\n
![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-3rd-Grade-Module-7-Lesson-12-Pattern-Sheet-Answer-Key-p-1.png\")
\nmultiply by 7 (6\u201310)
\nAnswer:
\n
![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-3-Module-7-Lesson-12-Pattern-Sheet-Answer-Key.png\")
<\/p>\nMultiply.
\n
\nmultiply by 7 (6\u201310)
\nAnswer:
\n<\/p>\n
Explanation:
\n7 \u00d7 6 = 42
\n7 \u00d7 7 = 49
\n7 \u00d7 8 = 56
\n7 \u00d7 9 = 63
\n7 \u00d7 10 = 70.<\/p>\n
Question 1.
\nMeasure and label the side lengths of the shapes below in centimeters. Then, 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-12-Problem-Set-Answer-Key-pr-1.png\")
\nPerimeter = _____cm +_____cm +_____cm +_____cm
\n= _______ cm
\nAnswer:
\nPerimeter of the ABCD given shape = Side + Side + Side + Side
\n= 2cm + 2cm + 2cm + 2cm
\n= 8cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-3-Module-7-Lesson-12-Pattern-Sheet-Answer-Key-1a.png\")
\nLength of the AB side of the ABCD shape = 2cm
\nLength of the BC side of the ABCD shape = 2cm
\nLength of the CD side of the ABCD shape = 2cm
\nLength of the DA side of the ABCD shape = 2cm
\nPerimeter of the given ABCD shape = Side + Side + Side + Side
\n= AB + BC + CD+ DA
\n= 2cm + 2cm + 2cm + 2cm
\n= 4cm + 2cm + 2cm
\n= 6cm + 2cm
\n= 8cm.<\/p>\n
<\/p>\n
b.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-2.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
\n= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
\n= 18cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-3-Module-7-Lesson-12-Pattern-Sheet-Answer-Key-1b.png\")
\nLength of the AB side of the given ABCDEF Hexagon shape = 3cm.
\nLength of the BC side of the given ABCDEF Hexagon shape = 3cm.
\nLength of the CD side of the given ABCDEF Hexagon shape = 3cm.
\nLength of the DE side of the given ABCDEF Hexagon shape = 3cm.
\nLength of the EF side of the given ABCDEF Hexagon shape = 3cm.
\nLength of the FA side of the given ABCDEF Hexagon shape = 3cm.
\nPerimeter of the given ABCDEF Hexagon shape = Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EF + FA
\n= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
\n= 6cm +3cm + 3cm + 3cm + 3cm
\n= 9cm +3cm + 3cm + 3cm
\n= 12cm +3cm + 3cm
\n= 15cm + 3cm
\n= 18cm.<\/p>\n
<\/p>\n
c.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-3.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
\n= 4cm + 4cm + 4cm + 4cm
\n= 16cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-3-Module-7-Lesson-12-Pattern-Sheet-Answer-Key-1c.png\")
\nLength of the AB side of the given ABCD Parallelogram shape = 4cm
\nLength of the BC side of the given ABCD Parallelogram shape = 4cm
\nLength of the CD\u00a0 side of the given ABCD Parallelogram shape = 4cm
\nLength of the DA side of the given ABCD Parallelogram shape = 4cm
\nPerimeter of the given ABCD Parallelogram shape = Side + Side + Side + Side
\n=AB + BC + CD +DA
\n= 4cm + 4cm + 4cm + 4cm
\n= 8cm + 4cm + 4cm
\n= 12cm + 4cm
\n=16cm.<\/p>\n
<\/p>\n
d.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-4.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the ABC Triangle = Side + Side + Side
\n= 5cm + 5cm + 5cm
\n= 15cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-1d.png\")
\nLength of the AB side of the ABC Triangle = 5cm
\nLength of the BC side of the ABC Triangle = 5cm
\nLength of the CA side of the ABC Triangle = 5cm
\nPerimeter of the ABC Triangle = Side + Side + Side
\n= AB + BC + CA
\n= 5cm + 5cm + 5cm
\n= 10cm + 5cm
\n= 15cm.<\/p>\n
e.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-5.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the given ABCDEF figure = Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE+ EF + FA
\n= 5.5cm+ 1cm + 3cm + 2cm + 3cm + 3cm
\n= 17.5cm<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-1e.png\")
\nLength of AB side if the given ABCDEF figure = 5.5cm
\nLength of BC side if the given ABCDEF figure = 1cm
\nLength of CD side if the given ABCDEF figure = 3cm
\nLength of DE side if the given ABCDEF figure = 2cm
\nLength of EF side if the given ABCDEF figure = 3cm
\nLength of FA side if the given ABCDEF figure = 3cm
\nPerimeter of the given ABCDEF figure = Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE+ EF + FA
\n= 5.5cm+ 1cm + 3cm + 2cm + 3cm + 3cm
\n= 6.5cm + 3cm + 2cm + 3cm + 3cm
\n= 9.5cm + 2cm + 3cm + 3cm
\n= 11.5cm + 3cm + 3cm
\n= 14.5cm + 3cm
\n= 17.5cm<\/p>\n
Question 2.
