\nPlot and connect the points A(3, 2), B(3, 7), and C(8, 2). Name the shape, and determine the area of the polygon.
\n
![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-1.png\")
\nAnswer:
\n
![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-2.png\")
\nRight Triangle<\/p>\n
Question 1.
\nPlot and connect the points A(3, 2), B(3, 7), and C(8, 2). Name the shape, and determine the area of the polygon.
\n
\nAnswer:
\n
\nRight Triangle<\/p>\n
A = \\(\\frac{1}{2}\\) bh Question 2. The shape is a triangle.<\/p>\n Area of Square Area of Triangle 1<\/p>\n A = \\(\\frac{1}{2}\\) bh Area of Triangle 2<\/p>\n A = \\(\\frac{1}{2}\\)bh Area of Triangle 3<\/p>\n A = \\(\\frac{1}{2}\\) bh Total Area of Triangle Question 3. This polygon has 4 sides and has no pairs of parallel sides. Area of Triangle 2 Total Area = 9 units2<\/sup> + 2 units2<\/sup> Question 4. Area of Shape 1 Area of Shape 2 and Shape 4 Because there are two of the same triangle, that makes a total of 8 units2<\/sup>.<\/p>\n Area of Shape 3 Question 5. One possible location of the other two vertices is (9, 2) and (9, 7). Using these coordinates will result in a distance, or side length, of 6 units. Since the height is 5 units, 5 units \u00d7 6 units = 30 units2<\/sup>.<\/p>\n Another possible location of the other two vertices is (-3, 2) and (-3, 7). Using these coordinates will result in a distance, or side length, of 6 units. Since the height is 5 units, 5 units \u00d7 6 units = 30 units2<\/sup>.<\/p>\n Exercise:<\/p>\n For Exercises 1 and 2, plot the points, name the shape, and determine the area of the shape. Then write an expression that could be used to determine the area of the figure. Explain how each part of the expression corresponds to the situation.<\/p>\n Question 1. This shape is a hexagon.<\/p>\n Area of Shape 1 Area of Shape 2 Area of Shape 3 Area of Shape 4 Area of Shape 5 Total Area = 4 units2<\/sup> + 4 units2<\/sup> + 27 units2<\/sup> + 7.5 units2<\/sup> + 9 units2<\/sup><\/p>\n Expression: The first term is the area of triangle 1 on the left. Question 2. This shape is a triangle.<\/p>\n Area of Outside Rectangle Area of Triangle 1<\/p>\n A = \\(\\frac{1}{2}\\) bh Area of Triangle 2<\/p>\n A = \\(\\frac{1}{2}\\)bh Area of Triangle 3 Total Area = 91 units2<\/sup> – 24.5 units2<\/sup> – 18 units2<\/sup> – 6.5 units2<\/sup> Expression:<\/p>\n (7)(13) – \\(\\frac{1}{2}\\) (7)(7) – \\(\\frac{1}{2}\\) (6)(6) – \\(\\frac{1}{2}\\) (13)(1)<\/p>\n The first term in the expression represents the area of the rectangle that goes around the outside of the triangle. Question 3. Question 4. Plot the points for each shape, determine the area of the polygon, and then write an expression that could be used to determine the area of the figure. Explain how each part of the expression corresponds to the situation.<\/p>\n Question 1. Area of Triangle 1 Area of Triangle 2 Area of Triangle 3 Pentagon total area = 15 units2<\/sup> + 22.5 units2<\/sup> + 22.5 units2<\/sup> Expression: Each term in the expression represents the area of one of the triangular pieces that fits inside the pentagon. They are all added together to form the complete figure.<\/p>\n Question 2. Area of Outside Rectangle Area of Top Triangle Area of Bottom Left Triangle Area of Bottom Right Triangle Area of center triangle = 77 units2<\/sup> – 14 units2<\/sup> – 14 units2<\/sup> – 16.5 units2<\/sup> Question 3. Area of Triangle on the Left Area of Triangle on the Right Expression: Question 4. Area of Rectangle Area of Triangle on Top Left Area of Triangle on Bottom Left Area of Triangle on Right Total Area = 96 units2<\/sup> – 20 units2<\/sup> – 8 units2<\/sup> – 24 units2<\/sup> Expression: (12)(8) – \\(\\frac{1}{2}\\) (4)(10) – \\(\\frac{1}{2}\\) (8)(2) – \\(\\frac{1}{2}\\) (4)(12) The two expressions are different because of the way we divided up the figure. In the first expression, we split the shape into two triangles that had to be added together to get the whole. In the second expression, we enclosed the triangle inside a new figure and then had to subtract the extra area.<\/p>\n Question 5. Question 6. Question 1. Methods to calculate the answer will vary.