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Solutions for Session 2, Part B
See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9
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Problem B1 | |
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Since 1 km = 100,000 cm, the scale 1 cm:250 km is equivalent to 1 cm:250 • 100,000 cm, or 1:25,000,000.
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Problem B2 | |
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The scale factor is 45:3. This can be simplified to 15:1 or expressed in other ways, such as 7.5:0.5 or 150:10.
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Problem B3 | |
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Answers will vary. Here's one example: Suppose that someone is exactly 6 ft. tall, with arms 3 ft. long. The scale factor will be 9 in.:6 ft. in order to leave 1 in. of border at the top and bottom. To simplify this, remember that 6 ft. = 72 in. The scale factor can then be expressed as 9:72 or 1:8. This person's arms would be 1/8 as long in the scale drawing; 1/8 of 3 ft. (36 in.) is 4.5 in.
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Problem B4 | |
a. | The farthest planet from the Sun, Pluto, has an average distance from the Sun of about 5.9 billion km. To fit on school grounds (say, within 100 m), this would require a scale factor of 100 m:5,900,000,000 km. Since 1 km = 1,000 m, the scale factor can be expressed as 100 m:5,900,000,000,000 m, or 1 m:59 billion m:
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b. | This scale would be hopelessly large for visualizing the difference in diameter sizes among planets, since the largest diameter (Jupiter's) is only about 143,000 km. A scale of 1 m:59 billion m would make Jupiter's diameter roughly 2.4 mm, extremely small. A better scale might be 1 m:590 million m, which would make Jupiter's diameter roughly 24 cm. The smallest planet, Pluto, would have a diameter of 3.8 mm on this scale, which is still small but certainly visible. |
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<< back to Problem B4 non-interactive version
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Problem B5 | |
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Most likely, they chose miles. The longest distance from the Sun, 5.9 • 109, becomes 1.6 km. This is just under one mile, so the scale was likely chosen to let the entire model fit within one mile. Using this scale, the smallest piece of data (Pluto's diameter) becomes 0.62 mm, which is very small but still visible.
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Problem B6 | |
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Answers may vary due to measurement. Here, answers are given to the nearest tenth of a centimeter:
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Side Lengths (S) in cm |
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Hypotenuse Length (H) in cm |
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Ratio S:S |
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Ratio H:S |
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1 |
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1.4 |
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1:1 |
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1.4:1 |
2 |
2.8 |
2:2 |
2.8:2 |
3 |
4.2 |
3:3 |
4.2:3 |
4 |
5.7 |
4:4 |
5.7:4 |
5 |
7.1 |
5:5 |
7.1:5 |
6 |
8.5 |
6:6 |
8.5:6 |
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<< back to Problem B6
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Problem B7 | |
a. | Not surprisingly, the ratio between the sides is constant at 1:1, since we worked exclusively with isosceles right triangles. The ratios between the hypotenuse and a side also seem to be about the same (this is evident if you divide the H:S ratios and write them as decimals). So there may be a constant ratio involved there as well. All the observations are between 1.4 and 1.425, so the constant ratio (if there is one) may be between these values. |
b. | We could multiply the length of the side by 1.41 (the average ratio) to get an approximate answer. We could also use the Pythagorean theorem (covered in the next section) to derive the measure of the hypotenuse. |
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Problem B8 | |
a. | Here is the completed table:
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Side Lengths (S) in cm |
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Hypotenuse Length (H) in cm |
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Ratio S:S |
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Pythagorean Ratio H:S |
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1 |
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1:1 |
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:1 |
2 |
2 |
2:2 |
2:2 |
3 |
3 |
3:3 |
3:3 |
4 |
4 |
4:4 |
4:4 |
5 |
5 |
5:5 |
5:5 |
6 |
6 |
6:6 |
6:6 |
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b. | The measures using the Pythagorean Theorem are more accurate. They show the constant ratio of :1 in all six cases. |
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Problem B9 | |
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An irrational number is a number that cannot be written as an exact ratio of two integers; is an example. In terms of measurement, the important implication (and one that the Greeks missed for centuries) is that there is no possible exact unit conversion between a whole number and an irrational number. If one measure is rational and another is irrational, they are incommensurate; that is, we can never say "A of these make B of these," where A and B are integers.
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