\nCarson draws two triangles to create the new shape shown below. Use a ruler to find the side lengths of Carson\u2019s shape in centimeters. Then, find the perimeter.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-6.png\")
\nAnswer:
\nPerimeter of ABCD Carson\u2019s shape = Side + Side + Side + Side
\n= AB + BC + CD+ DA
\n= 2cm + 2cm + 2cm + 2cm
\n=8cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-3-Module-7-Lesson-12-Pattern-Sheet-Answer-Key-2.png\")
<\/p>\n
Length of AB side of ABCD Carson\u2019s shape = 2cm
\nLength of BC side of ABCD Carson\u2019s shape = 2cm
\nLength of CD side of ABCD Carson\u2019s shape = 2cm
\nLength of DA side of ABCD Carson\u2019s shape = 2cm
\nPerimeter of ABCD Carson\u2019s shape = Side + Side + Side + Side
\n= AB + BC + CD+ DA
\n= 2cm + 2cm + 2cm + 2cm
\n=4cm + 2cm + 2cm
\n= 6cm + 2cm
\n=8cm.<\/p>\n
Question 3.
\nHugh and Daisy draw the shapes shown below. Measure and label the side lengths in centimeters. Whose shape has a greater perimeter? How do you know?
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-7.png\")
\nAnswer:
\nPerimeter of ABCDE Hugh’s shape = Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EA
\n= 3cm + 3cm + 3cm + 3cm + 3cm
\n= 15cm.
\nPerimeter of FGHI Daisy’s shape =\u00a0 Side + Side + Side + Side
\n= FG+ GH+ HI+ IF
\n=\u00a0 4cm + 5cm + 4cm + 3cm
\n= 16cm.
\nPerimeter of FGHI Daisy’s shape is greater than the Perimeter of ABCDE Hugh’s shape because the\u00a0 measurement value is greater than the other one’s shape.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-3...png\")
\nLength of the AB side of ABCDE Hugh’s shape = 3cm
\nLength of the BC side of ABCDE Hugh’s shape = 3cm
\nLength of the CD side of ABCDE Hugh’s shape = 3cm
\nLength of the DE side of ABCDE Hugh’s shape = 3cm
\nLength of the EA side of ABCDE Hugh’s shape = 3cm
\nPerimeter of ABCDE Hugh’s shape = Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EA
\n= 3cm + 3cm + 3cm + 3cm + 3cm
\n= 6cm + 3cm + 3cm + 3cm
\n= 9cm + 3cm + 3cm
\n= 12cm + 3cm
\n= 15cm.<\/p>\n
Length of the FG side of FGHI Daisy’s shape = 4cm
\nLength of the GH side of FGHI Daisy’s shape = 5cm
\nLength of the HI side of FGHI Daisy’s shape = 4cm
\nLength of the IF side of FGHI Daisy’s shape = 3cm
\nPerimeter of FGHI Daisy’s shape =\u00a0 Side + Side + Side + Side
\n= FG+ GH+ HI+ IF
\n=\u00a0 4cm + 5cm + 4cm + 3cm
\n= 9cm + 4cm +3cm
\n= 13cm + 3cm
\n= 16cm.<\/p>\n
Question 4.
