Illustrative Math Alignment: Grade 7 Unit 9

Putting it All Together

Lesson 5: How Crowded Is this Neighborhood?

Use the following Media4Math resources with this Illustrative Math lesson.

Title	Body	Curriculum Topic
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 Topic Ratios, Proportions, and Percents Description This example illustrates solving for b in a proportion where a is expressed as x + 2, and c and d are given constants (c = 5, d = 2). We set up the equation (x + 2) / b = 5 / 2 and solve for b, resulting in the expression b = (2(x + 2)) / 5. This problem demonstrates how to handle algebraic expressions in proportions.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 Topic Ratios, Proportions, and Percents Description This example demonstrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, one with side lengths of 6 and x, and the other with expressions of 3x + 2 and 22. The problem requires setting up a proportion to determine the value of x for which the parallelograms are similar: 6 / 22 = x / (3x + 2). Solving this equation leads to x = 3.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 Topic Ratios, Proportions, and Percents Description This example demonstrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, one with side lengths of 6 and x, and the other with expressions of 3x + 2 and 22. The problem requires setting up a proportion to determine the value of x for which the parallelograms are similar: 6 / 22 = x / (3x + 2). Solving this equation leads to x = 3.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 Topic Ratios, Proportions, and Percents Description This example demonstrates solving for c in a proportion where a = 9, b = 4, and d = 12. We set up the proportion 9 / 4 = c / 12 and solve for c, resulting in c = 27. This problem shows how to find an unknown value in the numerator of a proportion.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 Topic Ratios, Proportions, and Percents Description This example demonstrates solving for c in a proportion where a = 9, b = 4, and d = 12. We set up the proportion 9 / 4 = c / 12 and solve for c, resulting in c = 27. This problem shows how to find an unknown value in the numerator of a proportion.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 Topic Ratios, Proportions, and Percents Description This example illustrates solving for c in a proportion where a = 8, b = 3, and d is expressed as x + 3. We set up the equation 8 / 3 = c / (x + 3) and solve for c, resulting in the expression c = (8(x + 3)) / 3. This problem demonstrates how to handle algebraic expressions in the denominator of a proportion.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 Topic Ratios, Proportions, and Percents Description This example illustrates solving for c in a proportion where a = 8, b = 3, and d is expressed as x + 3. We set up the equation 8 / 3 = c / (x + 3) and solve for c, resulting in the expression c = (8(x + 3)) / 3. This problem demonstrates how to handle algebraic expressions in the denominator of a proportion.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7 Topic Ratios, Proportions, and Percents Description This example demonstrates solving for d in a proportion where a = 12, b = 5, and c = 48. We set up the proportion 12 / 5 = 48 / d and solve for d, resulting in d = 20. This problem shows how to find an unknown value in the denominator of a proportion when all other values are known.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7 Topic Ratios, Proportions, and Percents Description This example demonstrates solving for d in a proportion where a = 12, b = 5, and c = 48. We set up the proportion 12 / 5 = 48 / d and solve for d, resulting in d = 20. This problem shows how to find an unknown value in the denominator of a proportion when all other values are known.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8 Topic Ratios, Proportions, and Percents Description This example illustrates solving for d in a proportion where a = 11, b = 5, and c is expressed as x - 4. We set up the equation 11 / 5 = (x - 4) / d and solve for d, resulting in the expression d = (5(x - 4)) / 11. This problem demonstrates how to handle algebraic expressions in the numerator of a proportion.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8 Topic Ratios, Proportions, and Percents Description This example illustrates solving for d in a proportion where a = 11, b = 5, and c is expressed as x - 4. We set up the equation 11 / 5 = (x - 4) / d and solve for d, resulting in the expression d = (5(x - 4)) / 11. This problem demonstrates how to handle algebraic expressions in the numerator of a proportion.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9 Topic Ratios, Proportions, and Percents Description This example demonstrates solving a proportion problem using similar triangles. Two triangles are shown, with one having sides of 3 and 4, and the other having sides of 6 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 3 / 4 = 6 / x. Solving this equation leads to x = 8.	Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9	Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9 Topic Ratios, Proportions, and Percents Description This example demonstrates solving a proportion problem using similar triangles. Two triangles are shown, with one having sides of 3 and 4, and the other having sides of 6 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 3 / 4 = 6 / x. Solving this equation leads to x = 8.	Proportions
