{"id":78156,"date":"2022-06-13T21:31:41","date_gmt":"2022-06-13T16:01:41","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=78156"},"modified":"2022-05-10T14:14:15","modified_gmt":"2022-05-10T08:44:15","slug":"eureka-math-grade-7-module-3-lesson-18","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-7-module-3-lesson-18\/","title":{"rendered":"Eureka Math Grade 7 Module 3 Lesson 18 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 3 Lesson 18 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 3 Lesson 18 Example Answer Key<\/h3>\n

Example 1.
\nFind the area of the following semicircle. Use \u03c0 \u2248 \\(\\frac{22}{7}\\) .
\n\"Engage
\nAnswer:
\nIf the diameter of the circle is 14 cm, then the
\nradius is 7 cm. The area of the semicircle is half
\nof the area of the circular region.
\nA \u2248 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 (7 cm)2<\/sup>
\nA \u2248 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 49 cm2<\/sup>
\nA \u2248 77 cm2<\/sup><\/p>\n

What is the area of the quarter circle? Use \u03c0 \u2248 \\(\\frac{22}{7}\\) .
\n\"Engage
\nAnswer:
\nA \u2248 \\(\\frac{1}{4}\\) \u2219 \\(\\frac{22}{7}\\) (6 cm)2<\/sup>
\nA \u2248 \\(\\frac{1}{4}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 36 cm2<\/sup>
\nA \u2248 \\(\\frac{198}{7}\\) cm2<\/sup><\/p>\n

Example 2.
\nMarjorie is designing a new set of placemats for her dining room table. She sketched a drawing of the placement on graph paper. The diagram represents the area of the placemat consisting of a rectangle and two semicircles at either end. Each square on the grid measures 4 inches in length.
\nFind the area of the entire placemat. Explain your thinking regarding the solution to this problem.
\n\"Engage
\nAnswer:
\nThe length of one side of the rectangular section is 12 inches in length, while the width is 8 inches. The radius of the semicircular region is 4 inches. The area of the rectangular part is (8 in) \u2219 (12 in) = 96 in2<\/sup>. The total area must include the two semicircles on either end of the placemat. The area of the two semicircular regions is the same as the area of one circle with the same radius. The area of the circular region is A = \u03c0 \u2219 (4 in)2<\/sup> = 16\u03c0 in2<\/sup>. In this problem, using \u03c0 \u2248 3.14 makes more sense because there are no fractions in the problem. The area of the semicircular regions is approximately 50.24 in2<\/sup>. The total area for the placemat is the sum of the areas of the rectangular region and the two semicircular regions, which is approximately (96+50.24) in2<\/sup> = 146.24 in2<\/sup>.<\/p>\n

If Marjorie wants to make six placemats, how many square inches of fabric will she need? Assume there is no waste.
\nAnswer:
\nThere are 6 placemats that are each 146.24 in2<\/sup>, so the fabric needed for all is 6 \u2219 146.24 in2<\/sup> = 877.44 in2<\/sup>.<\/p>\n

Marjorie decides that she wants to sew on a contrasting band of material around the edge of the placemats. How much band material will Marjorie need?
\nAnswer:
\nThe length of the band material needed will be the sum of the lengths of the two sides of the rectangular region and the circumference of the two semicircles (which is the same as the circumference of one circle with the same radius).
\nP = (l+l+2\u03c0r)
\nP = (12+12+2 \u2219 \u03c0 \u2219 4) = 49.12
\nThe perimeter is 49.12 in2<\/sup>.<\/p>\n

Example 3.
\nThe circumference of a circle is 24\u03c0 cm. What is the exact area of the circle?
\nDraw a diagram to assist you in solving the problem.
\n\"Engage
\nWhat information is needed to solve the problem?
\nAnswer:
\nThe radius is needed to find the area of the circle. Let the radius be r cm. Find the radius by using the circumference formula.
\nC = 2\u03c0r
\n24\u03c0 = 2\u03c0r
\n(\\(\\frac{1}{2}\\) \u03c0)24\u03c0 = (\\(\\frac{1}{2}\\) \u03c0)2\u03c0r
\n12 = r
\nThe radius is 12 cm.<\/p>\n

