{"id":58879,"date":"2023-11-28T14:23:28","date_gmt":"2023-11-28T08:53:28","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=58879"},"modified":"2023-11-29T15:05:36","modified_gmt":"2023-11-29T09:35:36","slug":"eureka-math-grade-4-module-3-lesson-1","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-4-module-3-lesson-1\/","title":{"rendered":"Eureka Math Grade 4 Module 3 Lesson 1 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 1 Answer Key<\/h2>\n

Eureka Math Grade 4 Module 3 Lesson 1 Problem Set Answer Key<\/h3>\n

Question 1.
\nDetermine the perimeter and area of rectangles A and B.
\n\"Eureka
\na. A = _____63 cm2<\/sup>__________\u00a0 \u00a0 \u00a0A = ______54 cm2<\/sup>_________<\/p>\n

b. P = ______32 cm_________\u00a0 \u00a0 \u00a0P = ______30 cm_________
\nAnswer:
\n\"Eureka<\/p>\n

Answer:
\na. A = 63 cm2 <\/sup>, A = 54 cm2<\/sup>,<\/p>\n

Explanation:
\nAs shown in figure A length = 9 cm , breadth = 7 cm,
\nArea of rectangle is = l \u00d7 b = 9 cm \u00d7 7 cm = 63 square cm,
\nand in figure B length = 6 cm , breadth = 9 cm,
\nArea of rectangle is = l \u00d7 b = 6 cm \u00d7 9 cm = 54 square cm.<\/p>\n

b. P = 32 cm,\u00a0 P = 30 cm,<\/p>\n

Explanation:
\nAs shown in figure A length = 9 cm , breadth = 7 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2(9 cm + 7 cm) =
\n2 \u00d7 (16 cm) = 32 cm,
\nand in figure B length = 6 cm , breadth = 9 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (6 cm + 9 cm) =
\n2 \u00d7 (15 cm) = 30 cm.<\/p>\n

Question 2.
\nDetermine the perimeter and area of each rectangle.
\na.
\n\"Eureka
\nP = _____22 cm_______
\nA = _____30 cm2<\/sup>_______
\nAnswer:
\nPerimeter = 22 cm,
\nArea =\u00a030 cm2<\/sup>,<\/p>\n

Explanation:
\nGiven length = 5 cm and breadth = 6 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (5 cm + 6 cm) =
\n2 \u00d7 (11 cm) = 22 cm,
\nArea of rectangle is = l \u00d7 b = 5 cm \u00d7 6 cm = 30 square cm.<\/p>\n

b.
\n\"Eureka
\nP = ____22 cm_______
\nA = ____ 24 cm2<\/sup>________
\nAnswer:
\nPerimeter = 22 cm,
\nArea = 24 cm2<\/sup>,<\/p>\n

Explanation:
\nGiven length = 8 cm and breadth = 3 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (8 cm + 3 cm) =
\n2 \u00d7 (11 cm) = 22 cm,
\nArea of rectangle is = l \u00d7 b = 8 cm \u00d7 3 cm = 24 square cm.<\/p>\n

Question 3.
\nDetermine the perimeter of each rectangle.
\na.
\n\"Eureka
\nP = ____530 m________
\nAnswer:
\nPerimeter = 530 m,<\/p>\n

Explanation:
\nGiven length = 99 m and breadth = 166 m,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (99 m + 166 m)
\n= 2 \u00d7 ( 265 m) = 530 m.<\/p>\n

b.
\n\"Eureka
\nP = ____450 cm________
\nAnswer:
\nPerimeter =\u00a0 450 cm,<\/p>\n

Explanation:
\nGiven length = 75 cm and breadth = 1 m 50 cm,
\nwe know 1 m = 100 cm ,
\nso breadth =\u00a0 1 \u00d7 100 cm + 50 cm =150 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (75 cm + 150 cm)
\n= 2 \u00d7 (225 cm) = 450 cm.<\/p>\n

