\nThe 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.
\na.
\n
![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-17-Problem-Set-Answer-Key-pr-1.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 16 cm.<\/p>\n
Question 1.
\nThe 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.
\na.
\n
\nP =
\nAnswer:
\nPerimeter of the given figure = 16 cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/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.png\")
\nLength of the side AB in the given figure = 4 cm
\nLength of the side BC in the given figure = 2 cm
\nLength of the side CD in the given figure = 2 cm
\nLength of the side ED in the given figure = 1 cm
\nLength of the side EF in the given figure = 2 cm
\nLength of the side FA in the given figure = 3 cm
\nLength of the side GD in the given figure = 2 cm
\nPerimeter 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
\n= 4 cm + 2 cm + 2 cm + 1 cm + 2 cm + 3 cm + 2cm
\n= 6 cm + 2 cm + 1 cm + 2 cm + 3 cm + 2cm
\n= 8 cm + 1 cm + 2 cm + 3 cm + 2cm
\n= 9 cm + 2 cm + 3 cm + 2cm
\n= 11 cm + 3\u00a0cm + 2cm
\n= 14 cm + 2 cm
\n= 16 cm.<\/p>\n
<\/p>\n
b.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-17-Problem-Set-Answer-Key-pr-2.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 16ft.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/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-1.png\")
\nLength of the side AB in the given figure = 2ft
\nLength of the side BC in the given figure = 1ft
\nLength of the side CD in the given figure = 1ft
\nLength of the side DE in the given figure = 1ft
\nLength of the side EF in the given figure = 2ft
\nLength of the side GF in the given figure = 2ft
\nLength of the side GH in the given figure = 5ft
\nLength of the side HA in the given figure = 2ft
\nPerimeter 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
\n= 2ft + 1ft + 1ft + 1ft\u00a0 + 2ft + 2ft+ 5ft + 2ft
\n= 3ft + 1ft + 1ft\u00a0 + 2ft + 2ft+ 5ft + 2ft
\n= 4ft + 1ft\u00a0 + 2ft + 2ft+ 5ft + 2ft
\n= 5ft + 2ft + 2ft+ 5ft + 2ft
\n= 7ft + 2ft + 5ft + 2ft
\n= 9ft + 5ft + 2ft
\n= 14ft + 2ft
\n= 16ft.<\/p>\n
<\/p>\n
<\/p>\n
c.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-17-Problem-Set-Answer-Key-pr-3.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 24m.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/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.png\")
\nLength of the side AB in the given figure = 2m
\nLength of the side BC in the given figure = 2m
\nLength of the side CD in the given figure = 4m
\nLength of the side DE in the given figure = 2m
\nLength of the side EF in the given figure = 4m
\nLength of the side GF in the given figure = 2m
\nLength of the side GH in the given figure = 2m
\nLength of the side HA in the given figure = 6m
\nPerimeter 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
\n= 2m + 2m + 4m + 2m + 4m + 2m + 2m + 6m
\n= 4m + 4m + 2m + 4m + 2m + 2m + 6m
\n= 8m + 2m + 4m + 2m + 2m + 6m
\n= 10m + 4m + 2m + 2m + 6m
\n= 14m + 2m + 2m + 6m
\n= 16m + 2m + 6m
\n= 18m + 6m
\n= 24m.<\/p>\n
<\/p>\n
d.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-17-Problem-Set-Answer-Key-pr-4.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 26yd.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/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-1.png\")
\nLength of the side AB in the given figure = 7yd
\nLength of the side BC in the given figure = 2yd
\nLength of the side CD in the given figure = 2yd
\nLength of the side DE in the given figure = 4yd
\nLength of the side EF in the given figure = 2yd
\nLength of the side FG in the given figure = 2yd
\nLength of the side GH in the given figure = 1yd
\nLength of the side HI in the given figure = 2yd
\nLength of the side IJ in the given figure = 2yd
\nLength of the side JA in the given figure = 2yd
\nPerimeter 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
\n= 7yd + 2yd + 2yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
\n= 9yd + 2yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
\n= 11yd + 4yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
\n= 15yd + 2yd + 2yd + 1yd + 2yd + 2yd + 2yd
\n= 17yd + 2yd + 1yd + 2yd + 2yd + 2yd
\n= 19yd + 1yd + 2yd + 2yd + 2yd
\n= 20yd + 2yd + 2yd + 2yd
\n= 22yd + 2yd + 2yd
\n= 24yd + 2yd
\n= 26yd.<\/p>\n
<\/p>\n
Question 2.
