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A circle is inscribed in a square. What percentage of the area of the square is inside the circle?

Imagine that a giant hula hoop is fitted snugly around the Earth’s equator. The diameter of the hula hoop is 12,800 km. Next, imagine that the hula hoop is cut and its circumference is increased by 10 m. The hula hoop is adjusted around the equator so that every part of the hula hoop lies the same distance above the surface of the Earth. Would you be able to crawl under it? Walk under it standing upright? Drive a moving truck under it? Determine the new diameter of the hoop, and find out the distance between the Earth and the hula hoop.

**Problem H3**

A new car boasts a turning radius of 15 ft. This means that it can make a complete circle with a radius of 15 ft. and return to its original spot. The radius is measured from the center of the circle to the outside wheel. If the two front tires are 4.5 ft. apart, how much further do the outside tires have to travel than the inside tires to complete the circle?

Your dog is chained to a corner of the toolshed in your backyard. The chain measures 10 ft. in length. The toolshed is rectangular, with dimensions 6 ft. by 12 ft. Draw the picture showing the area the dog can reach while attached to the chain. Compute this area.

Draw a diagram of the shed and the possible areas that the dog could reach on its chain. Then divide the space into different sections and calculate the area of each section.

An annulus is the region bounded by two concentric circles.

- If the radius of the small circle is 10 cm and the radius of the large circle is 20 cm, what is the area of the annulus?
- A dartboard has four annular rings surrounding a bull’s-eye. The circles have radii 10, 20, 30, 40, and 50 cm. How do the areas of the annular rings compare? Suppose a dart is equally likely to hit any point on the board. Is the dart more likely to hit in the outermost ring or inside the region containing the bull’s-eye and the two innermost rings? Explain.

*
* Express the areas in terms of π and then compare them.

An oval track is made by erecting semicircles on each end of a 50 m-by-100 m rectangle. What is the length of the track? What is the area of the region enclosed by the track?

**Zebrowski, Ernest (1999). A History of the Circle (pp. 48-49). Piscataway, N.J.: Rutgers University Press.**

Reproduced with permission from the publisher. � 1999 by Rutgers University Press. All rights reserved.

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A History of the Circle

**Problem H1**

Area of circle: A_{C} = πr^{2}

Area of square: A_{S} = (2r)^{2} = 4r^{2}

A_{C}/A_{S} = πr^{2}/4r^{2} = π/4

The fractional part of the area is π/4. This is equal to 0.785, and therefore, expressed as a percentage, is approximately 78.5%.

**Problem H2**

Since the diameter of the hula hoop is 12,800 km, the circumference is approximately 40,212.386 km. If we cut the hula hoop and add 10 m (0.01 km), the circumference is now 40,212.396 km. The new diameter of the hula hoop is found by dividing 40,212.396 by π; it is 12,800.003 km. The difference between the two diameters is 0.003 km, or 3 m. Dividing this difference in half (since d = 2r) results in a 1.5-meter height change between the Earth and the hula hoop. You could easily crawl under it or walk under it in a crouched position, but you could not drive a truck under it!

The interesting fact about this problem is that the distance added to the diameter (and radius) is independent of the original diameter and circumference:

C + addition to C = π(d + addition to diameter) [new]

C = π • d [original]

–

addition to C = π • (addition to diameter)

or,

(addition to C)/π = addition to diameter

**Problem H3**

Since the radius of the circle formed by the outside tires is 15 ft., the radius formed by the inside tires is (15 – 4.5) = 10.5 ft. The circumference of the two circles can be calculated and subtracted:

15 • 2 • π – 10.5 • 2 • π = 4.5 • 2 • π,

which is approximately 28.3 ft. Note that the radius of the circle formed by the outside tires was not important to the final result (which only used the 4.5-foot difference). This means that the calculation is valid for any car whose wheels are 4.5 ft. apart.

**Problem H4**

The area is a three-quarters circle with radius 10 ft., plus a quarter-circle with radius 4 ft. (The dog can reach this area by stretching along the six-foot wall and then pointing into the exposed area.) The total area is

3/4 • π • (10)^{2} + 1/4 • π • (4^{2}) = 75 • π + 4 • π = 79 • π,

or approximately 248 ft^{2}.

**Problem H5**

- The area of the annulus is the difference between the circles’ areas. For these circles, the area is 300π cm
^{2}, or approximately 943 cm^{2}. - If the smallest circle has a radius of 10 cm, then the area of that bull’s-eye circle is 100π cm
^{2}. The area of the first annular ring is the 400π – 100π, or 300π cm^{2}. Since the second interior circle has a radius of 20 cm, we can find the area of it and then subtract the area of the bull’s-eye. Using this line of reasoning, the area of the second annular ring is 900π – 400π, or 500π cm^{2}; the area of the third annular ring is 1,600π – 900π, or 700π cm^{2}; and the area of the fourth (outer) annular ring is 2,500π – 1,600π, or 900π cm^{2}. The probability of a dart thrown at random hitting the outermost ring or a dart hitting the bull’s-eye and the two innermost rings is exactly the same; both regions have an area of 900π cm^{2}.

**Problem H6**

The length and area can be more easily calculated by isolating the circular sections, which then form a complete circle of radius 25 m.

Length: 2 • π • 25 + 2 • 100, or approximately 357 m

Area: π • 25^{2} + 100 • 50, or approximately 6,963 m^{2}