Question 4.
\nGiven the rectangle\u2019s area, find the unknown side length.
\na.
\n\"Eureka
\nx = ____10 cm________
\nAnswer:
\nUnknown side length = 10 cm,<\/p>\n

Explanation:
\nGiven width of rectangle as 8 cm and area as 80 square cm,
\nand unknow side length is x cm, we know area of rectangle is
\nlength \u00d7 breadth, So 80 sq cm = 8 cm \u00d7 x cm,
\nSo x cm = 80 sq cm \u00f7 8 cm = 10 cm,
\ntherefore unknown side length = 10 cm.<\/p>\n

b.
\n\"Eureka
\nx = ____7 cm________
\nAnswer:
\nUnknown side length = 7 cm,<\/p>\n

Explanation:
\nGiven width of rectangle as 7 cm and area as 49 square cm,
\nand unknow side length is x cm, we know area of rectangle is
\nlength \u00d7 breadth, So 49 sq cm = 7 cm \u00d7 x cm,
\nSo x cm = 49 sq cm \u00f7 7 cm = 7 cm,
\ntherefore unknown side length = 7 cm.<\/p>\n

Question 5.
\nGiven the rectangle\u2019s perimeter, find the unknown side length.
\na. P = 120 cm
\n\"Eureka
\nAnswer:
\nUnknown side length = 40 cm,<\/p>\n

Explanation:
\nGiven width of rectangle as 20 cm and perimeter as 120 cm,
\nand unknow side length is x cm, we know perimeter of
\nrectangle is 2 \u00d7 (length + breadth),
\nSo 120 cm = 2 \u00d7 (x cm + 20 cm),
\n120 cm \u00f7 2 = (x cm + 20 cm),
\n60 cm = (x cm + 20 cm),
\nSo x cm = 60 cm – 20 cm = 40 cm,
\ntherefore unknown side length = 40 cm.<\/p>\n

b. p = 1000 m
\n\"Eureka
\nAnswer:
\nUnknown side width = 250 m,<\/p>\n

Explanation:
\nGiven length of rectangle as 250 m and perimeter as 1000 m,
\nand unknow side width is x m, we know perimeter of
\nrectangle is 2 \u00d7 (length + breadth),
\nSo 1000 m = 2 \u00d7 (250 m + x m),
\n1000 m \u00f7 2 = (250 m + x m ),
\n500 m = 250 m + x m ,
\nSo x m = 500 m – 250 m = 250 m,
\ntherefore unknown side length = 250 m.<\/p>\n

Question 6.
\nEach of the following rectangles has whole number side lengths.
\nGiven the area and perimeter, find the length and width.
\na. P = 20 cm
\n\"Eureka
\nAnswer:
\nThe whole number side lengths are
\nIf the length of the rectangle is 4 cm then the width is 6 cm or
\nlength of the rectangle is 6 cm then the width is 4 cm,<\/p>\n

Explanation:
\nGiven area = 24 square cm and perimeter = 20 cm of rectangles,
\nlets take length as l and width as w and we know
\narea of rectangle = length \u00d7 width , 24 sq cm = l \u00d7 w and
\nperimeter = 2 \u00d7 (length + width) ,
\n20 cm = 2 \u00d7 ( l + w), l + w = 20 cm \u00f7 2 =10 cm,
\nso l = 10 cm – w, now 24 = (10 – w) \u00d7 w,
\nwe get w2<\/sup>– 10 w + 24 = 0,
\nSo w2<\/sup>– 6 w – 4 w + 24 = 0,
\nw(w – 6) – 4(w – 6) = 0, therefore w = 6 cm or w = 4 cm,
\nSo l = 10 cm – 6 cm = 4 cm or l = 10 cm\u00a0 – 4 cm = 6 cm,
\nTherfore, If length of rectangle is 4 cm then width is 6 cm or
\nlength of rectangle is 6 cm then width is 4 cm.<\/p>\n

b. P = 28 m
\n\"Eureka
\nAnswer:
\nThe whole number side lengths are
\nIf the length of the rectangle is 2 m then the width is 12 m or
\nThe length of the rectangle is 12 m the width is 2 m.<\/p>\n

