Circumference of Circles and Semicircles - Seventh Grade Mathematics

Circumference calculates the perimeter of circular boundaries, translating rotational revolutions into linear distances and modeling curved boundaries of composite semicircles.

The Circumference Formulas

Circumference is the linear perimeter of a circle. Because the universal definition of pi is circumference divided by diameter, multiplying both sides by diameter gives the primary circumference formula:

               THE CIRCUMFERENCE FORMULAS
                       C = pi * d
                       C = 2 * pi * r

Because the diameter d is equal to two times the radius (d = 2r), both formulas are completely identical. Use C = pi * d when diameter is given; use C = 2 * pi * r when radius is given.

Example 1: Find circumference of a circle with diameter 10 cm.
  Exact Form:        C = 10 * pi cm
  Approximate Form:  C = 10 * 3.14 = 31.4 cm

Example 2: Find circumference of a circle with radius 7 inches using 22/7.
  C = 2 * (22/7) * 7
  The factor of 7 cancels:  C = 2 * 22 = 44 inches

Using twenty-two sevenths when radius or diameter is a multiple of seven allows rapid mental computation without tedious decimal division.

Reverse Engineering: Finding Radius and Diameter from Circumference

When the circumference of a circular boundary is known, algebraic isolation reveals the underlying diameter and radius.

+-----------------------+-----------------------------+------------------------+
| Desired Dimension     | Algebraic Formula           | Operation Required     |
+-----------------------+-----------------------------+------------------------+
| Diameter (d)          | d = C / pi                  | Divide C by 3.14       |
| Radius (r)            | r = C / (2 * pi)            | Divide C by 6.28       |
+-----------------------+-----------------------------+------------------------+

Suppose a circular running track has an inside boundary measuring four hundred meters. To find the radius: r equals four hundred divided by the quantity two times 3.14. Two times 3.14 equals 6.28. Four hundred divided by 6.28 equals approximately sixty-three point sixty-nine meters.

Rotational Motion and Distance Traveled

A practical real-world application of circumference is rolling wheel physics. In one complete rotation (revolution) of a circular wheel, the vehicle moves forward along the ground by a linear distance exactly equal to the wheel's circumference.

Linear Distance = Number of Revolutions * Circumference
Distance = N * (pi * d)

Consider a bicycle whose wheels have a diameter of twenty-six inches. Circumference per revolution: twenty-six times 3.14 equals eighty-one point sixty-four inches. If the wheels rotate one hundred times: one hundred times eighty-one point sixty-four equals eight thousand one hundred sixty-four inches (or approximately six hundred eighty feet).

Perimeter of Semicircles

A semicircle is exactly half of a circle. Calculating the perimeter of a closed semicircle requires finding the curved arc length and adding the straight diameter base closing the shape.

               PERIMETER OF A CLOSED SEMICIRCLE
                     ...---...
                  .-'         '-.  <-- Curved Arc = (1/2) * pi * d
                .'               '.
               +-------------------+
               <--- Diameter (d) --->
               Perimeter = (1/2) * pi * d + d

A common mistake is calculating half of the circumference and stopping. Half of the circumference represents only the curved arc; a closed semicircular region requires adding the diameter base to form a closed perimeter.

Example: Perimeter of a closed semicircle with diameter 14 cm (using 22/7).
Curved Arc: (1/2) * pi * d = (1/2) * (22/7) * 14 = (1/2) * 44 = 22 cm
Straight Base: d = 14 cm
Total Perimeter: 22 + 14 = 36 cm

The perimeter of the closed semicircle is thirty-six centimeters.

Exercises

Work through each problem carefully in your notebook before reading the provided solutions.

Exercise 1. Calculate the exact circumference and the approximate circumference (using 3.14) of a circular trampoline with a radius of six feet.

Exercise 2. A circular dining table has a circumference of one hundred thirty-two inches. Using twenty-two sevenths as an approximation for pi, calculate the diameter of the table.

Exercise 3. A bicycle tire has a diameter of twenty inches. How many full revolutions does the tire make when traveling a distance of three thousand one hundred forty inches?

Exercise 4. A window is constructed in the shape of a closed semicircle with a straight bottom diameter measuring ten feet. Calculate the total perimeter of the window frame using 3.14 for pi.

Solutions

Solution to Exercise 1. Use the circumference formula C equals two times pi times r. The radius is six feet. In exact form: C equals two times pi times six, which equals twelve pi feet. In approximate form: multiply twelve by three point fourteen: twelve times three point fourteen equals thirty-seven point sixty-eight feet. The exact circumference is twelve pi feet, and the approximate circumference is thirty-seven point sixty-eight feet.

Solution to Exercise 2. Use the formula d equals C divided by pi. Substitute circumference C equals one hundred thirty-two inches and pi equals twenty-two sevenths: d equals one hundred thirty-two divided by twenty-two sevenths. Invert and multiply: one hundred thirty-two times seven twenty-seconds. Divide one hundred thirty-two by twenty-two to obtain six; then multiply six by seven to obtain forty-two. The diameter of the table is forty-two inches.

Solution to Exercise 3. First calculate the circumference of the bicycle tire: C equals pi times d. Multiply twenty by three point fourteen: twenty times three point fourteen equals sixty-two point eight inches per revolution. To find the number of revolutions, divide the total distance by the circumference per revolution: three thousand one hundred forty divided by sixty-two point eight. Multiplying numerator and denominator by ten gives thirty-one thousand four hundred divided by six hundred twenty-eight, which simplifies to fifty. The tire makes fifty full revolutions.

Solution to Exercise 4. The perimeter of a closed semicircle equals the curved semicircular arc plus the straight diameter base: perimeter equals one-half times pi times d plus d. The diameter is ten feet. Calculate the curved arc length: one-half times three point fourteen times ten equals one-half of thirty-one point four, which equals fifteen point seven feet. Add the straight diameter base of ten feet: fifteen point seven plus ten equals twenty-five point seven feet. The total perimeter of the window frame is twenty-five point seven feet.