\nNathan draws and labels the square and rectangle below. Find the perimeter of the new shape.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-17-Problem-Set-Answer-Key-pr-5.png\")
\nAnswer:
\nPerimeter of the\u00a0 ACDF\u00a0 new shape = 48cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/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.png\")
\nLength of the side of AB in the given figure = 6cm
\nLength of the side of BC in the given figure = 12cm
\nLength of the side of CD in the given figure = 6cm
\nLength of the side of DE in the given figure = 12cm
\nLength of the side of EF in the given figure = 6cm
\nLength of the side of FA in the given figure = 6cm
\nPerimeter 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
\n= 6cm + 12cm + 6cm + 12cm + 6cm + 6cm
\n= 18cm + 6cm + 12cm + 6cm + 6cm
\n= 24cm + 12cm + 6cm + 6cm
\n= 36cm + 6cm + 6cm
\n= 42cm + 6cm
\n= 48cm.<\/p>\n
<\/p>\n
Question 3.
\nLabel the unknown side lengths. Then, find the perimeter of the shaded rectangle.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-17-Problem-Set-Answer-Key-pr-6.png\")
\nAnswer:
\nPerimeter of the DFGC shaded rectangle = 26in.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/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.png\")
\nShaded rectangle = DFGC
\nLength of the side of AB of the given figure\u00a0 = 16in
\nLength of the side of BG of the given figure = 2in
\nLength of the side of GC of the given figure = EA – BG = 7in – 2in = 5in.
\nLength of the side of CD of the given figure= AB – DE = 16in – 8in = 8in.
\nLength of the side of DE of the given figure= 8in
\nLength of the side of EA of the given figure= 7in
\nLength of the side of DF of the given figure= 5in
\nLength of the side of FG of the given figure= 8in
\nPerimeter 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
\n= 8in + 5in + 8in + 5in
\n= 13in + 8in + 5in
\n= 21in + 5in
\n= 26in.<\/p>\n
<\/p>\n
<\/p>\n
Label the unknown side lengths. Then, find the perimeter of the shaded rectangle.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-3rd-Grade-Module-7-Lesson-17-Exit-Ticket-Answer-Key-t-1.png\")
\nAnswer:
\nPerimeter of the FGDE shaded rectangle = 30m.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Exit-Ticket-Answer-Key.png\")
\nShaded rectangle = FGDE
\nLength of the side AB in the given figure = 12m
\nLength of the side BC in the given figure = 14m
\nLength of the side CD in the given figure = 5m
\nLength of the side DE in the given figure = AB – CD = 12m – 5m = 7m.
\nLength of the side EF in the given figure = BC – FA = 14m – 6m = 8m.
\nLength of the side FA in the given figure = 6m
\nLength of the side FG in the given figure = 7m
\nLength of the side GD in the given figure = 8m
\nPerimeter of the FGDE shaded rectangle = Length of the side FG\u00a0 + Length of the side GD + Length of the side DE + Length of the side EF
\n= 7m + 8m + 7m + 8m
\n= 15m + 7m + 8m
\n= 22m + 8m
\n= 30m.<\/p>\n
<\/p>\n
<\/p>\n
Question 1.
\nThe 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.
\na.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-h-1.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 32m.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-1.png\")
\nLength of the side of AB in the given figure = 4m
\nLength of the side of BC in the given figure = 9m
\nLength of the side of CD in the given figure = 7m
\nLength of the side of DE in the given figure = 2m
\nLength of the side of EF in the given figure = 3m
\nLength of the side of FA in the given figure = 7m
\nPerimeter 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
\n= 4m + 9m + 7m + 2m + 3m + 7m
\n= 13m + 7m + 2m + 3m + 7m
\n= 20m + 2m + 3m + 7m
\n= 22m + 3m + 7m
\n= 25m + 7m
\n= 32m.<\/p>\n
<\/p>\n
<\/p>\n
b.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-h-2.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 34 cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-2.png\")
<\/p>\n
Length of the side of AB in the given figure = 2 cm
\nLength of the side of BC in the given figure = 4 cm
\nLength of the side of CD in the given figure = 4 cm
\nLength of the side of DE in the given figure = 3 cm
\nLength of the side of EF in the given figure = 2 cm
\nLength of the side of FG in the given figure = 5 cm
\nLength of the side of GH in the given figure = 8 cm
\nLength of the side of HA in the given figure = 6 cm
\nPerimeter 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
\n= 2 cm + 4 cm + 4 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
\n= 6 cm + 4 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
\n= 10 cm + 3 cm + 2 cm + 5 cm + 8 cm + 6 cm
\n= 13 cm + 2 cm + 5 cm + 8 cm + 6 cm
\n= 15 cm + 5 cm + 8 cm + 6 cm
\n= 20 cm + 8 cm + 6 cm
\n= 28 cm + 6 cm
\n= 34 cm.<\/p>\n
<\/p>\n
