lesson 3 homework 4.3

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Eureka Math Grade 4 Module 3 Lesson 1 Answer Key

Engage ny eureka math 4th grade module 3 lesson 1 answer key, eureka math grade 4 module 3 lesson 1 problem set answer key.

Answer: a. A = 63 cm 2 , A = 54 cm 2 ,

Explanation: As shown in figure A length = 9 cm , breadth = 7 cm, Area of rectangle is = l × b = 9 cm × 7 cm = 63 square cm, and in figure B length = 6 cm , breadth = 9 cm, Area of rectangle is = l × b = 6 cm × 9 cm = 54 square cm.

b. P = 32 cm,  P = 30 cm,

Explanation: As shown in figure A length = 9 cm , breadth = 7 cm, Perimeter of rectangle is =2 × (l + b) = 2(9 cm + 7 cm) = 2 × (16 cm) = 32 cm, and in figure B length = 6 cm , breadth = 9 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (6 cm + 9 cm) = 2 × (15 cm) = 30 cm.

Explanation: Given length = 5 cm and breadth = 6 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (5 cm + 6 cm) = 2 × (11 cm) = 22 cm, Area of rectangle is = l × b = 5 cm × 6 cm = 30 square cm.

Explanation: Given length = 8 cm and breadth = 3 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (8 cm + 3 cm) = 2 × (11 cm) = 22 cm, Area of rectangle is = l × b = 8 cm × 3 cm = 24 square cm.

Explanation: Given length = 99 m and breadth = 166 m, Perimeter of rectangle is =2 × (l + b) = 2 × (99 m + 166 m) = 2 × ( 265 m) = 530 m.

Explanation: Given length = 75 cm and breadth = 1 m 50 cm, we know 1 m = 100 cm , so breadth =  1 × 100 cm + 50 cm =150 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (75 cm + 150 cm) = 2 × (225 cm) = 450 cm.

Explanation: Given width of rectangle as 8 cm and area as 80 square cm, and unknow side length is x cm, we know area of rectangle is length × breadth, So 80 sq cm = 8 cm × x cm, So x cm = 80 sq cm ÷ 8 cm = 10 cm, therefore unknown side length = 10 cm.

Explanation: Given width of rectangle as 7 cm and area as 49 square cm, and unknow side length is x cm, we know area of rectangle is length × breadth, So 49 sq cm = 7 cm × x cm, So x cm = 49 sq cm ÷ 7 cm = 7 cm, therefore unknown side length = 7 cm.

Explanation: Given width of rectangle as 20 cm and perimeter as 120 cm, and unknow side length is x cm, we know perimeter of rectangle is 2 × (length + breadth), So 120 cm = 2 × (x cm + 20 cm), 120 cm ÷ 2 = (x cm + 20 cm), 60 cm = (x cm + 20 cm), So x cm = 60 cm – 20 cm = 40 cm, therefore unknown side length = 40 cm.

Explanation: Given length of rectangle as 250 m and perimeter as 1000 m, and unknow side width is x m, we know perimeter of rectangle is 2 × (length + breadth), So 1000 m = 2 × (250 m + x m), 1000 m ÷ 2 = (250 m + x m ), 500 m = 250 m + x m , So x m = 500 m – 250 m = 250 m, therefore unknown side length = 250 m.

Explanation: Given area = 24 square cm and perimeter = 20 cm of rectangles, lets take length as l and width as w and we know area of rectangle = length × width , 24 sq cm = l × w and perimeter = 2 × (length + width) , 20 cm = 2 × ( l + w), l + w = 20 cm ÷ 2 =10 cm, so l = 10 cm – w, now 24 = (10 – w) × w, we get w 2 – 10 w + 24 = 0, So w 2 – 6 w – 4 w + 24 = 0, w(w – 6) – 4(w – 6) = 0, therefore w = 6 cm or w = 4 cm, So l = 10 cm – 6 cm = 4 cm or l = 10 cm  – 4 cm = 6 cm, Therfore, If length of rectangle is 4 cm then width is 6 cm or length of rectangle is 6 cm then width is 4 cm.

Explanation: Given area = 24 square m and perimeter = 28 m of rectangles, lets take length as l and width as w and we know area of rectangle = length X width , 24 sq m = l X w and perimeter = 2 × (length + width) , 28 m = 2 × ( l + w), l + w = 28 m ÷ 2 =14 m, so l = 14 m – w, now 24 = (14 – w) × w, we get w 2 – 14 w + 24 = 0, So w 2 – 12 w – 2 w + 24 = 0, w(w – 12) – 2(w – 12) = 0, therefore w = 12 m or w = 2 m, So l = 14 m – 12 m = 2 m or l = 14 m – 2 m = 12 m, Therfore, If length of rectangle is 2 m then width is 12 m or length of rectangle is 12 m then width is 2 m.

Eureka Math Grade 4 Module 3 Lesson 1 Exit Ticket Answer Key

Explanation: Given length = 8 cm and breadth = 2 cm, Area of rectangle is = l × b = 8 cm × 2 cm = 16 square cm, Perimeter of rectangle is =2 × (l + b) = 2 × (8 cm + 2 cm) = 2 × (10 cm) = 20 cm.

