## 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 cm^{2}__________Â Â Â A = ______54 cm^{2}_________

b. P = ______32 cm_________Â Â Â P = ______30 cm_________

Answer:

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 Ã— b = 9 cm Ã— 7 cm = 63 square cm,

and in figure B length = 6 cm , breadth = 9 cm,

Area of rectangle is = l Ã— b = 6 cm Ã— 9 cm = 54 square cm.

b. P = 32 cm,Â P = 30 cm,

Explanation:

As shown in figure A length = 9 cm , breadth = 7 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2(9 cm + 7 cm) =

2 Ã— (16 cm) = 32 cm,

and in figure B length = 6 cm , breadth = 9 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (6 cm + 9 cm) =

2 Ã— (15 cm) = 30 cm.

Question 2.

Determine the perimeter and area of each rectangle.

a.

P = _____22 cm_______

A = _____30 cm^{2}_______

Answer:

Perimeter = 22 cm,

Area =Â 30 cm^{2},

Explanation:

Given length = 5 cm and breadth = 6 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (5 cm + 6 cm) =

2 Ã— (11 cm) = 22 cm,

Area of rectangle is = l Ã— b = 5 cm Ã— 6 cm = 30 square cm.

b.

P = ____22 cm_______

A = ____ 24 cm^{2}________

Answer:

Perimeter = 22 cm,

Area = 24 cm^{2},

Explanation:

Given length = 8 cm and breadth = 3 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (8 cm + 3 cm) =

2 Ã— (11 cm) = 22 cm,

Area of rectangle is = l Ã— b = 8 cm Ã— 3 cm = 24 square cm.

Question 3.

Determine the perimeter of each rectangle.

a.

P = ____530 m________

Answer:

Perimeter = 530 m,

Explanation:

Given length = 99 m and breadth = 166 m,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (99 m + 166 m)

= 2 Ã— ( 265 m) = 530 m.

b.

P = ____450 cm________

Answer:

Perimeter =Â 450 cm,

Explanation:

Given length = 75 cm and breadth = 1 m 50 cm,

we know 1 m = 100 cm ,

so breadth =Â 1 Ã— 100 cm + 50 cm =150 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (75 cm + 150 cm)

= 2 Ã— (225 cm) = 450 cm.

Question 4.

Given the rectangleâ€™s area, find the unknown side length.

a.

x = ____10 cm________

Answer:

Unknown side length = 10 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 Ã— breadth, So 80 sq cm = 8 cm Ã— x cm,

So x cm = 80 sq cm Ã· 8 cm = 10 cm,

therefore unknown side length = 10 cm.

b.

x = ____7 cm________

Answer:

Unknown side length = 7 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 Ã— breadth, So 49 sq cm = 7 cm Ã— x cm,

So x cm = 49 sq cm Ã· 7 cm = 7 cm,

therefore unknown side length = 7 cm.

Question 5.

Given the rectangleâ€™s perimeter, find the unknown side length.

a. P = 120 cm

Answer:

Unknown side length = 40 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 Ã— (length + breadth),

So 120 cm = 2 Ã— (x cm + 20 cm),

120 cm Ã· 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.

b. p = 1000 m

Answer:

Unknown side width = 250 m,

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 Ã— (length + breadth),

So 1000 m = 2 Ã— (250 m + x m),

1000 m Ã· 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.

Question 6.

Each of the following rectangles has whole number side lengths.

Given the area and perimeter, find the length and width.

a. P = 20 cm

Answer:

The whole number side lengths are

If the length of the rectangle is 4 cm then the width is 6 cm or

length of the rectangle is 6 cm then the width is 4 cm,

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 Ã— width , 24 sq cm = l Ã— w and

perimeter = 2 Ã— (length + width) ,

20 cm = 2 Ã— ( l + w), l + w = 20 cm Ã· 2 =10 cm,

so l = 10 cm – w, now 24 = (10 – w) Ã— 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 cmÂ – 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.

b. P = 28 m

Answer:

The whole number side lengths are

If the length of the rectangle is 2 m then the width is 12 m or

The length of the rectangle is 12 m the width is 2 m.

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 Ã— (length + width) ,

28 m = 2 Ã— ( l + w), l + w = 28 m Ã· 2 =14 m,

so l = 14 m – w, now 24 = (14 – w) Ã— 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

Question 1.

Determine the area and perimeter of the rectangle.

Answer:

Area = 16 cm^{2},

Perimeter = 20 cm,

Explanation:

Given length = 8 cm and breadth = 2 cm,

Area of rectangle is = l Ã— b = 8 cm Ã— 2 cm = 16 square cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (8 cm + 2 cm) =

2 Ã— (10 cm) = 20 cm.

Question 2.

Determine the perimeter of the rectangle.