Explanation:
\nGiven area = 24 square m and perimeter = 28 m of rectangles,
\nlets take length as l and width as w and we know
\narea of rectangle = length X width , 24 sq m = l X w and
\nperimeter = 2 \u00d7 (length + width) ,
\n28 m = 2 \u00d7 ( l + w), l + w = 28 m \u00f7 2 =14 m,
\nso l = 14 m – w, now 24 = (14 – w) \u00d7 w,
\nwe get w2<\/sup>– 14 w + 24 = 0,
\nSo w2<\/sup>– 12 w – 2 w + 24 = 0,
\nw(w – 12) – 2(w – 12) = 0, therefore w = 12 m or w = 2 m,
\nSo l = 14 m – 12 m = 2 m or l = 14 m – 2 m = 12 m,
\nTherfore, If length of rectangle is 2 m then width is 12 m or
\nlength of rectangle is 12 m then width is 2 m.<\/p>\n

Eureka Math Grade 4 Module 3 Lesson 1 Exit Ticket Answer Key<\/h3>\n

\"Eureka<\/p>\n

Question 1.
\nDetermine the area and perimeter of the rectangle.
\n\"Eureka
\nAnswer:
\nArea = 16 cm2<\/sup>,
\nPerimeter = 20 cm,<\/p>\n

Explanation:
\nGiven length = 8 cm and breadth = 2 cm,
\nArea of rectangle is = l \u00d7 b = 8 cm \u00d7 2 cm = 16 square cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (8 cm + 2 cm) =
\n2 \u00d7 (10 cm) = 20 cm.<\/p>\n

Question 2.
\nDetermine the perimeter of the rectangle.
\n\"Eureka
\nAnswer:
\nPerimeter = 892 m,<\/p>\n

Explanation:
\nGiven length = 347 m and breadth = 99 m,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (347 m + 99 m) =
\n2 \u00d7 (446 m) = 892 m.<\/p>\n

Eureka Math Grade 4 Module 3 Lesson 1 Homework Answer Key<\/h3>\n

\"Eureka<\/p>\n

Question 1.
\nDetermine the perimeter and area of rectangles A and B.
\n\"Eureka
\na. A = _____40 cm2<\/sup>__________\u00a0 \u00a0 A = ______35\u00a0cm2<\/sup>_________<\/p>\n

b. P = ______26 cm_________\u00a0 \u00a0 P = _______24 cm________
\nAnswer:
\n\"Eureka
\na. A = 40 cm2 <\/sup>, A = 35 cm2<\/sup>,<\/p>\n

Explanation:
\nAs shown in figure A length = 8 cm , breadth = 5 cm,
\nArea of rectangle is = l \u00d7 b = 8 cm \u00d7 5 cm = 40 square cm,
\nand in figure B length = 5 cm , breadth = 7 cm,
\nArea of rectangle is = l \u00d7 b = 5 cm \u00d7 7 cm = 35 square cm,<\/p>\n

b. P = 26 cm,\u00a0 P = 24 cm,<\/p>\n

Explanation:
\nAs shown in figure A length = 8 cm , breadth = 5 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2(8 cm + 5 cm) =
\n2 \u00d7 (13 cm) = 26 cm,
\nand in figure B length = 5 cm , breadth = 7 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (5 cm + 7 cm) =
\n2 \u00d7 (12 cm) = 24 cm.<\/p>\n

Question 2.
\nDetermine the perimeter and area of each rectangle.
\na.
\n\"Eureka
\nP = ____20 cm________
\nA = ___ 21 cm2<\/sup>____
\nAnswer:
\nPerimeter = 20 cm,
\nArea = 21 cm2<\/sup>,<\/p>\n