Math Example--Volume Concepts--Calculating Volume: Example 1	Math Example--Volume Concepts--Calculating Volume: Example 1 Topic Volume Description A rectangular prism with dimensions labeled: length = 30, width = 10, and height = 8. The image shows how to find the volume of the prism using the formula for volume of a rectangular prism. This image illustrates Example 1: The caption explains how to calculate the volume of the rectangular prism using the formula V = l * w * h. The given dimensions are substituted into the formula: V = 30 * 10 * 8 = 2400..	Volume
Math Example--Volume Concepts--Calculating Volume: Example 10	Math Example--Volume Concepts--Calculating Volume: Example 10 Topic Volume Description A green cylinder with a general radius y and height x. The radius is marked on the top surface, and the height is marked on the side. This image illustrates Example 10: The task is to find the volume of this cylinder. The volume formula V = πr2h is used, and substituting r = y and h = x, the volume is calculated as V = xy2π.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 11	Math Example--Volume Concepts--Calculating Volume: Example 11 Topic Volume Description A hollow green cylinder with an outer radius of 10 units, an inner radius of 9 units, and a height of 15 units. The radii are marked on the top surface, and the height is marked on the side. This image illustrates Example 11: The task is to find the volume of this hollow cylinder. The volume formula for a hollow cylinder V = πr12h1 - πr22h2 is used. Substituting values, the result is V = 285π.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 12	Math Example--Volume Concepts--Calculating Volume: Example 12 Topic Volume Description A hollow green cylinder with an outer radius y, an inner radius y - 1, and a height x. The radii are marked on the top surface, and the height is marked on the side. This image illustrates Example 12: The task is to find the volume of this hollow cylinder. Using V = π(r12h1 - r22h2), substituting values gives: V = πx(y2 - (y - 1)2= πx(2y - 1).	Volume
Math Example--Volume Concepts--Calculating Volume: Example 13	Math Example--Volume Concepts--Calculating Volume: Example 13 Topic Volume Description A rectangular-based pyramid is shown with dimensions: base length 10, base width 8, and height 30. The image demonstrates how to calculate the volume of this pyramid. This image illustrates Example 13: The caption provides a step-by-step solution for calculating the volume of a pyramid with a rectangular base using the formula V = (1/3) * Area of Base * h. Substituting values: V = (1/3) * 8 * 10 * 30 = 800.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 14	Math Example--Volume Concepts--Calculating Volume: Example 14 Topic Volume Description A general rectangular-based pyramid is shown with variables x, y, and z representing the base dimensions and height. This example shows how to calculate the volume of a pyramid using variables instead of specific numbers. This image illustrates Example 14: The caption explains how to calculate the volume of a pyramid with a rectangular base using the formula V = (1/3) * Area of Base * h, which simplifies to V = (1/3) * x * y * z.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 15	Math Example--Volume Concepts--Calculating Volume: Example 15 Topic Volume	Volume
Math Example--Volume Concepts--Calculating Volume: Example 16	Math Example--Volume Concepts--Calculating Volume: Example 16 Topic Volume Description A truncated rectangular-based pyramid is shown with variables x, y, and z representing dimensions. The smaller virtual pyramid has reduced dimensions by 3 units for both width and length and reduced height by z - 20. The image demonstrates how to calculate the volume in terms of variables. This image illustrates Example 16: The caption explains how to find the volume of a truncated pyramid using variables for both pyramids' dimensions. Formula: V = (1/3) * xy(z + 20) - (1/3) * (y - 3)(x - 3)(z), which simplifies to V = (1/3) * (xyz + 60x + 60y - 180).	Volume
