Illustrative Math Alignment: Grade 7 Unit 3

This is the transcript for the video of same title. Video contents: We visit Chaco Canyon in New Mexico to explore the circular kivas and in the process discover how circular buildings have been used to study the heavens.

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Topic

Geometry

Description

This example features a circle with radius x and a shaded sector with a central angle of 30 degrees. The task is to express the area of the shaded sector in terms of x. The solution involves using the central angle to find the fractional amount of the total area: Area = (30 / 360) * (π * x2) = π * x2 / 12.

Topic

Geometry

Description

This example presents a circle with radius x and a shaded sector with a central angle of 30 degrees. The task is to express the arc length of the shaded region in terms of x. The solution involves using the central angle to find the fractional amount of the circumference: Arc Length = (30 / 360) * (2 * π * x) = x / 6 * π.

Topic

Geometry

Description

This example features two concentric circles with radii of 5 and 4 units, and a shaded sector with a central angle of 30 degrees. The task is to calculate the area of the shaded region. The solution involves using the central angle to determine the fractional area difference between the larger and smaller circles: Area = (θ / 360) * (π * r12 - π * r22) = (30 / 360) * (π * 52 - π * 42) = π / 12 * (25 - 16) = 3π / 4.

Topic

Geometry

Description

This example presents two concentric circles with radii of 5 and 4 units, and a shaded sector with a central angle of 30 degrees. The task is to calculate the perimeter of the shaded region. The solution involves calculating arc lengths based on the central angle and adding straight-line segments between radii: Perimeter = (θ / 360) * (2 * π * r1 + 2 * π * r2) + 2 * (r1 - r2) = (30 / 360) * (2π * 5 + 2π * 4) + 2 * (5 - 4) = 3π / 2 + 2.

Topic

Geometry

Description

This example features two concentric circles with radii x and y, and a shaded sector with a central angle θ (theta). The task is to express the area of the shaded region in terms of x, y, and θ. The solution uses the central angle to find the fractional area difference between the larger and smaller circles: Area = (θ / 360) * (π * x2 - π * y2) = (θπ / 360) * (x2 - y2).

Topic

Geometry

Description

This example presents two concentric circles with radii x and y, and a shaded sector with a central angle θ (theta). The task is to express the perimeter of the shaded region in terms of x, y, and θ. The solution involves calculating arc lengths based on the central angle and adding straight-line segments between radii: Perimeter = (θ / 360) * (2πx + 2πy) + 2(x - y) = π/180 * θ(x + y) + 2(x - y).

Topic

Geometry

Description

This example features two concentric circles with radii 5 and y, and a shaded sector with a central angle of 30 degrees. The task is to express the area of the shaded region in terms of y. The solution involves calculating the difference between the areas of the larger and smaller circles, accounting for the central angle: Area = π/12 * (25 - y2).

Topic

Geometry

Description

This example features two concentric circles with radii 5 and y, and a shaded sector with a central angle of 30 degrees. The task is to calculate the perimeter of the shaded region. Given a central angle of 30 degrees, radius 5, and unknown radius y, the perimeter is calculated as: Perimeter = π/6 * (5 + y) + 2(5 - y).

Topic

Geometry

Description

This example presents two concentric circles with radii 4 and x, and a shaded sector with a central angle θ (theta). The task is to express the area of the shaded region in terms of x and θ. The solution uses the central angle to find the fractional area difference between the larger and smaller circles: Area = (θ / 360) * (π * x2 - π * 42) = (θ * π / 360) * (x2 - 16).

Topic

Geometry

Description

This example features two concentric circles with radii x and 4, and a shaded sector with a central angle θ (theta). The task is to express the perimeter of the shaded region in terms of x and θ. The solution involves calculating arc lengths based on the central angle and adding straight-line segments between radii: Perimeter = (θ / 360) * (2 * π * x + 2 * π * 4) + 2 * (x - 4) = (π / 180) * θ(x + 4) + 2(x - 4).

This is the transcript for the video of same title. Video contents: In this program we explore the properties of circles. We do this in the context of two real-world applications. In the first, we look at the design of the Roman Coliseum and explore how circular shapes could have been used to design this elliptical structure. In the second application we look at the Roman Pantheon, specifically its spherical dome, to see how the properties of chords and secants help clarify its unique design.

Topic

Geometry

Description

This example presents a circle with a radius of 5 units and a shaded sector with central angle θ (theta). The task is to express the arc length of the shaded region in terms of θ. The solution involves using the central angle to find the fractional amount of the circumference: Arc Length = (θ / 360) * (2 * π * 5) = θ / 18 * π.