Next, find the area.
\nAnswer:
\nA = \u03c0 r2<\/sup>
\nA = \u03c0(12)2<\/sup>
\nA = 144\u03c0
\nThe exact area of the circle is 144\u03c0 cm2<\/sup>.<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 18 Exercise Answer Key<\/h3>\n

Opening Exercise
\nDraw a circle with a diameter of 12 cm and a square with a side length of 12 cm on grid paper. Determine the area of the square and the circle.
\n\"Engage
\nAnswer:
\nArea of square: A = (12 cm)2<\/sup> = 144 cm2<\/sup>; Area of circle: A = \u03c0 \u2219 (6 cm)2<\/sup> = 36\u03c0 cm2<\/sup><\/p>\n

Brainstorm some methods for finding half the area of the square and half the area of the circle.
\nAnswer:
\nSome methods include folding in half and counting the grid squares and cutting each in half and counting the squares.<\/p>\n

Find the area of half of the square and half of the circle, and explain to a partner how you arrived at the area.
\nAnswer:
\nThe area of half of the square is 72 cm2<\/sup>. The area of half of the circle is 18\u03c0 cm2<\/sup>. Some students may count the squares; others may realize that half of the square is a rectangle with side lengths of 12 cm and 6 cm and use A = l \u2219 w to determine the area. Some students may fold the square vertically, and some may fold it horizontally. Some students will try to count the grid squares in the semicircle and find that it is easiest to take half of the area of the circle.<\/p>\n

What is the ratio of the new area to the original area for the square and for the circle?
\nAnswer:
\nThe ratio of the areas of the rectangle (half of the square) to the square is 72:144 or 1:2. The ratio for the areas of the circles is 18\u03c0:36\u03c0 or 1:2.<\/p>\n

Find the area of one – fourth of the square and one – fourth of the circle, first by folding and then by another method. What is the ratio of the new area to the original area for the square and for the circle?
\nAnswer:
\nFolding the square in half and then in half again will result in one – fourth of the original square. The resulting shape is a square with a side length of 6 cm and an area of 36 cm2<\/sup>. Repeating the same process for the circle will result in an area of 9\u03c0 cm2<\/sup>. The ratio for the areas of the squares is 36:144 or 1:4. The ratio for the areas of the circles is 9\u03c0:36\u03c0 or 1:4.<\/p>\n

Write an algebraic expression that expresses the area of a semicircle and the area of a quarter circle.
\nAnswer:
\nSemicircle: A = \\(\\frac{1}{2}\\) \u03c0r2<\/sup>; Quarter circle: A = \\(\\frac{1}{4}\\) \u03c0r2<\/sup><\/p>\n

Exercise 1.
\nFind the area of a circle with a diameter of 42 cm. Use \u03c0 \u2248 \\(\\frac{22}{7}\\) .
\nAnswer:
\nIf the diameter of the circle is 42 cm, then the radius is 21 cm .
\nA = \u03c0r2<\/sup>
\nA \u2248 \\(\\frac{22}{7}\\) (21 cm)2<\/sup>
\nA \u2248 1386 cm2<\/sup><\/p>\n

Exercise 2.
\nThe circumference of a circle is 9\u03c0 cm.
\na. What is the diameter?
\nAnswer:
\nIf C = \u03c0d, then 9\u03c0 cm = \u03c0d.
\nSolving the equation for the diameter, d, \\(\\frac{1}{\\pi}\\) \u2219 9\u03c0 cm = \\(\\frac{1}{\\pi}\\) \u03c0 \u2219 d.
\nSo, 9 cm = d.<\/p>\n

b. What is the radius?
\nAnswer:
\nIf the diameter is 9 cm, then the radius is half of that or \\(\\frac{9}{2}\\) cm.<\/p>\n

c. What is the area?
\nAnswer:
\nThe area of the circle is A = \u03c0 \u2219 (\\(\\frac{9}{2}\\) cm)2<\/sup>, so A = \\(\\frac{81}{4}\\) \u03c0 cm2<\/sup>.<\/p>\n

Exercise 3.
\nIf students only know the radius of a circle, what other measures could they determine? Explain how students would use the radius to find the other parts.
\nAnswer:
\nIf students know the radius, then they can find the diameter. The diameter is twice as long as the radius. The circumference can be found by doubling the radius and multiplying the result by \u03c0. The area can be found by multiplying the radius times itself and then multiplying that product by \u03c0.<\/p>\n