Math Example--Volume Concepts--Calculating Volume: Example 17	Math Example--Volume Concepts--Calculating Volume: Example 17 Topic Volume Description A green sphere with a radius labeled as 3. The image is part of a math example showing how to calculate the volume of a sphere. This image illustrates Example 17: The text describes finding the volume of a sphere. The formula used is V = (4/3) * π * r3, where r = 3. After substituting, the result is V = 36π. Volume is a fundamental concept in geometry that helps students understand the space occupied by three-dimensional objects. In this collection, each example uses various geometric shapes to calculate volume, showcasing real-life applications of volume in different shapes.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 18	Math Example--Volume Concepts--Calculating Volume: Example 18 Topic Volume Description A green sphere with a radius labeled as x. This image is part of a math example showing how to calculate the volume of a sphere using an unknown radius. This image illustrates Example 18: The text explains how to find the volume of a sphere with an unknown radius x. The formula used is V = (4/3) * π * r3, and substituting r = x gives V = (4/3) * x3 * π.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 19	Math Example--Volume Concepts--Calculating Volume: Example 19 Topic Volume Description A green cube with side length labeled as 7. The image illustrates how to calculate the volume of a cube with known side length. This image illustrates Example 19: The text describes finding the volume of a cube. The formula used is V = s3, where s = 7. After substituting, the result is V = 343. Volume is a fundamental concept in geometry that helps students understand the space occupied by three-dimensional objects. In this collection, each example uses various geometric shapes to calculate volume, showcasing real-life applications of volume in different shapes.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 2	Math Example--Volume Concepts--Calculating Volume: Example 2 Topic Volume Description A rectangular prism with dimensions labeled as x, y, and z. The image shows a general example of calculating the volume of a rectangular prism using variables instead of specific numbers. This image illustrates Example 2: The caption describes how to find the volume of a rectangular prism using variables for length (x), width (y), and height (z). The formula is given as V = x * y * z, but no specific values are provided.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 20	Math Example--Volume Concepts--Calculating Volume: Example 20 Topic Volume Description A green cube with side length labeled as x. This image is part of a math example showing how to calculate the volume of a cube using an unknown side length. This image illustrates Example 20: The text explains how to find the volume of a cube with an unknown side length x. The formula used is V = s3, and substituting s = x gives V = x3.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 21	Math Example--Volume Concepts--Calculating Volume: Example 21 Topic Volume Description A hollow cube with an outer edge of 9 and an inner hollow region with an edge of 7. The image shows how to calculate the volume by subtracting the volume of the inner cube from the outer cube. This image illustrates Example 21: Find the volume of a hollow cube. The formula used is V = s13 - s23, where s1 is the outer edge (9) and s2 is the inner edge (7). The solution calculates 9^3 - 7^3 = 386..	Volume
Math Example--Volume Concepts--Calculating Volume: Example 22	Math Example--Volume Concepts--Calculating Volume: Example 22 Topic Volume Description A hollow cube with an outer edge of x and an inner hollow region with an edge of x - 2. The image shows how to calculate the volume by subtracting the volume of the inner cube from the outer cube. This image illustrates Example 22: Find the volume of a hollow cube. The formula used is V = s13 - s23, where s1 = x and s2 = x - 2. Expanding and simplifying gives V = 6x2 - 12x + 8.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 23	Math Example--Volume Concepts--Calculating Volume: Example 23 Topic Volume Description A cone with a height of 12 and a radius of 4. The image shows how to calculate its volume using the cone volume formula (V = 1/3 * π * r2 * h). This image illustrates Example 23: Find the volume of a cone. The formula used is V = (1/3) * π * r2 * h, where r = 4 and h = 12. Substituting these values gives V = (1/3) * π * (42) * 12 = 64π.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 24	Math Example--Volume Concepts--Calculating Volume: Example 24 Topic Volume Description A cone with a height labeled as y and a radius labeled as x. The image shows how to calculate its volume using the cone volume formula (V = 1/3 * π * r2 * h). This image illustrates Example 24: Find the volume of a cone. The formula used is V = (1/3) * π * r2 * h, where r = x and h = y. Substituting these variables gives V = (x^2 * y)/3 * π.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 3	Math Example--Volume Concepts--Calculating Volume: Example 3 Topic Volume Description A hollow rectangular prism with outer dimensions: length = 60, width = 20, and height = 16. The inner hollow part has dimensions: length = 60, width = 18, and height = 14. The image shows how to subtract volumes to find the hollow volume. This image illustrates Example 3: The caption explains how to calculate the volume of a hollow rectangular prism by subtracting the inner volume from the outer volume. V = (60 * 20 * 16) - (60 * 18 * 14) = 4080.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 4	Math Example--Volume Concepts--Calculating Volume: Example 4 Topic Volume Description A hollow rectangular prism with outer dimensions labeled as x, y, and z, and inner hollow dimensions labeled as x - 2 and y - 2. The image shows a symbolic calculation for finding the hollow volume using variables. This image illustrates Example 4: The caption describes how to calculate the volume of a hollow rectangular prism by subtracting the inner volume from the outer volume using variables: V = xyz - z(y - 2)(x - 2) = 2z(y + x - 2).	Volume