Answer:

Perimeter = 892 m,

Explanation:

Given length = 347 m and breadth = 99 m,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (347 m + 99 m) =

2 Ã— (446 m) = 892 m.

### Eureka Math Grade 4 Module 3 Lesson 1 Homework Answer Key

Question 1.

Determine the perimeter and area of rectangles A and B.

a. A = _____40 cm^{2}__________Â Â A = ______35Â cm^{2}_________

b. P = ______26 cm_________Â Â P = _______24 cm________

Answer:

a. A = 40 cm^{2 }, A = 35 cm^{2},

Explanation:

As shown in figure A length = 8 cm , breadth = 5 cm,

Area of rectangle is = l Ã— b = 8 cm Ã— 5 cm = 40 square cm,

and in figure B length = 5 cm , breadth = 7 cm,

Area of rectangle is = l Ã— b = 5 cm Ã— 7 cm = 35 square cm,

b. P = 26 cm,Â P = 24 cm,

Explanation:

As shown in figure A length = 8 cm , breadth = 5 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2(8 cm + 5 cm) =

2 Ã— (13 cm) = 26 cm,

and in figure B length = 5 cm , breadth = 7 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (5 cm + 7 cm) =

2 Ã— (12 cm) = 24 cm.

Question 2.

Determine the perimeter and area of each rectangle.

a.

P = ____20 cm________

A = ___ 21 cm^{2}____

Answer:

Perimeter = 20 cm,

Area = 21 cm^{2},

Explanation:

Given length = 7 cm and breadth = 3 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (7 cm + 3 cm) =

2 Ã— (10 cm) = 20 cm,

Area of rectangle is = l Ã— b = 7 cm Ã— 3 cm = 21 square cm,

b.

P = ___26 cm_________

A = ______36 cm^{2}______

Answer:

Perimeter = 26 cm,

Area = 36 cm^{2},

Explanation:

Given length = 9 cm and breadth = 4 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (9 cm + 4 cm) =

2 Ã— (13 cm) = 26 cm,

Area of rectangle is = l Ã— b = 9 cm Ã— 4 cm = 36 square cm,

Question 3.

Determine the perimeter of each rectangle.

a.

P = _____850 m_______

Answer:

Perimeter = 850 m,

Explanation:

Given length = 76 m and breadth = 149 m,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (76 m + 149 cm) =

2 Ã— ( 425 m) = 850 m.

b.

P = ____510 cm________

Answer:

Perimeter =Â 510 cm,

Explanation:

Given length = 45 cm and breadth = 2 m 10 cm,

we know 1 m = 100 cm ,

so breadth =Â 2 Ã— 100 cm + 10 cm = 200 cm + 10 cm = 210 cm,

Perimeter of rectangle is =2 Ã— (l + b) = 2 Ã— (45 cm + 210 cm) =

2 Ã— (255 cm) = 510 cm.

Question 4.

Given the rectangleâ€™s area, find the unknown side length.

a.

x = _____10 cm_______

Answer:

Unknown side length = 10 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 Ã— breadth, So 60 sq cm = 6 cm Ã— x cm,

So x cm = 60 sq cm Ã· 6 cm = 10 cm,

therefore unknown side length = 10 cm.

b.

x = _____5 m_______

Answer:

Unknown side length = 5 m,

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 Ã— breadth, So 25 sq m = 5 m Ã— x m,

So x m = 25 sq m Ã· 5 m = 5 m,

therefore unknown side length = 5 m.

Question 5.

Given the rectangleâ€™s perimeter, find the unknown side length.

a. P = 180 cm

x = ____40 cm________

Answer:

Unknown side length = 40 cm,

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 Ã— (length + breadth),

So 180 cm = 2 Ã— (x cm + 40 cm),

180 cm Ã· 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.

b. P = 1,000 m

x = ____350 m________

Answer:

Unknown side width = 350 m,

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 Ã· 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.

Question 6.

Each of the following rectangles has whole number side lengths. Given the area and perimeter, find the length and width.

a. A = 32 square cm

P = 24 cm

Answer:

The whole number side lengths are

If the length of the rectangle is 4 cm then the width is 8 cm or

length of the rectangle is 8 cm then the width is 4 cm,

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 Ã— (length + width) ,

24 cm = 2 Ã— ( l + w), l + w = 24 cm Ã· 2 =12 cm,

so l = 12 cm – w, now 32 = (12 – w) Ã— 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 cmÂ – 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.

b. A = 36 square m

P = 30 m

Answer:

The whole number side lengths are

If the length of the rectangle is 3 m then the width is 12 m or

The length of the rectangle is 12 m the width is 3 m.

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 Ã— (length + width) ,

30 m = 2 Ã— ( l + w), l + w = 30 m Ã· 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.