Explanation:
\nGiven length = 7 cm and breadth = 3 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (7 cm + 3 cm) =
\n2 \u00d7 (10 cm) = 20 cm,
\nArea of rectangle is = l \u00d7 b = 7 cm \u00d7 3 cm = 21 square cm,<\/p>\n

b.
\n\"Eureka
\nP = ___26 cm_________
\nA = ______36 cm2<\/sup>______
\nAnswer:
\nPerimeter = 26 cm,
\nArea = 36 cm2<\/sup>,<\/p>\n

Explanation:
\nGiven length = 9 cm and breadth = 4 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (9 cm + 4 cm) =
\n2 \u00d7 (13 cm) = 26 cm,
\nArea of rectangle is = l \u00d7 b = 9 cm \u00d7 4 cm = 36 square cm,<\/p>\n

Question 3.
\nDetermine the perimeter of each rectangle.
\na.
\n\"Eureka
\nP = _____850 m_______
\nAnswer:
\nPerimeter = 850 m,<\/p>\n

Explanation:
\nGiven length = 76 m and breadth = 149 m,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (76 m + 149 cm) =
\n2 \u00d7 ( 425 m) = 850 m.<\/p>\n

b.
\n\"Eureka
\nP = ____510 cm________
\nAnswer:
\nPerimeter =\u00a0 510 cm,<\/p>\n

Explanation:
\nGiven length = 45 cm and breadth = 2 m 10 cm,
\nwe know 1 m = 100 cm ,
\nso breadth =\u00a0 2 \u00d7 100 cm + 10 cm = 200 cm + 10 cm = 210 cm,
\nPerimeter of rectangle is =2 \u00d7 (l + b) = 2 \u00d7 (45 cm + 210 cm) =
\n2 \u00d7 (255 cm) = 510 cm.<\/p>\n

Question 4.
\nGiven the rectangle\u2019s area, find the unknown side length.
\na.
\n\"Eureka
\nx = _____10 cm_______
\nAnswer:
\nUnknown side length = 10 cm,<\/p>\n

Explanation:
\nGiven width of rectangle as 6 cm and area as 60 square cm,
\nand unknow side length is x cm, we know area of rectangle is
\nlength \u00d7 breadth, So 60 sq cm = 6 cm \u00d7 x cm,
\nSo x cm = 60 sq cm \u00f7 6 cm = 10 cm,
\ntherefore unknown side length = 10 cm.<\/p>\n

b.
\n\"Eureka
\nx = _____5 m_______
\nAnswer:
\nUnknown side length = 5 m,<\/p>\n

Explanation:
\nGiven width of rectangle as 5 m and area as 25 square m,
\nand unknown side length is x m, we know the area of the rectangle is length \u00d7 breadth, So 25 sq m = 5 m \u00d7 x m,
\nSo x m = 25 sq m \u00f7 5 m = 5 m,
\ntherefore unknown side length = 5 m.<\/p>\n

Question 5.
\nGiven the rectangle\u2019s perimeter, find the unknown side length.
\na. P = 180 cm
\n\"Eureka
\nx = ____40 cm________
\nAnswer:
\nUnknown side length = 40 cm,<\/p>\n

Explanation:
\nGiven width of rectangle as 40 cm and perimeter as 180 cm,
\nand unknow side length is x cm, we know perimeter of
\nrectangle is 2 \u00d7 (length + breadth),
\nSo 180 cm = 2 \u00d7 (x cm + 40 cm),
\n180 cm \u00f7 2 = (x cm + 40 cm),
\n90 cm = (x cm + 40 cm),
\nSo x cm = 90 cm – 40 cm = 50 cm,
\ntherefore unknown side length = 50 cm.<\/p>\n

b. P = 1,000 m
\n\"Eureka
\nx = ____350 m________
\nAnswer:
\nUnknown side width = 350 m,<\/p>\n

