Learn on PengiYoshiwara Core MathChapter 3: Measurement

Lesson 3.4: Circles and Spheres

In this Grade 8 lesson from Yoshiwara Core Math, students learn to calculate the circumference and area of circles using the formulas C = π × d and A = π × r², and extend those concepts to find the volume of a sphere. The lesson also covers key circle terminology including radius, diameter, and circumference, and introduces π as an irrational number. Students apply these formulas to real-world problems involving circular and spherical objects.

Section 1

📘 Circles and Spheres

New Concept

This lesson introduces key geometric formulas. You'll learn to calculate the circumference and area of circles, and the volume of spheres, using the radius, diameter, and the constant π\pi.

What’s next

Next, you'll apply these formulas through guided examples and practice cards to calculate area, circumference, and volume.

Section 2

Circumference of a Circle

Property

The distance from the center of a circle to any point on the circle itself is called the radius of the circle.
The diameter of a circle is the length of a line segment joining two points on the circle and passing through the center. Thus, the diameter of a circle is twice its radius.
The perimeter of a circle is called its circumference.
The circumference CC of a circle is given by

C=π×dC = \pi \times d

where dd is the diameter of the circle. The Greek letter π\pi (pi) stands for an irrational number: π=3.141592654...\pi = 3.141592654 ...

Examples

  • A circular pool has a diameter of 10 meters. Its circumference is C=π×1031.42C = \pi \times 10 \approx 31.42 meters.
  • A bicycle wheel has a radius of 14 inches. Its diameter is 2×14=282 \times 14 = 28 inches, so its circumference is C=π×2887.96C = \pi \times 28 \approx 87.96 inches.
  • If a running track has a circumference of 400 meters, its diameter can be found by d=Cπ=400π127.32d = \frac{C}{\pi} = \frac{400}{\pi} \approx 127.32 meters.

Explanation

Circumference is the special name for a circle's perimeter. It's the distance around the circle's edge. This distance is always a little more than 3 times the circle's diameter, a constant ratio we call pi (π\pi).

Section 3

Area of a Circle

Property

The area enclosed by a circle is given by

Area=π×r2\text{Area} = \pi \times r^2

where rr is the radius of the circle. An exponent tells us how many times the base occurs as a factor in a product. So r2r^2 means r×rr \times r. We compute powers before we compute the rest of a product.

Examples

  • A pizza has a radius of 7 inches. Its area is A=π×72=49π153.94A = \pi \times 7^2 = 49\pi \approx 153.94 square inches.
  • A circular garden has a diameter of 20 feet. Its radius is 10 feet, so its area is A=π×102=100π314.16A = \pi \times 10^2 = 100\pi \approx 314.16 square feet.
  • A circular rug has an area of 50 square feet. Its radius is found by solving r=Aπ=50π15.913.99r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{50}{\pi}} \approx \sqrt{15.91} \approx 3.99 feet.

Explanation

The area of a circle tells you the amount of space inside it. To find it, you first square the radius (multiply it by itself), and then multiply that result by pi (π\pi). This gives the total surface coverage.

Section 4

Volume of a Sphere

Property

The volume of a sphere is given by

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

where rr is the radius of the sphere. Recall that r3r^3, which we read as 'rr cubed,' means r×r×rr \times r \times r.

Examples

  • A gumball has a radius of 1 centimeter. Its volume is V=43π(1)3=43π4.19V = \frac{4}{3} \pi (1)^3 = \frac{4}{3}\pi \approx 4.19 cubic centimeters.
  • A soccer ball has a diameter of 22 cm, so its radius is 11 cm. Its volume is V=43π(11)3=43π(1331)5575.28V = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi (1331) \approx 5575.28 cubic centimeters.
  • A spherical ornament has a volume of 36π36\pi cubic inches. To find its radius, solve 36π=43πr336\pi = \frac{4}{3}\pi r^3, which simplifies to 27=r327 = r^3, so the radius is r=273=3r = \sqrt[3]{27} = 3 inches.

Explanation

Volume measures the space inside a 3D shape, like a ball or a planet. For a sphere, you cube the radius (multiply it by itself three times), then multiply by pi (π\pi), and finally multiply by the fraction 43\frac{4}{3}.

Section 5

Circle and Sphere Formulas

Property

The diameter of a circle is twice its radius.

D=2×rD = 2 \times r

Circumference of a circle

C=π×DorC=2×π×rC = \pi \times D \quad \text{or} \quad C = 2 \times \pi \times r

Area of a circle

A=π×r2A = \pi \times r^2

Volume of a sphere

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

Examples

  • For a circle with a radius of 5 cm, the circumference is C=2π(5)=10π31.42C = 2\pi(5) = 10\pi \approx 31.42 cm and the area is A=π(5)2=25π78.54A = \pi(5)^2 = 25\pi \approx 78.54 square cm.
  • A sphere has a diameter of 12 inches, so its radius is 6 inches. Its volume is V=43π(6)3=43π(216)=288π904.78V = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi \approx 904.78 cubic inches.
  • A circular plate has a circumference of 30 inches. Its diameter is d=30π9.55d = \frac{30}{\pi} \approx 9.55 inches, and its radius is approximately 4.7754.775 inches.

