Eureka Math Grade 3 Module 7 Lesson 17 Answer Key

Engage NY Eureka Math 3rd Grade Module 7 Lesson 17 Answer Key

Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key

Question 1.
The shapes below are made up of rectangles. Label the unknown side lengths. Then, write and solve an equation to find the perimeter of each shape.
a.
Engage NY Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key pr 1
P =
Answer:
Perimeter of the given figure = 16 cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key-1a
Length of the side AB in the given figure = 4 cm
Length of the side BC in the given figure = 2 cm
Length of the side CD in the given figure = 2 cm
Length of the side ED in the given figure = 1 cm
Length of the side EF in the given figure = 2 cm
Length of the side FA in the given figure = 3 cm
Length of the side GD in the given figure = 2 cm
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side ED + Length of the side EF + Length of the side FA + Length of the side GD
= 4 cm + 2 cm + 2 cm + 1 cm + 2 cm + 3 cm + 2cm
= 6 cm + 2 cm + 1 cm + 2 cm + 3 cm + 2cm
= 8 cm + 1 cm + 2 cm + 3 cm + 2cm
= 9 cm + 2 cm + 3 cm + 2cm
= 11 cm + 3 cm + 2cm
= 14 cm + 2 cm
= 16 cm.

 

b.
Engage NY Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key pr 2
P =
Answer:
Perimeter of the given figure = 16ft.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key-1b
Length of the side AB in the given figure = 2ft
Length of the side BC in the given figure = 1ft
Length of the side CD in the given figure = 1ft
Length of the side DE in the given figure = 1ft
Length of the side EF in the given figure = 2ft
Length of the side GF in the given figure = 2ft
Length of the side GH in the given figure = 5ft
Length of the side HA in the given figure = 2ft
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side DE + Length of the side EF + Length of the side GF + Length of the side GH + Length of the side HA
= 2ft + 1ft + 1ft + 1ft  + 2ft + 2ft+ 5ft + 2ft
= 3ft + 1ft + 1ft  + 2ft + 2ft+ 5ft + 2ft
= 4ft + 1ft  + 2ft + 2ft+ 5ft + 2ft
= 5ft + 2ft + 2ft+ 5ft + 2ft
= 7ft + 2ft + 5ft + 2ft
= 9ft + 5ft + 2ft
= 14ft + 2ft
= 16ft.

 

 

c.
Engage NY Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key pr 3
P =
Answer:
Perimeter of the given figure = 24m.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key-1c
Length of the side AB in the given figure = 2m
Length of the side BC in the given figure = 2m
Length of the side CD in the given figure = 4m
Length of the side DE in the given figure = 2m
Length of the side EF in the given figure = 4m
Length of the side GF in the given figure = 2m
Length of the side GH in the given figure = 2m
Length of the side HA in the given figure = 6m
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side DE + Length of the side EF + Length of the side GF + Length of the side GH + Length of the side HA
= 2m + 2m + 4m + 2m + 4m + 2m + 2m + 6m
= 4m + 4m + 2m + 4m + 2m + 2m + 6m
= 8m + 2m + 4m + 2m + 2m + 6m
= 10m + 4m + 2m + 2m + 6m
= 14m + 2m + 2m + 6m
= 16m + 2m + 6m
= 18m + 6m
= 24m.

 

d.
Engage NY Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key pr 4
P =
Answer:
Perimeter of the given figure = 26yd.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key-1d
Length of the side AB in the given figure = 7yd
Length of the side BC in the given figure = 2yd
Length of the side CD in the given figure = 2yd
Length of the side DE in the given figure = 4yd
Length of the side EF in the given figure = 2yd
Length of the side FG in the given figure = 2yd
Length of the side GH in the given figure = 1yd
Length of the side HI in the given figure = 2yd
Length of the side IJ in the given figure = 2yd
Length of the side JA in the given figure = 2yd
Perimeter of the given figure = Length of the side AB + Length of the side BC + Length of the side CD + Length of the side DE + Length of the side EF + Length of the side FG + Length of the side GH + Length of the side HI + Length of the side IJ + Length of the side JA
= 7yd + 2yd + 2yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 9yd + 2yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 11yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 15yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 17yd + 2yd + 1yd + 2yd + 2yd + 2yd
= 19yd + 1yd + 2yd + 2yd + 2yd
= 20yd + 2yd + 2yd + 2yd
= 22yd + 2yd + 2yd
= 24yd + 2yd
= 26yd.