<\/p>\n #1 Area of shape a Area of shape b Total Area = 12 units2<\/sup> + 4 units2<\/sup> = 16 units2<\/sup> Expression: (2)(6) + latex]\\frac{1}{2}[\/latex] (4)(2)<\/p>\n The first term represents the area of the rectangle on the left, which makes up part of the figure. #2 Area of outside rectangle Area of shape c Area of shape d Area of shape e Total Area = 42 units2<\/sup> – 9 units2<\/sup> – 10.5 units2<\/sup> – 3 units2<\/sup> Expression: The first term in the expression is the area of a rectangle that goes around the triangle. Engage NY Eureka Math Grade 6 Module 5 Lesson 8 Answer Key Eureka Math Grade 6 Module 5 Lesson 8 Example Answer Key Question 1. Plot and connect the points A(3, 2), B(3, 7), and C(8, 2). Name the shape, and determine the area of the polygon. Answer: Right Triangle A = bh A = … 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\/82411"}],"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=82411"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82411\/revisions"}],"predecessor-version":[{"id":137318,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82411\/revisions\/137318"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=82411"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=82411"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=82411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nA = \\(\\frac{1}{2}\\) (5 units)(5 units)
\nA = \\(\\frac{1}{2}\\) (25 units2<\/sup>)
\nA = 12.5 units2<\/sup><\/p>\n
\nPlot and connect the points E(-8, 8), F(-2, 5), and G(-7, 2). Then give the best name for the polygon, and determine the area.
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-3.png\")
<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-4.png\")
<\/p>\n
\nA = s2<\/sup>
\nA = (6 units)2<\/sup>
\nA = 36 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) (1 unit) (6 units)
\nA = \\(\\frac{1}{2}\\) (6 units2<\/sup>)
\nA = 3 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) (6 units)(3 units)
\nA = \\(\\frac{1}{2}\\) (18 units2<\/sup>)
\nA = 9 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) (5 units) (3 units)
\nA = \\(\\frac{1}{2}\\) (15 units2<\/sup>)
\nA = 7.5 units2<\/sup><\/p>\n
\nA = 36 units2<\/sup> – 3 units2<\/sup> – 9 units2<\/sup> – 7.5 units2<\/sup>
\nA = 16.5 units2<\/sup><\/p>\n
\nPlot and connect the following points: K(-9, -7), L(-4, -2), M(-1, -5), and N(-5, -5). Give the best name for the polygon, and determine the area.
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-5.png\")
<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-6.png\")
<\/p>\n
\nTherefore, the best name for this shape is a quadrilateral.
\nTo determine the area, I will separate the shape into two
\ntriangles.
\nArea of Triangle 1
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (6 units) (3 units)
\nA = \\(\\frac{1}{2}\\) (18 units2<\/sup>)
\nA = 9 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (2 units) (2 units)
\nA = \\(\\frac{1}{2}\\) (4 units2<\/sup>)
\nA = 2 units2<\/sup><\/p>\n
\nTotal Area = 11 units2<\/sup><\/p>\n
\nPlot and connect the following points: P(1, -4), Q(5, -2), R(9, -4), S(7, -8), and T(3, -8). Give the best name for the polygon, and determine the area.
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-7.png\")
<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-8.png\")
\nTo determine the area, I will decompose this pentagon into four smaller shapes.<\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (8 units)(2 units)
\nA = \\(\\frac{1}{2}\\) (16 units2<\/sup>)
\nA = 8 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 units)(2 units)
\nA = \\(\\frac{1}{2}\\) (8 units2<\/sup>)
\nA = 4 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 units) (4 units)
\nA = 16 units2<\/sup>
\nTotal Area = 8 units2<\/sup> + 8 units2<\/sup> + 16 units2<\/sup>
\nTotal Area = 32 units2<\/sup><\/p>\n
\nTwo of the coordinates of a rectangle are A (3, 7) and B (3, 2). The rectangle has an area of 30 square units. Give the possible locations of the other two vertices by identifying their coordinates. (Use the coordinate plane to draw and check your answer.)