\nAndrea measures one side length of the square below and says she can find the perimeter with that measurement. Explain Andrea\u2019s thinking. Then, find the perimeter in centimeters.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-pr-8.png\")
\nAnswer:
\nAndrea\u2019s figure is a Square. So, her figure’s\u00a0 all sides are going to be equal. Yes, she is correct in her thinking.
\nPerimeter of ABCD Andrea\u2019s Square = Side + Side + Side + Side
\n= 4cm + 4cm + 4cm+ 4cm
\n= 16cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Problem-Set-Answer-Key-4.png\")
\nAndrea\u2019s figure is a Square. So, her figure’s\u00a0 all sides are going to be equal. Yes, she is correct in her thinking, ” measuring one side length of the square below, she can find the perimeter with that measurement.”
\nLength of AB side of Andrea\u2019s Square = 4cm
\nPerimeter of ABCD Andrea\u2019s Square = Side + Side + Side + Side
\n= 4cm + 4cm + 4cm+ 4cm
\n= 8cm + 4cm + 4cm
\n=12cm + 4cm
\n= 16cm.<\/p>\n
Measure and label the side lengths of the shape below in centimeters. Then, find the perimeter.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-3rd-Grade-Module-7-Lesson-12-Exit-Ticket-Answer-Key-t-1.png\")
\nPerimeter = __________________________________________
\n= _______ cm
\nAnswer:
\nPerimeter of the given ABCDEFGHIJK Shape =\u00a0 Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side
\n= AB + BC+ CD + DE + EF + FG + GH + HI + IJ + JK +KL + LA
\n= 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 32cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Exit-Ticket-Answer-Key-1.png\")
\nLength of the AB side of the given ABCDEFGHIJK Shape = 4cm
\nLength of the BC side of the given ABCDEFGHIJK Shape = 2cm
\nLength of the CD side of the given ABCDEFGHIJK Shape = 2cm
\nLength of the DE side of the given ABCDEFGHIJK Shape = 4cm
\nLength of the EF side of the given ABCDEFGHIJK Shape = 2cm
\nLength of the FG side of the given ABCDEFGHIJK Shape = 2cm
\nLength of the GH side of the given ABCDEFGHIJK Shape = 4cm
\nLength of the HI side of the given ABCDEFGHIJK Shape = 2cm
\nLength of the IJ side of the given ABCDEFGHIJK Shape =\u00a0 2cm
\nLength of the JK side of the given ABCDEFGHIJK Shape = 4cm
\nLength of the KL side of the given ABCDEFGHIJK Shape = 2cm
\nLength of the LA side of the given ABCDEFGHIJK Shape = 2cm
\nPerimeter of the given ABCDEFGHIJK Shape =\u00a0 Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side + Side
\n= AB + BC+ CD + DE + EF + FG + GH + HI + IJ + JK +KL + LA
\n= 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 6cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 8cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 12cm + 2cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 14cm + 2cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 16cm + 4cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 20 cm + 2cm + 2cm + 4cm + 2cm + 2cm
\n= 22cm + 2cm + 4cm + 2cm + 2cm
\n= 24cm + 4cm + 2cm + 2cm
\n= 28cm + 2cm + 2cm
\n= 30cm + 2cm
\n= 32cm.<\/p>\n
Question 1.
\nMeasure and label the side lengths of the shapes below in centimeters. Then, find the perimeter of each shape.