Explanation

This card is a quick reference for all the key formulas for circles and spheres. Notice how the radius (rr) or diameter (DD) is the key measurement you need to find circumference, area, or volume.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 3: Measurement

  1. Lesson 1

    Lesson 3.1: Volume and Surface Area

  2. Lesson 2

    Lesson 3.2: Exponents

  3. Lesson 3

    Lesson 3.3: Units of Measure

  4. Lesson 4Current

    Lesson 3.4: Circles and Spheres

  5. Lesson 5

    Lesson 3.5: Large Numbers

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Circles and Spheres

New Concept

This lesson introduces key geometric formulas. You'll learn to calculate the circumference and area of circles, and the volume of spheres, using the radius, diameter, and the constant π\pi.

What’s next

Next, you'll apply these formulas through guided examples and practice cards to calculate area, circumference, and volume.

Section 2

Circumference of a Circle

Property

The distance from the center of a circle to any point on the circle itself is called the radius of the circle.
The diameter of a circle is the length of a line segment joining two points on the circle and passing through the center. Thus, the diameter of a circle is twice its radius.
The perimeter of a circle is called its circumference.
The circumference CC of a circle is given by

C=π×dC = \pi \times d

where dd is the diameter of the circle. The Greek letter π\pi (pi) stands for an irrational number: π=3.141592654...\pi = 3.141592654 ...

Examples

  • A circular pool has a diameter of 10 meters. Its circumference is C=π×1031.42C = \pi \times 10 \approx 31.42 meters.
  • A bicycle wheel has a radius of 14 inches. Its diameter is 2×14=282 \times 14 = 28 inches, so its circumference is C=π×2887.96C = \pi \times 28 \approx 87.96 inches.
  • If a running track has a circumference of 400 meters, its diameter can be found by d=Cπ=400π127.32d = \frac{C}{\pi} = \frac{400}{\pi} \approx 127.32 meters.

Explanation

Circumference is the special name for a circle's perimeter. It's the distance around the circle's edge. This distance is always a little more than 3 times the circle's diameter, a constant ratio we call pi (π\pi).

Section 3

Area of a Circle

Property

The area enclosed by a circle is given by

Area=π×r2\text{Area} = \pi \times r^2

where rr is the radius of the circle. An exponent tells us how many times the base occurs as a factor in a product. So r2r^2 means r×rr \times r. We compute powers before we compute the rest of a product.

Examples

  • A pizza has a radius of 7 inches. Its area is A=π×72=49π153.94A = \pi \times 7^2 = 49\pi \approx 153.94 square inches.
  • A circular garden has a diameter of 20 feet. Its radius is 10 feet, so its area is A=π×102=100π314.16A = \pi \times 10^2 = 100\pi \approx 314.16 square feet.
  • A circular rug has an area of 50 square feet. Its radius is found by solving r=Aπ=50π15.913.99r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{50}{\pi}} \approx \sqrt{15.91} \approx 3.99 feet.

Explanation

The area of a circle tells you the amount of space inside it. To find it, you first square the radius (multiply it by itself), and then multiply that result by pi (π\pi). This gives the total surface coverage.

Section 4

Volume of a Sphere

Property

The volume of a sphere is given by

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

where rr is the radius of the sphere. Recall that r3r^3, which we read as 'rr cubed,' means r×r×rr \times r \times r.

Examples

  • A gumball has a radius of 1 centimeter. Its volume is V=43π(1)3=43π4.19V = \frac{4}{3} \pi (1)^3 = \frac{4}{3}\pi \approx 4.19 cubic centimeters.
  • A soccer ball has a diameter of 22 cm, so its radius is 11 cm. Its volume is V=43π(11)3=43π(1331)5575.28V = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi (1331) \approx 5575.28 cubic centimeters.
  • A spherical ornament has a volume of 36π36\pi cubic inches. To find its radius, solve 36π=43πr336\pi = \frac{4}{3}\pi r^3, which simplifies to 27=r327 = r^3, so the radius is r=273=3r = \sqrt[3]{27} = 3 inches.

Explanation

Volume measures the space inside a 3D shape, like a ball or a planet. For a sphere, you cube the radius (multiply it by itself three times), then multiply by pi (π\pi), and finally multiply by the fraction 43\frac{4}{3}.

Section 5

Circle and Sphere Formulas

Property

The diameter of a circle is twice its radius.

D=2×rD = 2 \times r

Circumference of a circle

C=π×DorC=2×π×rC = \pi \times D \quad \text{or} \quad C = 2 \times \pi \times r

Area of a circle

A=π×r2A = \pi \times r^2

Volume of a sphere

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

Examples

  • For a circle with a radius of 5 cm, the circumference is C=2π(5)=10π31.42C = 2\pi(5) = 10\pi \approx 31.42 cm and the area is A=π(5)2=25π78.54A = \pi(5)^2 = 25\pi \approx 78.54 square cm.
  • A sphere has a diameter of 12 inches, so its radius is 6 inches. Its volume is V=43π(6)3=43π(216)=288π904.78V = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi \approx 904.78 cubic inches.
  • A circular plate has a circumference of 30 inches. Its diameter is d=30π9.55d = \frac{30}{\pi} \approx 9.55 inches, and its radius is approximately 4.7754.775 inches.

Explanation

This card is a quick reference for all the key formulas for circles and spheres. Notice how the radius (rr) or diameter (DD) is the key measurement you need to find circumference, area, or volume.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 3: Measurement

  1. Lesson 1

    Lesson 3.1: Volume and Surface Area

  2. Lesson 2

    Lesson 3.2: Exponents

  3. Lesson 3

    Lesson 3.3: Units of Measure

  4. Lesson 4Current

    Lesson 3.4: Circles and Spheres

  5. Lesson 5

    Lesson 3.5: Large Numbers