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-9.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Example-Answer-Key-10.png\")
<\/p>\nEureka Math Grade 6 Module 5 Lesson 8 Exercise Answer Key<\/h3>\n
![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Exercise-Answer-Key-11.png\")
<\/p>\n
\nA(4, 6), B(8, 6), C(10, 2), D(8, -3), E(5, -3), and F(2, 2)
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Exercise-Answer-Key-12.png\")
<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Exercise-Answer-Key-13.png\")
<\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (2 units)(4 units)
\nA = \\(\\frac{1}{2}\\) (8 units2)
\nA = 4 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = (1 unit) (4 units)
\nA = 4 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = (3 units) (9 units)
\nA = 27 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (3 units)(5 units)
\nA = \\(\\frac{1}{2}\\) (15 units2)
\nA = 7.5 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (2 units)(9 units)
\nA = \\(\\frac{1}{2}\\) (18 units2)
\nA = 9 units2<\/sup><\/p>\n
\n\\(\\frac{1}{2}\\) (2) (4) + (1) (4) + (3) (9) + \\(\\frac{1}{2}\\) (3)(5) +\\(\\frac{1}{2}\\) (2)(9).
\nEach term represents the area of a section of the hexagon. They must be added together to get the total.<\/p>\n
\nThe second term is the area of rectangle 2.
\nThe third term is the area of the large rectangle 3.
\nThe fourth term is the area of triangle 4 on the left.
\nThe fifth term is the area of triangle 5 on the right.<\/p>\n
\nX (-9, 6), Y (-2, -1), and Z (-8, -7)
\nAnswer:
\n<\/p>\n
\nA = lw
\nA = (7 units) (13 units)
\nA = 91 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) (7 units)(7 units)
\nA = \\(\\frac{1}{2}\\) (49 units2<\/sup>)
\nA = 24.5 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) (6 units)(6 units)
\nA = \\(\\frac{1}{2}\\) (36 units2<\/sup>)
\nA = 18 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (13 units)(1 unit)
\nA = \\(\\frac{1}{2}\\) (13 units2<\/sup>)
\nA = 6.5 units2<\/sup><\/p>\n
\nTotal Area = 42 units2<\/sup><\/p>\n
\nThe next three terms represent the areas that need to be subtracted from the rectangle so that we are only left with the given triangle.
\nThe second term is the area of the top right triangle.
\nThe third term is the area of the bottom right triangle.
\nThe fourth term is the area of the triangle on the left.<\/p>\n
\nA rectangle with vertices located at (-3, 4) and (5, 4) has an area of 32 square units. Determine the location of the other two vertices.
\nAnswer:
\nThe other two points could be located at (-3, 8) and (5, 8) or (-3, 0) and (5, 0).<\/p>\n
\nChallenge: A triangle with vertices located at (-2, -3) and (3, -3) has an area of 20 square units. Determine one possible location of the other vertex.
\nAnswer:
\nAnswers will vary. Possible solutions include points that are 8 units from the base. (-2, 5) or (3, -11).<\/p>\nEureka Math Grade 6 Module 5 Lesson 8 Problem Set Answer Key<\/h3>\n
\nA(1, 3), B(2, 8), C(8, 8), D(10, 3), and E(5, -2)
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Problem-Set-Answer-Key-15.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Problem-Set-Answer-Key-16.png\")
<\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (6 units) (5 units)
\nA = \\(\\frac{1}{2}\\) (30 units2<\/sup>)
\nA = 15 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (9 units)(5 units)
\nA = \\(\\frac{1}{2}\\) (45 units2<\/sup>)
\nA = 22.5 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (9 units)(5 units)
\nA = \\(\\frac{1}{2}\\) (45 units2<\/sup>)
\nA = 22.5 units2<\/sup><\/p>\n
\nTotal Area = 60 units2<\/sup><\/p>\n
\n\\(\\frac{1}{2}\\) (6)(5) + \\(\\frac{1}{2}\\) (9)(5) + \\(\\frac{1}{2}\\) (9)(5)<\/p>\n
\nX(-10, 2), Y(-3, 6), and Z(-6, -5)
\n
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Problem-Set-Answer-Key-17.png\")
<\/p>\n
\nA = lw
\nA = (11 units) (7 units)
\nA = 77 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (7 units)(4 units)
\nA = \\(\\frac{1}{2}\\) (28 units2<\/sup>)
\nA = 14 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 units)(7 units)
\nA = \\(\\frac{1}{2}\\) (28 units2<\/sup>)
\nA = 14 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (3 units)(11 units)
