfy the Pythagorean theorem.<\/p>\n

Question 2.
\nWhat would the length of the hypotenuse need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
\nAnswer:
\nLet c mm represent the length of the hypotenuse.
\nThen,
\n72<\/sup> + 72<\/sup> = c2<\/sup>
\n49 + 49 = c2<\/sup>
\n98 = c2<\/sup>
\n\\(\\sqrt{98}\\) = c
\nThe hypotenuse would need to be \\(\\sqrt{98}\\) mm for the triangle with sides of 7 mm and 7 mm to be a right triangle.<\/p>\n

Question 3.
\nIf one of the leg lengths is 7 mm, what would the other leg length need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
\nAnswer:
\nLet a mm represent the length of one leg.
\nThen,
\na2<\/sup> + 72<\/sup> = 102<\/sup>
\na2<\/sup> + 49 = 100
\na2<\/sup> + 49 – 49 = 100 – 49
\na2<\/sup> = 51
\na = \\(\\sqrt{51}\\)
\nThe leg length would need to be \\(\\sqrt{51}\\) mm so that the triangle with one leg length of 7 mm and the hypotenuse of 10 mm is a right triangle.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 16 Answer Key Eureka Math Grade 8 Module 7 Lesson 16 Classwork Answer Key Classwork Proof of the Converse of the Pythagorean Theorem Answer: \u2192 What do we know or not know about each of these triangles? In the first triangle, ABC, we know that a2 … 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\/83644"}],"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=83644"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83644\/revisions"}],"predecessor-version":[{"id":137279,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83644\/revisions\/137279"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=83644"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=83644"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=83644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}