Geometry Demystified, 2nd edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 11
1. In free space, light rays travel at approximately 300,000 kilometers per second. In 1 minute (60 seconds), a light beam will go 300,000 x 60, or 18,000,000, kilometers. That's the distance equivalent of 1 minute. The correct choice is D.
2. Let's convert the speed of light in free space to a meters-per-second figure. To do that, we multiply 300,000 kilometers per second by 1000 to get 300,000,000 meters per second. In order to find out how long it takes a ray of light to travel 150 meters in free space, we must divide 150 by 300,000,000, which gives us 0.0000005. That's 500 nanoseconds (500 billionths of a second, when we use the American terminology and say that a billionth equals a thousandth of a millionth). The answer is A.
3. In the "real world," no object can follow a path that strays outside of a light cone such as the one shown in Fig. 11-12. If an object did stray outside the light cone, it would have to travel faster than the speed of light in free space. According to Albert Einstein's special theory of relativity, that can't happen. (Some experimental physicists claim to have observed subatomic particles that travel faster than the speed of light in free space, but the observations remain controversial as of January 2012 when I wrote this file.) In Fig. 11-12, paths Q and R remain inside the light cone, but paths P and S stray outside it. Therefore, according to the theory of special relativity, only paths Q and R can exist in the "real world." The correct choice is B.
4. When we observe a regular polygon on the surface of a sphere, we'll always find that the interior angles have greater measures than they do for a regular polygon with the same number of sides on a flat plane. A regular hexagon on a flat plane has interior angles that each measure 120º. Therefore, on a sphere, a regular hexagon must have interior angles that measure more than 120º. (The extent of the difference between the plane angles and the spherical angles depends on the size of the hexagon relative to the sphere. As the hexagon grows in size on the surface of the sphere, the interior-angle measures increase.)The correct choice is C.
5. Suppose that the circle, as shown in Fig. 11-13, lay on a flat plane instead of a sphere. In order to find the circumference, we would multiply 2π times the radius of 200 units, getting 400π units. But the circle lies on a sphere, and we measure the radius over the sphere's surface, not from the circle's true center in Euclidean three-space (a point that lies inside the sphere). The true radius is therefore smaller than the radius as defined over the sphere's surface. for that reason, we can conclude that the circle's circumference is less than 400π units. The answer is B.

6. To calculate the distance (let's call it d)between the origin and a point in Cartesian four-space, we square each one of the four individual coordinates, then add the squares, and finally take the square root of the result. When we carry out that arithmetic routine for the point (1,1,1,1), we get

d = (12 + 12 + 12 + 12)1/2
= (1 + 1 + 1 + 1)1/2
= 41/2
= 2

The correct choice is C.

7. Let's start by converting the dimensions of the rectangular prism from meters to kilometers, as follows:

  • It measures 60 meters high, equal to 0.060 kilometers
  • It measures 120 meters wide, equal to 0.120 kilometers
  • It measures 300 meters deep, equal to 0.300 kilometers

The prism's volume (in three dimensions) equals 0.060 x 0.120 x 0.300 = 0.00216 cubic kilometers. We know that the object "lasts" for 0.01 second. That's the equivalent of 0.01 x 300,000 = 3000 kilometers, because light travels 300,000 kilometers per second in free space. Now we can calculate the hypervolume of the prism, over the course of its "lifetime," as 0.00216 x 3000 = 6.48 quartic kilometer-equivalents. The answer is A.

8. We can convert the dimensions of the rectangular prism from meters to microsecond-equivalents, noting that light travels 300 meters in one microsecond. We proceed as follows:

  • It measures 60 meters high, equivalent to 60/300 or 0.20 microseconds
  • It measures 120 meters wide, equivalent to 120/300 or 0.40 microseconds
  • It measures 300 meters deep, equivalent to 300/300 or 1.00 microseconds

Again, we're told that the object "lasts" for 0.01 second, which equals 10,000 microseconds. Now we can calculate the hypervolume as 0.20 x 0.40 x 1.00 x 10,000 = 800 quartic microsecond-equivalents. The correct choice is D.
9. We can count the number of edges in Fig. 11-14 to determine the number of line-segment edges that a tesseract has. That task might prove difficult and confusing if we look at the figure and count by "brute force," so let's break the process down. An ordinary three-dimensional cube has 12 edges, and Fig. 11-14 has two "concentric cubes" for a total of 24 edges in both cubes. Each cube has eight vertices, so it takes eight more line segments to connect the vertices of the "inner cube" with the vertices of the "outer cube" in the rendition of Fig. 11-14. The total number of line-segment edges in the tesseract therefore equals 24 + 8, or 32. The correct choice is C.
10. In order to calculate the hypervolume of a tesseract, we must determine the fourth power of the length of any single edge. Conversely, if we want to calculate the length of any single edge, we must take the fourth root of the hypervolume. We're told that the tesseract in Fig. 11-14 has a hypervolume of 4096 quartic units. The fourth root of 4096 equals 8. Therefore, each line-segment edge in this tesseract measures 8 units long. The correct choice is D.