\nA = \\(\\frac{1}{2}\\) (33 units2<\/sup>)
\nA = 16.5 units2<\/sup><\/p>\n
\nArea of center triangle = 32.5 units2<\/sup>
\nExpression:
\n(11)(7) – \\(\\frac{1}{2}\\) (7)(4) – \\(\\frac{1}{2}\\) (4)(7) – \\(\\frac{1}{2}\\) (3)(11)
\nThe first term in the expression represents the area of the rectangle that would enclose the triangle. Then the three terms after represent the areas of the triangles that need to be removed from the area of the rectangle to find the area of the given triangle.<\/p>\n
\nE(5, 7), F(9, -5), and G(1, -3)
\n
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Problem-Set-Answer-Key-18.png\")
<\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (11 units)(4 units)
\nA = 22 units<\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (11 units)(4 units)
\nA = 22 units2<\/sup>
\nTotal Area = 22 units2<\/sup> + 22 units2<\/sup> = 44 units2<\/sup><\/p>\n
\n\\(\\frac{1}{2}\\) (11)(4) + \\(\\frac{1}{2}\\) (11)(4)
\nEach term in the expression represents the area of a triangle that makes up the total area. The first term is the area of the triangle on the left, and the second term is the area of the triangle on the right.<\/p>\n
\nFind the area of the triangle In Problem 3 using a different method. Then, compare the expressions that can be used for both solutions in Problems 3 and 4.
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Problem-Set-Answer-Key-19.png\")
<\/p>\n
\nA = lw
\nA = (12 units) (8 units)
\nA = 96 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 units)(10 units)
\nA = 20 units2<\/sup><\/p>\n
\nA =\\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (8 units)(2 units)
\nA = 8 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 units)(12 units)
\nA = 24 units2<\/sup><\/p>\n
\nTotal Area = 44 units2<\/sup><\/p>\n
\nThe first term in the expression is the area of a rectangle around the outside of the figure. Then we subtracted all of the extra areas with the next three terms.<\/p>\n
\nTwo vertices of a rectangle are (8, -5) and (8, 7). If the area of the rectangle is 72 square units, name the possible location of the other two vertices.
\nAnswer:
\n(2, -5) and (2, 7) or (14, -5) and (14, 7)<\/p>\n
\nA triangle with two vertices located at (5, -8) and (5, 4) has an area of 48 square units. Determine one possible location of the other vertex.
\nAnswer:
\nAnswers will vary. Possible solutions include points that are 8 units from the base such as (13, -2) or (-3, -2).<\/p>\nEureka Math Grade 6 Module 5 Lesson 8 Exit Ticket Answer Key<\/h3>\n
\nDetermine the area of both polygons on the coordinate plane, and explain why you chose the methods you used. Then write an expression that could be used to determine the area of the figure. Explain how each part of the expression corresponds to the situation.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Exit-Ticket-Answer-Key-20.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-8-Exit-Ticket-Answer-Key-21.png\")
<\/p>\n
\nA = lw
\nA = (2 units) (6 units)
\nA = 12 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 units)(2 units)
\nA = \\(\\frac{1}{2}\\) (8 units2<\/sup>)
\nA = 4 units2<\/sup><\/p>\n
\nExplanations will vary depending on the method chosen.<\/p>\n
\nThe second term represents the area of the triangle on the right that completes the figure.<\/p>\n
\nA = lw
\nA = (7 units) (6 units)
\nA = 42 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (6 units)(3 units)
\nA = \\(\\frac{1}{2}\\) (18 units2<\/sup>)
\nA = 9 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (7 units)(3 units)
\nA = \\(\\frac{1}{2}\\) (21 units2<\/sup>)
\nA = 10.5 units2<\/sup><\/p>\n
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (1 units)(6 units)
\nA = \\(\\frac{1}{2}\\) (6 units2<\/sup>)
\nA = 3 units2<\/sup><\/p>\n
\nTotal Area = 19.5 units2<\/sup>
\nExplanations will vary depending on method chosen.<\/p>\n
\n(7)(6) – \\(\\frac{1}{2}\\) (6)(3) – \\(\\frac{1}{2}\\) (7)(3) – \\(\\frac{1}{2}\\) (1)(6)<\/p>\n
\nEach of the other terms represents the triangles that need to be subtracted from the rectangle so that we are left with just the figure in the center.<\/p>\n","protected":false},"excerpt":{"rendered":"