This is the transcript for the video of same title. Video contents: The Roman Coliseum is a large elliptical structure. Yet, the Romans likely used circular arcs to build it. This segment explores the properties of circles and shows how arcs can be used to create elliptical shapes.

This is the transcript for the video of same title. Video contents: The Roman Pantheon is a domed structure that shows a keen awareness of the position of the sun throughout the year. The source of light from the top of the dome allows for the exploration of chords, inscribed angles, central angles, and intercepted arcs.

This is the Teacher's Guide that accompanies Geometry Applications: Circles.

Related Resources

To see resources related to this topic click on the Related Resources tab above.

This set of tutorials provides 23 examples of solving for the area and circumferences of circles and sections of circles.

Related Resources

To see resources related to this topic click on the Related Resources tab above.

The Roman Pantheon is a domed structure that shows a keen awareness of the position of the sun throughout the year. The source of light from the top of the dome allows for the exploration of chords, inscribed angles, central angles, and intercepted arcs.

The Roman Coliseum is a large elliptical structure. Yet, the Romans likely used circular arcs to build it. This segment explores the properties of circles and shows how arcs can be used to create elliptical shapes.

We visit Chaco Canyon in New Mexico to explore the circular kivas and in the process discover how circular buildings have been used to study the heavens.

In this program we explore the properties of circles. We do this in the context of two real-world applications. In the first, we look at the design of the Roman Coliseum and explore how circular shapes could have been used to design this elliptical structure. In the second application we look at the Roman Pantheon, specifically its spherical dome, to see how the properties of chords and secants help clarify its unique design.

In this TI-Nspire Activity, use the Geometry and Graphing Tools to explore the ratio of Circumference to Radius as a linear function. 

Note: The Preview is a Google Slide Show and the download is a PPT. Subscribers to Media4Math can download resources.

In this Algebra Application, students study the direction between diameter and circumference of a circle. Through measurement and data gathering students analyze the line of best fit and explore ways of calculating pi. The math topics covered include: Mathematical modeling, Linear functions, Data gathering and analysis, Ratios, Direct variation. This is a great back-to-school activity for middle school or high school students. This is also a great crossover activity that ties algebra and geometry.

December 2022. In this issue of Math in the News we look at the Orion rocket's return to Earth from its lunar mission. This provides an opportunity to explore different types of orbits. 

Note: The download is a PPT file.

Related Resources

To see resources related to this topic click on the Related Resources tab above.

Topic

Geometry

Description

This example features a circle with a radius of 5 units. The objective is to calculate the circumference of the circle using the given radius. The solution involves applying the circumference formula: C = 2 * π * r = 2 * π * (5) = 10π.

Circular area and circumference calculations are fundamental in geometry. These examples provide students with practical applications of theoretical concepts, helping them understand the relationship between a circle's radius and its circumference.

We visit Chaco Canyon in New Mexico to explore the circular kivas and in the process discover how circular buildings have been used to study the heavens.

The Roman Coliseum is a large elliptical structure. Yet, the Romans likely used circular arcs to build it. This segment explores the properties of circles and shows how arcs can be used to create elliptical shapes.

The Roman Pantheon is a domed structure that shows a keen awareness of the position of the sun throughout the year. The source of light from the top of the dome allows for the exploration of chords, inscribed angles, central angles, and intercepted arcs.

January 2014. In this issue of Math in the News we examine the Polar Vortex. We examine the physics of the cyclonic winds that make up the Polar Vortex and under what conditions the current expansion of Arctic weather affects a large part of the United States.

Related Resources

To see resources related to this topic click on the Related Resources tab above.

The Roman Coliseum is a large elliptical structure. Yet, the Romans likely used circular arcs to build it. This segment explores the properties of circles and shows how arcs can be used to create elliptical shapes. Note: The download for this resources is the Promethean Flipchart.

To access the full video [Geometry Applications: Circles, Segment 2: Circles and Arcs]: https://www.media4math.com/library/geometry-applications-circles-segment-2-circles-and-arcs

This video includes a video transcript: https://media4math.com/library/video-transcript-geometry-applications-circles-segment-2-circles-and-arcs

The Roman Pantheon is a domed structure that shows a keen awareness of the position of the sun throughout the year. The source of light from the top of the dome allows for the exploration of chords, inscribed angles, central angles, and intercepted arcs. Note: The download for this resources is the Promethean Flipchart.