\na.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-1.png\")
\nPerimeter = _____cm +_____cm +_____cm
\n= _______ cm
\nAnswer:
\nPerimeter of ABC triangle = Side + Side + Side
\n= AB + BC + CA
\n= 3cm + 5cm + 4cm
\n= 12cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-1a.png\")
\nLength of the AB side of ABC triangle = 3cm
\nLength of the BC side of ABC triangle = 5cm
\nLength of the CA side of ABC triangle = 4cm
\nPerimeter of ABC triangle = Side + Side + Side
\n= AB + BC + CA
\n= 3cm + 5cm + 4cm
\n= 8cm + 4cm
\n= 12cm.<\/p>\n
b.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-2.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of ABCD rectangle = Side + Side + Side + Side
\n= AB + BC + CD + DA
\n= 6cm + 4cm + 6cm + 4cm
\n= 20cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-1b.png\")
\nLength of the AB side of ABCD rectangle = 6cm
\nLength of the BC side of ABCD rectangle = 4cm
\nLength of the CD side of ABCD rectangle = 6cm
\nLength of the DA side of ABCD rectangle = 4cm
\nPerimeter of ABCD rectangle = Side + Side + Side + Side
\n= AB + BC + CD + DA
\n= 6cm + 4cm + 6cm + 4cm
\n= 10cm + 6cm + 4cm
\n= 16cm + 4cm
\n= 20cm.<\/p>\n
<\/p>\n
c.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-3.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the ABCD Quadrilateral = Side + Side + Side + Side
\n= AB + BC + CD + DA
\n= 3cm + 4cm + 5cm + 4cm
\n= 16cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-1c.png\")
\nLength of the AB side in the ABCD Quadrilateral = 3cm
\nLength of the BC side in the ABCD Quadrilateral = 4cm
\nLength of the CD side in the ABCD Quadrilateral = 5cm
\nLength of the DA side in the ABCD Quadrilateral = 4cm
\nPerimeter of the ABCD Quadrilateral = Side + Side + Side + Side
\n= AB + BC + CD + DA
\n= 3cm + 4cm + 5cm + 4cm
\n= 7cm + 5cm + 4cm
\n= 12cm + 4cm
\n= 16cm.<\/p>\n
<\/p>\n
d.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-4.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the ABCD Parallelogram = Side + Side + Side + Side
\n= AB + BC + CD + DA
\n= 5cm + 5cm + 5cm + 5cm
\n= 20cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-1d.png\")
\nLength of the AB side of the ABCD parallelogram = 5cm
\nLength of the BC side of the ABCD parallelogram = 5cm
\nLength of the CD side of the ABCD parallelogram = 5cm
\nLength of the DA side of the ABCD parallelogram = 5cm
\nPerimeter of the ABCD Parallelogram = Side + Side + Side + Side
\n= AB + BC + CD + DA
\n= 5cm + 5cm + 5cm + 5cm
\n= 10cm + 5cm + 5cm
\n= 15cm + 5cm
\n= 20cm.<\/p>\n
<\/p>\n
e.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-5.png\")
\nPerimeter = _____________________
\n= _______ cm
\nAnswer:
\nPerimeter of the given ABCDEFGH figure = Side + Side + Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EF + FG + GH + HA
\n= 2cm + 2cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
\n= 24cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-1e.png\")
\nLength of AB side of the given ABCDEFGH figure = 2cm
\nLength of BC side of the given ABCDEFGH figure = 2cm
\nLength of CD side of the given ABCDEFGH figure = 3.5cm
\nLength of DE side of the given ABCDEFGH figure = 2cm
\nLength of EF side of the given ABCDEFGH figure = 2cm
\nLength of FG side of the given ABCDEFGH figure = 2.5cm
\nLength of GH side of the given ABCDEFGH figure = 7.5cm
\nLength of HA side of the given ABCDEFGH figure = 2.5cm
\nPerimeter of the given ABCDEFGH figure = Side + Side + Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EF + FG + GH + HA
\n= 2cm + 2cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
\n= 4cm + 3.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
\n= 7.5cm + 2cm + 2cm + 2.5cm + 7.5cm + 2.5cm
\n= 9.5cm + 2cm + 2.5cm + 7.5cm + 2.5cm
\n= 11.5cm + 2.5cm + 7.5cm + 2.5cm
\n= 14cm + 7.5cm + 2.5cm
\n= 21.5cm + 2.5cm
\n= 24cm.<\/p>\n
<\/p>\n
Question 2.
