{"id":81326,"date":"2022-06-08T19:44:49","date_gmt":"2022-06-08T14:14:49","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=81326"},"modified":"2022-05-10T14:18:33","modified_gmt":"2022-05-10T08:48:33","slug":"eureka-math-grade-6-module-5-lesson-4","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-6-module-5-lesson-4\/","title":{"rendered":"Eureka Math Grade 6 Module 5 Lesson 4 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Grade 6 Module 5 Lesson 4 Answer Key<\/h2>\n

Eureka Math Grade 6 Module 5 Lesson 4 Exercise Answer Key<\/h3>\n

Opening Exercise:<\/p>\n

Draw and label the altitude of each triangle below.<\/p>\n

Question a.
\n\"Eureka
\nAnswer:
\n\"Eureka<\/p>\n

Question b.
\n\"Eureka
\nAnswer:
\n\"Eureka<\/p>\n

Question c.
\n\"Eureka
\nAnswer:
\n\"Eureka<\/p>\n

Exploratory Challenge\/Exercises 1 – 5:<\/p>\n

Question 1.
\nUse rectangle X and the triangle with the altitude inside (triangle X) to show that the area formula for the triangle is A = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height.<\/p>\n

a. Step One: Find the area of rectangle X.
\nAnswer:
\nA = 3 in. \u00d7 2.5 in. = 7.5 in2<\/sup><\/p>\n

b. Step Two: What is half the area of rectangle X?
\nAnswer:
\nHalf of the area of the rectangle is 7.5 in2<\/sup> \u00f7 2= 3.75 in2<\/sup><\/p>\n

c. Step Three: Prove, by decomposing triangle X, that it is the same as half of rectangle X. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle X. What conclusions can you make about the triangle\u2019s area compared to the rectangle\u2019s area?
\nAnswer:
\nBecause the triangle fits inside half of the rectangle, we know the triangle\u2019s area is half of the rectangle’s area.<\/p>\n

Question 2.
\nUse rectangle Y and the triangle with a side that \u00a1s the altitude (triangle Y) to show the area formula for the triangle is A = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height.<\/p>\n

a. Step One: Find the area of rectangle Y.
\nAnswer:
\nA = 3 in. \u00d7 3 in. = 9 in2<\/sup><\/p>\n

b. Step Two: What is half the area of rectangle Y?
\nAnswer:
\nHalf the area of the rectangle is 9 in2<\/sup> \u00f7 2 = 4.5 in2<\/sup>.<\/p>\n

c. Step Three: Prove, by decomposing triangle Y, that it is the same as half of rectangle Y. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Y. What conclusions can you make about the triangle\u2019s area compared to the rectangle\u2019s area?
\nAnswer:
\nThe right triangle also fits in exactly half of the rectangle, so the triangle\u2019s area is \u2022 Students may struggle once again half the size of the rectangle\u2019s area. with this step since they have yet to see an obtuse<\/p>\n

Question 3.
\nUse rectangle Z and the triangle with the altitude outside (triangle Z) to show that the area angle. formula for the triangle is A = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height.<\/p>\n

a. Step One: Find the area of rectangle Z.
\nAnswer:
\nA= 3 in. \u00d7 2.5 in.=7.5 in2<\/sup><\/p>\n

b. Step Two: What is half the area of rectangle Z?
\nAnswer:
\nHalf of the area of the rectangle is 7.5 in2<\/sup> \u00f72 = 3.75 in2<\/sup>.<\/p>\n

c. Step Three: Prove, by decomposing triangle Z, that it is the same as half of rectangle Z. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Z. What conclusions can you make about the triangle\u2019s area compared to the rectangle\u2019s area?
\nAnswer:
\nSimilar to the other two triangles, when the altitude is outside the triangle, the area of the triangle is exactly half of the area of the rectangle.<\/p>\n

Question 4.
\nWhen finding the area of a triangle, does it matter where the altitude is located?
\nAnswer:
\nIt does not matter where the altitude is located. To find the area of a triangle, the formula is always
\nA = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height.<\/p>\n

Question 5.
\nHow can you determine which part of the triangle is the base and which is the height?
\nAnswer:
\nThe base and the height of any triangle form a right angle because the altitude is always perpendicular to the base.<\/p>\n

Exercises 6 – 8:<\/p>\n

Calculate the area of each triangle. Figures are not drawn to scale.<\/p>\n

Question 6.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) \u00d7 (24 in.) (8 in.) = 96 in2<\/sup><\/p>\n

Question 7.
\n\"Eureka
\nAnswer:
\n\"Eureka<\/p>\n

Question 8.
\nDraw three triangles (acute, right, and obtuse) that have the same area. Explain how you know they have the same area.
\nAnswer:
\nAnswers will vary.<\/p>\n