To access the full video [Geometry Applications: Circles, Segment 3: Chords and Inscribed Angles]: https://media4math.com/library/geometry-applications-circles-segment-3-chords-and-inscribed-angles

This video includes a Video Transcript: https://www.media4math.com/library/video-transcript-geometry-applications-circles-segment-3-chords-and-inscribed-angles

Topic

Geometric Models

Topic

Geometric Models

This is part of a collection of math clip art images that show Venn Diagrams.

This is part of a collection of math clip art images that show Venn Diagrams.

Topic

Geometry

Description

This example presents a circle with a radius of 5 units. The task is to calculate the area of the circle using the given radius. The solution involves substituting the radius value into the area formula: A = π * r^2 = π * (5)^2 = 25π.

Understanding circular area and circumference is integral to mastering geometry. Concepts such as calculating areas and circumferences of circles are fundamental, and exercises like these examples not only provide practice but also deepen the understanding of theoretical concepts in a practical way.

In this program we explore the properties of circles. We do this in the context of two real-world applications. In the first, we look at the design of the Roman Coliseum and explore how circular shapes could have been used to design this elliptical structure. In the second application we look at the Roman Pantheon, specifically its spherical dome, to see how the properties of chords and secants help clarify its unique design.

Topic

Geometry

Description

This example introduces a circle with an unknown radius represented by x. The task is to express the area of the circle in terms of x. The solution demonstrates how to use the area formula with a variable radius: A = π * r2 = π * (x)2 = π * x2.

Working with variable expressions in geometry helps students transition from concrete to abstract thinking. This example bridges the gap between numerical calculations and algebraic representations, a crucial skill in advanced mathematics.

Topic

Geometry

Description

This example presents a circle with an unknown radius represented by x. The task is to express the circumference of the circle in terms of x. The solution demonstrates how to use the circumference formula with a variable radius: C = 2 * π * r = 2 * π * x = 2πx.

Understanding circular area and circumference is crucial in geometry. This example helps students transition from concrete numerical values to abstract algebraic expressions, fostering a deeper comprehension of the relationship between a circle's radius and its circumference.

Topic

Geometry

Description

This example features two concentric circles with radii of 5 and 4 units. The task is to calculate the area of the shaded region between these circles. The solution involves subtracting the area of the smaller circle from the area of the larger circle: A = π * (5)2 - π * (4)2 = 25π - 16π = 9π.

Concentric circles and shaded regions introduce students to more complex geometric concepts. This example builds upon basic circular area calculations, encouraging students to think about the relationships between different circles and how to find areas of composite shapes.

Topic

Geometry

Description

This example presents two concentric circles with radii of 5 and y units. The objective is to express the area of the shaded region between these circles in terms of y. The solution involves subtracting the area of the smaller circle from the area of the larger circle: A = π * (5)2 - π * y2 = 25π - πy2 = π(25 - y2).

Topic

Geometry

Description

This example features two concentric circles with radii x and y. The task is to express the area of the shaded region between these circles in terms of x and y. The solution involves subtracting the area of the smaller circle from the area of the larger circle: A = π * x2 - π * y2 = π(x2 - y2).

Topic

Geometry

Description

This example presents a circle with a radius of 5 units and a shaded sector with a central angle of 30 degrees. The task is to calculate the area of the shaded sector. The solution involves using the central angle to find the fractional amount of the total area: A = (30 / 360) * π * 52 = 25π / 12.

Topic

Geometry

Description

This example features a circle with a radius of 5 units and a shaded arc corresponding to a central angle of 30 degrees. The task is to calculate the length of the shaded arc. The solution involves using the central angle to find the fractional amount of the circumference: Arc Length = (30 / 360) * 2 * π * 5 = 5π / 6.

Topic

Geometry

Description

This example presents a circle with radius x and a shaded sector with central angle θ (theta). The task is to express the area of the shaded sector in terms of x and θ. The solution involves using the central angle to find the fractional amount of the total area: A = (θ / 360) * π * x2 = πx2θ / 360.

Topic

Geometry

Description

This example presents a circle with radius x and a shaded sector with central angle θ (theta). The task is to express the arc length of the shaded region in terms of x and θ. The solution involves using the central angle to find the fractional amount of the circumference: Arc Length = (θ / 360) * (2 * π * x) = (θ * x / 180) * π.

Topic

Geometry

Description

This example features a circle with a radius of 5 units and a shaded sector with central angle θ (theta). The task is to express the area of the shaded sector in terms of θ. The solution involves using the central angle to find the fractional amount of the total area: Area = (θ / 360) * (π * 52) = 25π / 360 * θ = 5π / 72 * θ.