Exercise 4.
\nFind the area in the rectangle between the two quarter circles if AF = 7 ft, FB = 9 ft, and HD = 7 ft. Use \u03c0 \u2248 \\(\\frac{22}{7}\\) . Each quarter circle in the top – left and lower – right corners have the same radius.
\n\"Engage
\nAnswer:
\nThe area between the quarter circles can be found by subtracting the area of the two quarter circles from the area of the rectangle. The area of the rectangle is the product of the length and the width. Side AB has a length of 16 ft and Side AD has a length of 14 ft. The area of the rectangle is
\nA = 16 ft \u2219 14 ft = 224 ft2<\/sup>. The area of the two quarter circles is the same as the area of a semicircle, which is half the area of a circle. A = \\(\\frac{1}{2}\\) \u03c0r2<\/sup>.
\nA \u2248 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 (7 ft)2<\/sup>
\nA \u2248 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 49 ft2<\/sup>
\nA \u2248 77 ft2<\/sup>
\nThe area between the two quarter circles is 224 ft2<\/sup> – 77 ft2<\/sup> = 147 ft2<\/sup>.<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 18 Problem Set Answer Key<\/h3>\n

Question 1.
\nMark created a flower bed that is semicircular in shape. The diameter of the flower bed is 5 m .
\na. What is the perimeter of the flower bed? (Approximate \u03c0 to be 3.14.)
\n\"Eureka
\nAnswer:
\nThe perimeter of this flower bed is the sum of the diameter and one – half the circumference of a circle with the same diameter.
\nP = diameter+\\(\\frac{1}{2}\\) \u03c0 \u2219 diameter
\nP \u2248 5 m+\\(\\frac{1}{2}\\) \u2219 3.14 \u2219 5 m
\nP \u2248 12.85 m<\/p>\n

b. What is the area of the flower bed? (Approximate \u03c0 to be 3.14.)
\nAnswer:
\nA = \\(\\frac{1}{2}\\) \u03c0 (2.5 m)2<\/sup>
\nA = \\(\\frac{1}{2}\\) \u03c0 (6.25 m2<\/sup>)
\nA \u2248 0.5 \u2219 3.14 \u2219 6.25 m2<\/sup>
\nA \u2248 9.8 m2<\/sup><\/p>\n

Question 2.
\nA landscape designer wants to include a semicircular patio at the end of a square sandbox. She knows that the area of the semicircular patio is 25.12 cm2<\/sup>.
\na. Draw a picture to represent this situation.
\nAnswer:
\n\"Eureka<\/p>\n

b. What is the length of the side of the square?
\nAnswer:
\nIf the area of the patio is 25.12 cm2<\/sup>, then we can find the radius by solving the equation A = \\(\\frac{1}{2}\\) \u03c0r2<\/sup> and substituting the information that we know. If we approximate \u03c0 to be 3.14 and solve for the radius, r, then
\n25.12 cm2<\/sup> \u2248 \\(\\frac{1}{2}\\) \u03c0r2<\/sup>
\n\\(\\frac{2}{1}\\) \u2219 25.12 cm2<\/sup> \u2248 \\(\\frac{2}{1}\\) \u2219 \\(\\frac{1}{2}\\) \u03c0r2<\/sup>
\n50.24 cm2<\/sup> \u2248 3.14r2<\/sup>
\n\\(\\frac{1}{3.14}\\) \u2219 50.24 cm2<\/sup> \u2248 \\(\\frac{1}{3.14}\\) \u2219 3.14r2<\/sup>
\n16 cm2<\/sup> \u2248 r2<\/sup>
\n4 cm \u2248 r
\nThe length of the diameter is 8 cm; therefore, the length of the side of the square is 8 cm.<\/p>\n

