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Solutions for Session 7 Homework
See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6
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Problem H1 | |
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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
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Problem H2 | |
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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:
or,
(addition to C)/ = addition to diameter
<< back to Problem H2
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Problem H3 | |
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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
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Problem H4 | |
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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.
<< back to Problem H4
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Problem H5 | |
a. | The area of the annulus is the difference between the circles' areas. For these circles, the area is 300 cm2, or approximately 943 cm2. |
b. | 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
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Problem H6 | |
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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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