c.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-h-3.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 40in.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-1c.png\")
\nLength of the side of AB in the given figure = 12in
\nLength of the side of BC in the given figure = 2in
\nLength of the side of CD in the given figure = 4in
\nLength of the side of DE in the given figure = 6in
\nLength of the side of EF in the given figure = 4in
\nLength of the side of FG in the given figure = 6in
\nLength of the side of GH in the given figure = 4in
\nLength of the side of HA in the given figure = 2in
\nPerimeter of the given figure = Length of the side of AB\u00a0 + 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
\n= 12in + 2in + 4in + 6in + 4in + 6in + 4in + 2in
\n= 14in + 4in + 6in + 4in + 6in + 4in + 2in
\n= 18in + 6in + 4in + 6in + 4in + 2in
\n= 24in + 4in + 6in + 4in + 2in
\n= 28in + 6in + 4in + 2in
\n= 34in + 4in + 2in
\n= 38in + 2in
\n= 40in.<\/p>\n
<\/p>\n
d.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-h-4.png\")
\nP =
\nAnswer:
\nPerimeter of the given figure = 30ft.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-1d.png\")
<\/p>\n
Length of the side of AB in the given figure = 8ft
\nLength of the side of BC in the given figure = 3ft
\nLength of the side of CD in the given figure = 3ft
\nLength of the side of DE in the given figure = 1ft
\nLength of the side of EF in the given figure = 3ft
\nLength of the side of FG in the given figure = 3ft
\nLength of the side of GH in the given figure = 2ft
\nLength of the side of HA in the given figure = 7ft
\nPerimeter of the given figure = Length of the side of AB + Length of the side of BC + Length of the side of CD\u00a0 + 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
\n= 8ft + 3ft + 3ft + 1ft + 3ft + 3ft + 2ft + 7ft
\n= 11ft + 3ft + 1ft + 3ft + 3ft + 2ft + 7ft
\n= 14ft + 1ft + 3ft + 3ft + 2ft + 7ft
\n= 15ft + 3ft + 3ft + 2ft + 7ft
\n= 18ft + 3ft + 2ft + 7ft
\n= 21ft + 2ft + 7ft
\n= 23ft + 7ft
\n= 30ft.<\/p>\n
<\/p>\n
<\/p>\n
Question 2.
\nSari draws and labels the squares and rectangle below. Find the perimeter of the new shape.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-h-5.png\")
\nAnswer:
\nPerimeter of the new shape = 72cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-2...png\")
<\/p>\n
Length of the side of AB in the given figure = 6cm
\nLength of the side of BC in the given figure = 18cm
\nLength of the side of CD in the given figure = 6cm
\nLength of the side of DE in the given figure = 6cm
\nLength of the side of EF in the given figure = 6cm
\nLength of the side of FG in the given figure = 18cm
\nLength of the side of GH in the given figure = 6cm
\nLength of the side of HA in the given figure = 6cm
\nPerimeter 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
\n= 6cm + 18cm + 6cm + 6cm + 6cm + 18cm + 6cm + 6cm
\n= 24cm + 6cm + 6cm + 6cm + 18cm + 6cm + 6cm
\n= 30cm + 6cm + 6cm + 18cm + 6cm + 6cm
\n= 36cm + 6cm + 18cm + 6cm + 6cm
\n= 42cm + 18cm + 6cm + 6cm
\n= 60cm + 6cm + 6cm
\n= 66cm + 6cm
\n= 72cm.<\/p>\n
<\/p>\n
Question 3.
\nLabel the unknown side lengths. Then, find the perimeter of the shaded rectangle.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-h-6.png\")
\nAnswer:
\nPerimeter of the shaded rectangle = 37in.<\/p>\n
Explanation:<\/p>\n
![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-17-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-17-Homework-Answer-Key-3.png\")
<\/p>\n
Shaded rectangle =\u00a0 BCDG
\nLength of the side of AB in the given figure = 5in
\nLength of the side of BC in the given figure =\u00a0 EF – AB = 18in – 5in = 13in
\nLength of the side of CD in the given figure = FA – DE = 8in – 2in = 6in
\nLength of the side of DE in the given figure = 2in
\nLength of the side of EF in the given figure = 18in
\nLength of the side of FA in the given figure = 8in
\nLength of the side of GB in the given figure = 6in
\nLength of the side of GD in the given figure = 13in
\nPerimeter 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
\n= 13in + 6in + 13in + 6in
\n= 19in + 13in + 6in
\n= 31in + 6in
\n= 37in.<\/p>\n
<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
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. P = … Read more<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/55873"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=55873"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/55873\/revisions"}],"predecessor-version":[{"id":92446,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/55873\/revisions\/92446"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=55873"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=55873"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=55873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}