Math Example--Volume Concepts--Calculating Volume: Example 5	Math Example--Volume Concepts--Calculating Volume: Example 5 Topic Volume Description The image shows a triangular prism with dimensions labeled as base (7), height (10), and length (25). It is part of an example on how to calculate the volume of a solid triangular prism. This image illustrates Example 5: "Find the volume of this triangular prism." The solution involves substituting the given measurements into the volume formula for a triangular prism: V = 1/2 * b * h * l = 1/2 * 7 * 10 * 25 = 875.	Volume
Math Example--Volume Concepts--Calculating Volume: Example 6	Math Example--Volume Concepts--Calculating Volume: Example 6 Topic Volume Description The image depicts a triangular prism with dimensions labeled as x, y, and z. The example demonstrates how to calculate the volume using a general formula for a triangular prism. This image illustrates Example 6: "Find the volume of this triangular prism." The solution uses the formula V = 1/2 * b * h * l, which is simplified to V = 1/2 * x * y * z..	Volume
Math Example--Volume Concepts--Calculating Volume: Example 7	Math Example--Volume Concepts--Calculating Volume: Example 7 Topic Volume Description The image shows a hollow triangular prism with outer dimensions labeled as base (10), height (7), and length (35), and inner dimensions labeled as base (8) and height (5). The example calculates the volume by subtracting the hollow region from the full prism. This image illustrates Example 7: "Find the volume of this hollow triangular prism." The solution calculates the full volume using V = 1/2 * b1 * h1 * l1 - 1/2 * b2 * h2 * l2, which simplifies to V = 1/2 * 10 * 7 * 35 - 1/2 * 8 * 5 * 35 = 525..	Volume
Math Example--Volume Concepts--Calculating Volume: Example 8	Math Example--Volume Concepts--Calculating Volume: Example 8 Topic Volume Description This image shows a hollow triangular prism with outer dimensions labeled as x, y, and z, and inner dimensions reduced by 2 units each. It demonstrates how to calculate the volume by subtracting the hollow region from the full prism. This image illustrates Example 8: "Find the volume of this hollow triangular prism." The solution uses V = 1/2 * b1 * h1 * l1 - 1/2 * b2 * h2 * l2, which simplifies to V = z(xy - (x - 2)(y - 2)) = z(x + y - 2)..	Volume
Math Example--Volume Concepts--Calculating Volume: Example 9	Math Example--Volume Concepts--Calculating Volume: Example 9 Topic Volume Description A green cylinder with a radius of 10 units and a height of 8 units. The radius is marked on the top surface, and the height is marked on the side. This image illustrates Example 9: The task is to find the volume of the cylinder. The volume formula V = πr2h is used. Substituting the values r = 10 and h = 8, the volume is calculated as V= 800π.	Volume
MATH EXAMPLES--Teacher's Guide: Polygon Classification	MATH EXAMPLES--Teacher's Guide: Polygon Classification This set of tutorials provides 36 examples of how to classify polygons based on side and angle characteristics. This is part of a collection of teacher's guides. To see the complete collection of teacher's guides, click on this link. Note: The download is a PDF file. Related Resources To see resources related to this topic click on the Related Resources tab above.	Definition of a Polygon
MATH EXAMPLES--Teacher's Guide: Quadrilateral Area and Perimeter	MATH EXAMPLES--Teacher's Guide: Quadrilateral Area and Perimeter This Teacher's Guide provides an overview of the 24 worked-out examples that show how to calculate the area and perimeter of different quadrilaterals. This is part of a collection of teacher's guides. To see the complete collection of teacher's guides, click on this link. Note: The download is a PDF file. Related Resources To see resources related to this topic click on the Related Resources tab above.	Area and Perimeter of Quadrilaterals
MATH EXAMPLES--Teacher's Guide: Surface Area	MATH EXAMPLES--Teacher's Guide: Surface Area This Teacher's Guide provides an overview of the 24 worked-out examples that show how to calculate the surface area of different three-dimensional figures. This is part of a collection of teacher's guides. To see the complete collection of teacher's guides, click on this link. Note: The download is a PDF file. Related Resources To see resources related to this topic click on the Related Resources tab above.	Surface Area

Illustrative Math Alignment: Grade 7 Unit 9

Putting it All Together

Lesson 5: How Crowded Is this Neighborhood?

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