 

Question 2.
Nathan draws and labels the square and rectangle below. Find the perimeter of the new shape.
Engage NY Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key pr 5
Answer:
Perimeter of the  ACDF  new shape = 48cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key-2
Length of the side of AB in the given figure = 6cm
Length of the side of BC in the given figure = 12cm
Length of the side of CD in the given figure = 6cm
Length of the side of DE in the given figure = 12cm
Length of the side of EF in the given figure = 6cm
Length of the side of FA in the given figure = 6cm
Perimeter of the ACDF new shape = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FA
= 6cm + 12cm + 6cm + 12cm + 6cm + 6cm
= 18cm + 6cm + 12cm + 6cm + 6cm
= 24cm + 12cm + 6cm + 6cm
= 36cm + 6cm + 6cm
= 42cm + 6cm
= 48cm.

 

Question 3.
Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.
Engage NY Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key pr 6
Answer:
Perimeter of the DFGC shaded rectangle = 26in.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Problem Set Answer Key-3
Shaded rectangle = DFGC
Length of the side of AB of the given figure  = 16in
Length of the side of BG of the given figure = 2in
Length of the side of GC of the given figure = EA – BG = 7in – 2in = 5in.
Length of the side of CD of the given figure= AB – DE = 16in – 8in = 8in.
Length of the side of DE of the given figure= 8in
Length of the side of EA of the given figure= 7in
Length of the side of DF of the given figure= 5in
Length of the side of FG of the given figure= 8in
Perimeter of the DFGC shaded rectangle = Length of the side of FG + Length of the side of GC + Length of the side of CD + Length of the side of DF
= 8in + 5in + 8in + 5in
= 13in + 8in + 5in
= 21in + 5in
= 26in.

 

 

Eureka Math 3rd Grade Module 7 Lesson 17 Exit Ticket Answer Key

Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.
Eureka Math 3rd Grade Module 7 Lesson 17 Exit Ticket Answer Key t 1
Answer:
Perimeter of the FGDE shaded rectangle = 30m.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math 3rd Grade Module 7 Lesson 17 Exit Ticket Answer Key
Shaded rectangle = FGDE
Length of the side AB in the given figure = 12m
Length of the side BC in the given figure = 14m
Length of the side CD in the given figure = 5m
Length of the side DE in the given figure = AB – CD = 12m – 5m = 7m.
Length of the side EF in the given figure = BC – FA = 14m – 6m = 8m.
Length of the side FA in the given figure = 6m
Length of the side FG in the given figure = 7m
Length of the side GD in the given figure = 8m
Perimeter of the FGDE shaded rectangle = Length of the side FG  + Length of the side GD + Length of the side DE + Length of the side EF
= 7m + 8m + 7m + 8m
= 15m + 7m + 8m
= 22m + 8m
= 30m.

 

 

Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key

Question 1.
The shapes below are made up of rectangles. Label the unknown side lengths. Then, write and solve an equation to find the perimeter of each shape.
a.
Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key h 1
P =
Answer:
Perimeter of the given figure = 32m.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key-1
Length of the side of AB in the given figure = 4m
Length of the side of BC in the given figure = 9m
Length of the side of CD in the given figure = 7m
Length of the side of DE in the given figure = 2m
Length of the side of EF in the given figure = 3m
Length of the side of FA in the given figure = 7m
Perimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FA
= 4m + 9m + 7m + 2m + 3m + 7m
= 13m + 7m + 2m + 3m + 7m
= 20m + 2m + 3m + 7m
= 22m + 3m + 7m
= 25m + 7m
= 32m.

 

 

b.
Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key h 2
P =
Answer:
Perimeter of the given figure = 34 cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key-2

Length of the side of AB in the given figure = 2 cm
Length of the side of BC in the given figure = 4 cm
Length of the side of CD in the given figure = 4 cm
Length of the side of DE in the given figure = 3 cm
Length of the side of EF in the given figure = 2 cm
Length of the side of FG in the given figure = 5 cm
Length of the side of GH in the given figure = 8 cm
Length of the side of HA in the given figure = 6 cm
Perimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 2 cm + 4 cm + 4 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 6 cm + 4 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 10 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 13 cm + 2 cm + 5 cm + 8 cm + 6 cm
= 15 cm + 5 cm + 8 cm + 6 cm
= 20 cm + 8 cm + 6 cm
= 28 cm + 6 cm
= 34 cm.