\nMelinda draws two trapezoids to create the hexagon shown below. Use a ruler to find the side lengths of Melinda\u2019s hexagon in centimeters. Then, find the perimeter.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-6.png\")
\nAnswer:
\nPerimeter of ABCDEF Melinda\u2019s hexagon = Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EF +FA
\n= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
\n= 18cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-2.png\")
\nLength of the AB side of ABCDEF Melinda\u2019s hexagon = 3cm
\nLength of the BC side of ABCDEF Melinda\u2019s hexagon = 3cm
\nLength of the CD side of ABCDEF Melinda\u2019s hexagon = 3cm
\nLength of the DE side of ABCDEF Melinda\u2019s hexagon = 3cm
\nLength of the EF side of ABCDEF Melinda\u2019s hexagon = 3cm
\nLength of the FA side of ABCDEF Melinda\u2019s hexagon = 3cm
\nPerimeter of the ABCDEF Melinda\u2019s hexagon = Side + Side + Side + Side + Side + Side
\n= AB + BC + CD + DE + EF +FA
\n= 3cm + 3cm + 3cm + 3cm + 3cm + 3cm
\n= 6cm + 3cm + 3cm + 3cm + 3cm
\n= 9cm + 3cm + 3cm + 3cm
\n= 12cm + 3cm + 3cm
\n= 15cm + 3cm
\n= 18cm.<\/p>\n
Question 3.
\nVictoria and Eric draw the shapes shown below. Eric says his shape has a greater perimeter because it has more sides than Victoria\u2019s shape. Is Eric right? Explain your answer.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-7.png\")
\nAnswer:
\nYes, Eric is correct because DEFG Square has more sides than the ABC Victoria’s triangle shape.
\nPerimeter of DEFG Eric’s Square shape is greater than the Perimeter of ABC Victoria’s triangle shape.
\nPerimeter of ABC Victoria’s triangle shape = 12cm<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-3.png\")
\nLength of AB Victoria’s triangle shape = 3cm
\nLength of BC Victoria’s triangle shape = 5cm
\nLength of CA Victoria’s triangle shape =4cm
\nPerimeter of ABC Victoria’s triangle shape = Side + Side + Side
\n= 3cm +\u00a0 4cm + 5cm
\n= 7cm + 5cm
\n= 12cm.<\/p>\n
Length of DEFG Eric’s Square shape = 4cm
\nPerimeter of DEFG Eric’s Square shape = Side \u00d7 Side
\n= 4cm \u00d7 4cm
\n= 16cm.
\nPerimeter of Eric’s Square shape is greater than the Perimeter of Victoria’s triangle shape.<\/p>\n
<\/p>\n
Question 4.
\nJamal uses his ruler and a right angle tool to draw the rectangle shown below. He says the perimeter of his rectangle is 32 centimeters. Do you agree with Jamal? Why or why not?
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-h-8.png\")
\nAnswer:
\nNO, I disagree with Jamal. Jamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.<\/p>\n
Explanation:
\n![\"Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Eureka-Math-3rd-Grade-Module-7-Lesson-12-Answer-Key-Eureka-Math-Grade-3-Module-7-Lesson-12-Homework-Answer-Key-4.png\")
\nLength of the ABCD Rectangle Jamal drawn = 7.5cm
\nWidth of the ABCD Rectangle Jamal drawn = 4cm
\nPerimeter of ABCD Rectangle Jamal drawn = Length \u00d7 Width
\n= 7.5 \u00d7 4
\n= 30 cm.
\nJamal answer is incorrect because actual Perimeter of his ABCD rectangle is 30cm not 32cm.<\/p>\n","protected":false},"excerpt":{"rendered":"
Engage NY Eureka Math 3rd Grade Module 7 Lesson 12 Answer Key Eureka Math Grade 3 Module 7 Lesson 12 Pattern Sheet Answer Key Multiply. multiply by 7 (6\u201310) Answer: Explanation: 7 \u00d7 6 = 42 7 \u00d7 7 = 49 7 \u00d7 8 = 56 7 \u00d7 9 = 63 7 \u00d7 10 = … 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\/55712"}],"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=55712"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/55712\/revisions"}],"predecessor-version":[{"id":92458,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/55712\/revisions\/92458"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=55712"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=55712"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=55712"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}