Illustrative Math Alignment: Grade 7 Unit 3

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 Basics

Definition

The distance around the edge of a circle.

Description

The circumference is the distance around the edge of a circle. This concept is fundamental in understanding the properties of circles and their measurements. The formula for the circumference is C = 2πr, where r is the radius of the circle. Understanding the circumference is crucial for solving problems related to circle geometry, such as finding the length of an arc or the area of a sector.

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.

Topic

Circles

Definition

The circumference of a circle is the distance around the circle, calculated as C = 2πr.

Description

The circumference is a fundamental concept in geometry, representing the perimeter of a circle. It is widely used in fields such as engineering, design, and manufacturing, where precise measurements of circular objects are required. The formula 

C = 2πr 

Topic

Circles

Definition

The ratio of the circumference to the diameter of a circle is the constant π.

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.

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.

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).

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 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 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 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 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 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 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 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.

In this Slide Show, apply concepts of linear functions to the context of circumference vs. diameter.

Note: The download is a PPT file.

Related Resources

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.

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.

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 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 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 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 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 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 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 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 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 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.

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.

The formula for the Circumference of a Circle.

Related Resources

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