Question 3.
\nA window manufacturer designed a set of windows for the top of a two – story wall. If the window is comprised of 2 squares and 2 quarter circles on each end, and if the length of the span of windows across the bottom is 12 feet, approximately how much glass will be needed to complete the set of windows?
\n\"Eureka
\nAnswer:
\nThe area of the windows is the sum of the areas of the two quarter circles and the two squares that make up the bank of windows. If the span of windows is 12 feet across the bottom, then each window is 3 feet wide on the bottom. The radius of the quarter circles is 3 feet, so the area for one quarter circle window is A = \\(\\frac{1}{4}\\) \u03c0 \u2219 (3 ft)2<\/sup>, or A \u2248 7.065 ft2<\/sup>. The area of one square window is A = (3 ft)2<\/sup>, or 9 ft2<\/sup>. The total area is A = 2(area of quarter circle)+2(area of square), or A \u2248 (2 \u2219 7.065 ft2<\/sup> )+(2 \u2219 9 ft2<\/sup> ) \u2248 32.13 ft2<\/sup>.<\/p>\n

Question 4.
\nFind the area of the shaded region. (Approximate \u03c0 to be \\(\\frac{22}{7}\\) .)
\n\"Eureka
\nAnswer:
\nA = \\(\\frac{1}{4}\\) \u03c0(12 in)2<\/sup>
\nA = \\(\\frac{1}{4}\\) \u03c0 \u2219 144 in2<\/sup>
\nA \u2248 \\(\\frac{1}{4}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 144 in2<\/sup>
\nA \u2248 792\/7 in2<\/sup> or 113.1 in2<\/sup><\/p>\n

Question 5.
\nThe figure below shows a circle inside of a square. If the radius of the circle is 8 cm, find the following and explain your solution.
\n\"Eureka
\na. The circumference of the circle
\nAnswer:
\nC = 2\u03c0 \u2219 8 cm
\nC = 16\u03c0 cm<\/p>\n

b. The area of the circle
\nAnswer:
\nA = \u03c0 \u2219 (8 cm)2<\/sup>
\nA = 64 \u03c0 cm2<\/sup><\/p>\n

c. The area of the square
\nAnswer:
\nA = 16 cm \u2219 16 cm
\nA = 256 cm2<\/sup><\/p>\n

Question 6.
\nMichael wants to create a tile pattern out of three quarter circles for his kitchen backsplash. He will repeat the three quarter circles throughout the pattern. Find the area of the tile pattern that Michael will use. Approximate \u03c0 as 3.14.
\n\"Eureka
\nAnswer:
\nThere are three quarter circles in the tile design. The area of one quarter circle multiplied by 3 will result in the total area.
\nA = \\(\\frac{1}{4}\\) \u03c0 \u2219 (16 cm)2<\/sup>
\nA \u2248 \\(\\frac{1}{4}\\) \u2219 3.14 \u2219 256 cm2<\/sup>
\nA \u2248 200.96 cm2<\/sup><\/p>\n

A \u2248 3 \u2219 200.96 cm2<\/sup>
\nA \u2248 602.88 cm2<\/sup>
\nThe area of the tile pattern is approximately 602.88 cm2<\/sup>.<\/p>\n

Question 7.
\nA machine shop has a square metal plate with sides that measure 4 cm each. A machinist must cut four semicircles with a radius of \\(\\frac{1}{2}\\) cm and four quarter circles with a radius of 1 cm from its sides and corners. What is the area of the plate formed? Use \\(\\frac{22}{7}\\) to approximate \u03c0.
\n\"Eureka
\nAnswer:
\nThe area of the metal plate is determined by subtracting the four quarter circles (corners) and the four half – circles (on each side) from the area of the square. Area of the square: A = (4 cm)2<\/sup> = 16 cm2<\/sup>.
\nThe area of four quarter circles is the same as the area of a circle with a radius of
\n1 cm: A \u2248 \\(\\frac{22}{7}\\) (1 cm)2<\/sup> \u2248 \\(\\frac{22}{7}\\) cm2<\/sup>.
\nThe area of the four semicircles with radius \\(\\frac{1}{2}\\) cm is
\nA \u2248 4 \u2219 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 (\\(\\frac{1}{2}\\) cm)2<\/sup>
\nA \u2248 4 \u2219 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 \\(\\frac{1}{4}\\) cm2<\/sup> \u2248 11\/7 cm2<\/sup>.
\nThe area of the metal plate is
\nA \u2248 16 cm2<\/sup> – \\(\\frac{22}{7}\\) cm2<\/sup> – \\(\\frac{11}{7}\\) cm2<\/sup> \u2248 \\(\\frac{79}{7}\\) cm2<\/sup><\/p>\n