Explanation: Given length = 347 m and breadth = 99 m, Perimeter of rectangle is =2 × (l + b) = 2 × (347 m + 99 m) = 2 × (446 m) = 892 m.

Eureka Math Grade 4 Module 3 Lesson 1 Homework Answer Key

Explanation: As shown in figure A length = 8 cm , breadth = 5 cm, Area of rectangle is = l × b = 8 cm × 5 cm = 40 square cm, and in figure B length = 5 cm , breadth = 7 cm, Area of rectangle is = l × b = 5 cm × 7 cm = 35 square cm,

b. P = 26 cm,  P = 24 cm,

Explanation: As shown in figure A length = 8 cm , breadth = 5 cm, Perimeter of rectangle is =2 × (l + b) = 2(8 cm + 5 cm) = 2 × (13 cm) = 26 cm, and in figure B length = 5 cm , breadth = 7 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (5 cm + 7 cm) = 2 × (12 cm) = 24 cm.

Explanation: Given length = 7 cm and breadth = 3 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (7 cm + 3 cm) = 2 × (10 cm) = 20 cm, Area of rectangle is = l × b = 7 cm × 3 cm = 21 square cm,

Explanation: Given length = 9 cm and breadth = 4 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (9 cm + 4 cm) = 2 × (13 cm) = 26 cm, Area of rectangle is = l × b = 9 cm × 4 cm = 36 square cm,

Explanation: Given length = 76 m and breadth = 149 m, Perimeter of rectangle is =2 × (l + b) = 2 × (76 m + 149 cm) = 2 × ( 425 m) = 850 m.

Explanation: Given length = 45 cm and breadth = 2 m 10 cm, we know 1 m = 100 cm , so breadth =  2 × 100 cm + 10 cm = 200 cm + 10 cm = 210 cm, Perimeter of rectangle is =2 × (l + b) = 2 × (45 cm + 210 cm) = 2 × (255 cm) = 510 cm.

Explanation: Given width of rectangle as 6 cm and area as 60 square cm, and unknow side length is x cm, we know area of rectangle is length × breadth, So 60 sq cm = 6 cm × x cm, So x cm = 60 sq cm ÷ 6 cm = 10 cm, therefore unknown side length = 10 cm.

Explanation: Given width of rectangle as 5 m and area as 25 square m, and unknown side length is x m, we know the area of the rectangle is length × breadth, So 25 sq m = 5 m × x m, So x m = 25 sq m ÷ 5 m = 5 m, therefore unknown side length = 5 m.

Explanation: Given width of rectangle as 40 cm and perimeter as 180 cm, and unknow side length is x cm, we know perimeter of rectangle is 2 × (length + breadth), So 180 cm = 2 × (x cm + 40 cm), 180 cm ÷ 2 = (x cm + 40 cm), 90 cm = (x cm + 40 cm), So x cm = 90 cm – 40 cm = 50 cm, therefore unknown side length = 50 cm.

Explanation: Given length of rectangle as 150 m and perimeter as 1000 m, and unknow side width is x m, we know perimeter of rectangle is 2 X (length + breadth), So 1000 m = 2 X (150 m + x m), 1000 m ÷ 2 = (150 m + x m ), 500 m = 150 m + x m , So x m = 500 m – 150 m = 350 m, therefore unknown side length = 350 m.

Explanation: Given area = 32 square cm and perimeter = 24 cm of rectangles, lets take length as l and width as w and we know area of rectangle = length X width , 32 sq cm = l X w and perimeter = 2 × (length + width) , 24 cm = 2 × ( l + w), l + w = 24 cm ÷ 2 =12 cm, so l = 12 cm – w, now 32 = (12 – w) × w, we get w 2 – 12 w + 32 = 0, So w 2 – 8 w – 4 w + 32 = 0, w(w – 8) – 4(w – 8) = 0, therefore w = 8 cm or w = 4 cm, So l = 12 cm – 8 cm = 4 cm or l = 12 cm  – 4 cm = 8 cm, Therfore, If length of recatangle is 4 cm then width is 8 cm or length of rectangle is 8 cm then width is 4 cm.

Explanation: Given area = 36 square m and perimeter = 30 m of rectangles, lets take length as l and width as w and we know area of rectangle = length X width , 36 sq m = l X w and perimeter = 2 × (length + width) , 30 m = 2 × ( l + w), l + w = 30 m ÷ 2 =15 m, so l = 15 m – w, now 36 = (15 – w) X w, we get w 2 – 15 w + 36 = 0, So w 2 – 12 w – 3 w + 24 = 0, w(w – 12) – 3(w – 12) = 0, therefore w = 12 m or w = 3 m, So l = 15 m – 12 m = 3 m or l = 15 m – 3 m = 12 m, Therfore, If length of rectangle is 3 m then width is 12 m or length of rectangle is 12 m then width is 3 m.

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