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Learning Math Home
Measurement Session 7: Solutions
Session 7 Part A Part B Homework
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Session 7 Materials:

A B 


Solutions for Session 7 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6

Problem H1

Area of circle: AC = r2
Area of square: AS = (2r)2 = 4r2
AC/AS = r2/4r2 = /4

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

<< back to Problem H1


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 ; 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:


(addition to C)/ = addition to diameter

<< back to Problem H2


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 • - 10.5 • 2 • = 4.5 • 2 • ,

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.

<< back to Problem H3


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 • • (10)2 + 1/4 • • (42) = 75 • + 4 • = 79 • ,

or approximately 248 ft2.

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Problem H5


The area of the annulus is the difference between the circles' areas. For these circles, the area is 300 cm2, or approximately 943 cm2.


If the smallest circle has a radius of 10 cm, then the area of that bull's-eye circle is 100 cm2. The area of the first annular ring is the 400 - 100, or 300 cm2. 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 900 - 400, or 500 cm2; the area of the third annular ring is 1,600 - 900, or 700 cm2; and the area of the fourth (outer) annular ring is 2,500 - 1,600, or 900 cm2. 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 900 cm2.

<< back to Problem H5


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 • • 25 + 2 • 100, or approximately 357 m
Area: • 252 + 100 • 50, or approximately 6,963 m2

<< back to Problem H6


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