 

c.
Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key h 3
P =
Answer:
Perimeter of the given figure = 40in.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key-1c
Length of the side of AB in the given figure = 12in
Length of the side of BC in the given figure = 2in
Length of the side of CD in the given figure = 4in
Length of the side of DE in the given figure = 6in
Length of the side of EF in the given figure = 4in
Length of the side of FG in the given figure = 6in
Length of the side of GH in the given figure = 4in
Length of the side of HA in the given figure = 2in
Perimeter of the given figure = Length of the side of AB  + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 12in + 2in + 4in + 6in + 4in + 6in + 4in + 2in
= 14in + 4in + 6in + 4in + 6in + 4in + 2in
= 18in + 6in + 4in + 6in + 4in + 2in
= 24in + 4in + 6in + 4in + 2in
= 28in + 6in + 4in + 2in
= 34in + 4in + 2in
= 38in + 2in
= 40in.

 

d.
Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key h 4
P =
Answer:
Perimeter of the given figure = 30ft.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key-1d

Length of the side of AB in the given figure = 8ft
Length of the side of BC in the given figure = 3ft
Length of the side of CD in the given figure = 3ft
Length of the side of DE in the given figure = 1ft
Length of the side of EF in the given figure = 3ft
Length of the side of FG in the given figure = 3ft
Length of the side of GH in the given figure = 2ft
Length of the side of HA in the given figure = 7ft
Perimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD  + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 8ft + 3ft + 3ft + 1ft + 3ft + 3ft + 2ft + 7ft
= 11ft + 3ft + 1ft + 3ft + 3ft + 2ft + 7ft
= 14ft + 1ft + 3ft + 3ft + 2ft + 7ft
= 15ft + 3ft + 3ft + 2ft + 7ft
= 18ft + 3ft + 2ft + 7ft
= 21ft + 2ft + 7ft
= 23ft + 7ft
= 30ft.

 

 

Question 2.
Sari draws and labels the squares and rectangle below. Find the perimeter of the new shape.
Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key h 5
Answer:
Perimeter of the new shape = 72cm.

Explanation:
Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key-2..

Length of the side of AB in the given figure = 6cm
Length of the side of BC in the given figure = 18cm
Length of the side of CD in the given figure = 6cm
Length of the side of DE in the given figure = 6cm
Length of the side of EF in the given figure = 6cm
Length of the side of FG in the given figure = 18cm
Length of the side of GH in the given figure = 6cm
Length of the side of HA in the given figure = 6cm
Perimeter of the new shape = Length of the side of AB + Length of the side of BC + Length of the side of CD + Length of the side of DE + Length of the side of EF + Length of the side of FG + Length of the side of GH + Length of the side of HA
= 6cm + 18cm + 6cm + 6cm + 6cm + 18cm + 6cm + 6cm
= 24cm + 6cm + 6cm + 6cm + 18cm + 6cm + 6cm
= 30cm + 6cm + 6cm + 18cm + 6cm + 6cm
= 36cm + 6cm + 18cm + 6cm + 6cm
= 42cm + 18cm + 6cm + 6cm
= 60cm + 6cm + 6cm
= 66cm + 6cm
= 72cm.

 

Question 3.
Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.
Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key h 6
Answer:
Perimeter of the shaded rectangle = 37in.

Explanation:

Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka Math Grade 3 Module 7 Lesson 17 Homework Answer Key-3

Shaded rectangle =  BCDG
Length of the side of AB in the given figure = 5in
Length of the side of BC in the given figure =  EF – AB = 18in – 5in = 13in
Length of the side of CD in the given figure = FA – DE = 8in – 2in = 6in
Length of the side of DE in the given figure = 2in
Length of the side of EF in the given figure = 18in
Length of the side of FA in the given figure = 8in
Length of the side of GB in the given figure = 6in
Length of the side of GD in the given figure = 13in
Perimeter of the shaded rectangle = Length of the side of BC+ Length of the side of CD + Length of the side of GD + Length of the side of GB
= 13in + 6in + 13in + 6in
= 19in + 13in + 6in
= 31in + 6in
= 37in.

 

 

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