Explanation:
\nGiven length of rectangle as 150 m and perimeter as 1000 m,
\nand unknow side width is x m, we know perimeter of
\nrectangle is 2 X (length + breadth),
\nSo 1000 m = 2 X (150 m + x m),
\n1000 m \u00f7 2 = (150 m + x m ),
\n500 m = 150 m + x m ,
\nSo x m = 500 m – 150 m = 350 m,
\ntherefore unknown side length = 350 m.<\/p>\n

Question 6.
\nEach of the following rectangles has whole number side lengths. Given the area and perimeter, find the length and width.
\na. A = 32 square cm
\nP = 24 cm
\n\"Eureka
\nAnswer:
\nThe whole number side lengths are
\nIf the length of the rectangle is 4 cm then the width is 8 cm or
\nlength of the rectangle is 8 cm then the width is 4 cm,<\/p>\n

Explanation:
\nGiven area = 32 square cm and perimeter = 24 cm of rectangles,
\nlets take length as l and width as w and we know
\narea of rectangle = length X width , 32 sq cm = l X w and
\nperimeter = 2 \u00d7 (length + width) ,
\n24 cm = 2 \u00d7 ( l + w), l + w = 24 cm \u00f7 2 =12 cm,
\nso l = 12 cm – w, now 32 = (12 – w) \u00d7 w,
\nwe get w2<\/sup>– 12 w + 32 = 0,
\nSo w2<\/sup>– 8 w – 4 w + 32 = 0,
\nw(w – 8) – 4(w – 8) = 0, therefore w = 8 cm or w = 4 cm,
\nSo l = 12 cm – 8 cm = 4 cm or l = 12 cm\u00a0 – 4 cm = 8 cm,
\nTherfore, If length of recatangle is 4 cm then width is 8 cm or
\nlength of rectangle is 8 cm then width is 4 cm.<\/p>\n

b. A = 36 square m
\nP = 30 m
\n\"Eureka
\nAnswer:
\nThe whole number side lengths are
\nIf the length of the rectangle is 3 m then the width is 12 m or
\nThe length of the rectangle is 12 m the width is 3 m.<\/p>\n

Explanation:
\nGiven area = 36 square m and perimeter = 30 m of rectangles,
\nlets take length as l and width as w and we know
\narea of rectangle = length X width , 36 sq m = l X w and
\nperimeter = 2 \u00d7 (length + width) ,
\n30 m = 2 \u00d7 ( l + w), l + w = 30 m \u00f7 2 =15 m,
\nso l = 15 m – w, now 36 = (15 – w) X w,
\nwe get w2<\/sup>– 15 w + 36 = 0,
\nSo w2<\/sup>– 12 w – 3 w + 24 = 0,
\nw(w – 12) – 3(w – 12) = 0, therefore w = 12 m or w = 3 m,
\nSo l = 15 m – 12 m = 3 m or l = 15 m – 3 m = 12 m,
\nTherfore, If length of rectangle is 3 m then width is 12 m or length of rectangle is 12 m then width is 3 m.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 1 Answer Key Eureka Math Grade 4 Module 3 Lesson 1 Problem Set Answer Key Question 1. Determine the perimeter and area of rectangles A and B. a. A = _____63 cm2__________\u00a0 \u00a0 \u00a0A = ______54 cm2_________ b. P = ______32 cm_________\u00a0 \u00a0 \u00a0P = ______30 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":171300,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/ccssanswers.com\/wp-content\/uploads\/2022\/06\/Eureka-Math-Grade-4-Module-3-Lesson-1-Answer-Key.png","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/58879"}],"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=58879"}],"version-history":[{"count":3,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/58879\/revisions"}],"predecessor-version":[{"id":171304,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/58879\/revisions\/171304"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media\/171300"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=58879"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=58879"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=58879"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}