Eureka Math Grade 6 Module 5 Lesson 4 Problem Set Answer Key<\/h3>\n

Calculate the area of each figure below. Figures are not drawn to scale.<\/p>\n

Question 1.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (21 in.) (8 in.) = 84 in2<\/sup><\/p>\n

Question 2.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (72 m) (21 m) = 756 m2<\/sup><\/p>\n

Question 3.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (75.8 km) (29.2 km) = 1,106.68 km2<\/sup><\/p>\n

Question 4.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (5 m) (12 m) = 30 m2<\/sup>
\nA = \\(\\frac{1}{2}\\) (7 m) (29 m) = 101.5 m2<\/sup>
\nA = (12 m) (19 m) = 228 m2<\/sup>
\nA = 30 m2<\/sup> + 30 m2<\/sup> + 101.5 m2<\/sup> + 228 m2<\/sup>
\nA = 389.5 m2<\/sup><\/p>\n

Question 5.
\nThe Andersons are going on a long sailing trip during the summer. However, one of the sails on their sailboat ripped, and they have to replace it. The sail is pictured below. If the sailboat sails are on sale for $2 per square foot, how much will the new sail cost?
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) bh
\n= \\(\\frac{1}{2}\\) (8 ft.) (12 ft.)
\n= 48 ft2<\/sup>
\n\\(\\frac{2 \\text { dollars }}{\\mathrm{ft}^{2}}\\) \u00d7 48 ft2<\/sup> = 96 dollars(or $96)
\nThe cost of the new sail is $96.<\/p>\n

Question 6.
\nDarnell and Donovan are both trying to calculate the area of an obtuse triangle. Examine their calculations below.
\n\"Eureka<\/p>\n\n\n\n\n
Darnell\u2019s Work<\/td>\nDonovan\u2019s Work<\/td>\n<\/tr>\n
A = \\(\\frac{1}{2}\\) \u00d7 3 in. \u00d7 4 in.<\/p>\n

A = 6 in2<\/sup><\/td>\n

A = \\(\\frac{1}{2}\\) \u00a0\u00d7 12 in. \u00d7 4 in.<\/p>\n

A = 24 in2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Which student calculated the area correctly? Explain why the other student is not correct.
\nAnswer:
\nDonovan calculated the area correctly. Although Darnell did use the altitude of the triangle, he used the length between the altitude and the base rather than the length of the actual base.<\/p>\n

Question 7.
\nRussell calculated the area of the triangle below. His work is shown.
\n\"Eureka
\nA = \\(\\frac{1}{2}\\) bh
\nA = 150.5 cm2<\/sup>
\nAlthough Russell was told his work is correct, he had a hard time explaining why it is correct. Help Russell explain why his calculations are correct.
\nAnswer:
\nThe formula for the area of triangle is A = bit. Russell followed this formula because 7 cm is the height of the triangle, and 43 cm is the base of the triangle.<\/p>\n

Question 8.
\nThe larger triangle below has a base of 10.14 m; the gray triangle has an area of 40.325 m2<\/sup>.
\n\"Eureka<\/p>\n

a. Determine the area of the larger triangle if it has a height of 12.2 m.
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (10.14 m) (12.2 m)
\n= 61.854 m2<\/sup><\/p>\n

b. Let A be the area of the unshaded (white) triangle in square meters. Write and solve an equation to determine the value of A, using the areas of the larger triangle and the gray triangle.
\nAnswer:
\n40.325m2<\/sup> + A = 61.854m2<\/sup>
\n40.325m2<\/sup> + A – 40.325 m2<\/sup> = 61.854 m2<\/sup> – 40.325 m2<\/sup>
\nA = 21.529 m2<\/sup><\/p>\n

Eureka Math Grade 6 Module 5 Lesson 4 Exit Ticket Answer Key<\/h3>\n

Find the area of each triangle. Figures are not drawn to scale.<\/p>\n

Question 1.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (12.6 cm) (16.8 cm) = 105.84 cm2<\/sup><\/p>\n

Question 2.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (28 in.) (15 in.) = 210 in2<\/sup><\/p>\n

Question 3.
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{2}\\) (12 ft.) (21 ft.) = 126 ft2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math Grade 6 Module 5 Lesson 4 Answer Key Eureka Math Grade 6 Module 5 Lesson 4 Exercise Answer Key Opening Exercise: Draw and label the altitude of each triangle below. Question a. Answer: Question b. Answer: Question c. Answer: Exploratory Challenge\/Exercises 1 – 5: Question 1. Use rectangle X and the … 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\/81326"}],"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=81326"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/81326\/revisions"}],"predecessor-version":[{"id":137559,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/81326\/revisions\/137559"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=81326"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=81326"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=81326"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}