Question 8.
\nA graphic artist is designing a company logo with two concentric circles (two circles that share the same center but have different radii). The artist needs to know the area of the shaded band between the two concentric circles. Explain to the artist how he would go about finding the area of the shaded region.
\n\"Eureka
\nAnswer:
\nThe artist should find the areas of both the larger and smaller circles. Then, the artist should subtract the area of the smaller circle from the area of the larger circle to find the area between the two circles. The area of the larger circle is
\nA = \u03c0 \u2219 (9 cm)\\(\\frac{22}{7}\\) or 81\u03c0 cm\\(\\frac{22}{7}\\) .
\nThe area of the smaller circle is
\nA = \u03c0(5 cm)\\(\\frac{22}{7}\\) or 25\u03c0 cm\\(\\frac{22}{7}\\) .
\nThe area of the region between the circles is 81\u03c0 cm\\(\\frac{22}{7}\\) – 25\u03c0 cm\\(\\frac{22}{7}\\) = 56\u03c0 cm\\(\\frac{22}{7}\\) . If we approximate \u03c0 to be 3.14, then A \u2248 175.84 cm\\(\\frac{22}{7}\\) .<\/p>\n

Question 9.
\nCreate your own shape made up of rectangles, squares, circles, or semicircles, and determine the area and perimeter.
\nAnswer:
\nStudent answers may vary.<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 18 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nKen\u2019s landscape gardening business creates odd – shaped lawns that include semicircles. Find the area of this semicircular section of the lawn in this design. Use \\(\\frac{22}{7}\\) for \u03c0.
\n\"Eureka
\nAnswer:
\nIf the diameter is 5 m, then the radius is \\(\\frac{5}{2}\\) m. Using the formula for area of a semicircle, A = \\(\\frac{1}{2}\\) \u03c0r2<\/sup>, A \u2248 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 (\\(\\frac{5}{2}\\) m)2<\/sup>. Using the order of operations,
\nA \u2248 \\(\\frac{1}{2}\\) \u2219 \\(\\frac{22}{7}\\) \u2219 \\(\\frac{25}{4}\\) m2<\/sup> \u2248 \\(\\frac{550}{56}\\) m2<\/sup> \u2248 9.8 m2<\/sup>.<\/p>\n

Question 2.
\nIn the figure below, Ken\u2019s company has placed sprinkler heads at the center of the two small semicircles. The radius of the sprinklers is 5 ft. If the area in the larger semicircular area is the shape of the entire lawn, how much of the lawn will not be watered? Give your answer in terms of \u03c0 and to the nearest tenth. Explain your thinking.
\n\"Eureka
\nAnswer:
\nThe area not covered by the sprinklers would be the area between the larger semicircle and the two smaller ones. The area for the two semicircles is the same as the area of one circle with the same radius of 5 ft. The area not covered by the sprinklers can be found by subtracting the area of the two smaller semicircles from the area of the large semicircle.<\/p>\n

Area Not Covered = Area of large semicircle – Area of two smaller semicircles
\nA = \\(\\frac{1}{2}\\) \u03c0 \u2219 (10 ft)2<\/sup> – (2 \u2219 (\\(\\frac{1}{2}\\) (\u03c0 \u2219 (5 ft)2<\/sup> )))
\nA = \\(\\frac{1}{2}\\) \u03c0 \u2219 100 ft2<\/sup> – \u03c0 \u2219 25 ft2<\/sup>
\nA = 50\u03c0 ft2<\/sup> – 25\u03c0 ft2<\/sup> = 25\u03c0 ft2<\/sup>
\nLet \u03c0 \u2248 3.14
\nA \u2248 78.5 ft2<\/sup>
\nThe sprinklers will not cover 25\u03c0 ft2<\/sup> or 78.5 ft2<\/sup> of the lawn.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 3 Lesson 18 Answer Key Eureka Math Grade 7 Module 3 Lesson 18 Example Answer Key Example 1. Find the area of the following semicircle. Use \u03c0 \u2248 . Answer: If the diameter of the circle is 14 cm, then the radius is 7 cm. The area of … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/78156"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=78156"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/78156\/revisions"}],"predecessor-version":[{"id":137773,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/78156\/revisions\/137773"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=78156"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=78156"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=78156"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}