This lesson is very similar to Lesson 9-2 in that the focus is on vocabulary. The difference is that today's lesson is not an activity, but a traditional worksheet.
Indeed, students are asked to calculate slant height in two of the problems on this worksheet. As we already know, this requires the Pythagorean Theorem. A right triangle can be formed with the slant height as the hypotenuse and the altitude height as one leg. The other leg is unnamed in the U of Chicago text, but notice that it's actually the apothem of the regular polygon base.
In fact, we discussed this in previous years -- many Geometry texts define "apothem," but not the U of Chicago text. Indeed, Lesson 8-6 explains the trapezoid area formula as follows:
"There is no know general formula for the area of a polygon even if you know all the lengths of its sides and the measures of its angles. But if a polygon can be split into triangles with altitudes or sides of the same length, then there can be a formula. One kind of polygon that can be split in this way is the trapezoid."
Well, another polygon that can be split in this way is the regular polygon. Indeed, all the altitudes and sides are of the same length, because the triangles are congruent. The common altitude of these triangles is the apothem of the regular polygon. (It has come to my attention that apothems do appear in the modern Third Edition of the text. This is in the new Lesson 8-7 on Special Right Triangles, since 30-60-90 and 45-45-90 triangles are used to find the apothems of equilateral triangles, squares, and regular hexagons.)
But let's get back on track with pyramids and cones. Again, we look at Euclid's definitions:
Notice that Euclid's definition of "pyramid" isn't that much different from the U of Chicago's. But Euclid's cones, like his cylinders, are solids of revolution. Therefore they are limited to right cones and right cylinders. The U of Chicago generalizes this with a definition of cone not unlike the definition of pyramid. This allows cones to be oblique as well as right, and pyramids and cones are examples of conic surfaces (the surfaces of conic solids).
Oh, and I also notice one more difference between the old Second and new Third Editions. The point at the top of the pyramid is called the vertex in the Second Edition and the apex in the Third. Euclid, meanwhile, never names the top point of the pyramid.
Let's look at the next proposition for us to prove:
This is basically the Two Perpendiculars Theorem of Lesson 3-5, except that the two given lines are perpendicular to the same plane, not merely the same line. The line version of Two Perpendiculars appears as the last step of the proof.
Let's modernize Euclid's proof:
Given: Line AB perp. plane P, line CD perp. plane P (B, D in plane P)
Prove: AB | | CD
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. Draw DE in plane P such that 2. Point-Line-Plane, part b (Ruler/Protractor Postulates)
DE perp BD, DE = AB
3. AB perp. BD, AB perp. BE, 3. Definition of line perpendicular to plane
CD perp. BD, CD perp. DE
4. BD = BD 4. Reflexive Property of Congruence
5. Triangle ABD = Triangle EDB 5. SAS Congruence Theorem [steps 2,3,4]
6. AD = BE, Angle ABD = EDB 6. CPCTC
7. AE = AE 7, Reflexive Property of Congruence
8. Triangle ABE = Triangle EDA 8. SSS Congruence Theorem [steps 2,6,7]
9. Angle ABE = Angle EDA 9. CPCTC
10. DE perp. DB, DE perp. DA 10. Definition of perpendicular lines [steps 6,9]
11. Lines BD, DA, DC coplanar 11. Proposition 5 from Friday (DE perp. to all, steps 2,10)
12. Lines BD, AB, DC coplanar 12. Point-Line-Plane, part e (points A, B)
13. AB | | CD 13. Two Perpendiculars Theorem [step 2]
Some steps look strange here, such as Step 12. Here, Euclid uses the "proposition" (actually a postulate) that all triangles lie in a plane, so the plane containing two sides of Triangle ABD (namely BD and DA) must contain the third (AB). Here, we use our Point-Line-Plane Postulate part e. This is because Step 11 establishes that a single plane (not plane P, by the way -- we can call it plane Q if we want) contains the lines BD, DA, and DC -- that is, points A and B lie in plane Q. Thus by part e, the entire line AB must also lie in plane Q.
The reason the proof is so convoluted goes back to the original Two Perpendiculars Theorem:
If two coplanar lines l and m are each perpendicular to the same line, then they are parallel to each other. [emphasis mine]
Notice that key word coplanar. Most of the time we use the Two Perpendiculars Theorem, we take it for granted that the lines in question are coplanar. But in this proof, the bulk of the proof is just to establish that AB and CD are coplanar! After all, the two perpendicular statements (AB perp. BD, CD perp. BD) were established all the way back in Step 2, so we could skip directly from Step 2 to Step 13 if we didn't have to prove that the lines are coplanar.
On the other hand, it is truly necessary to prove that all three lines (the two lines and the transversal) are perpendicular. Otherwise, we could prove, for example, that DB | | DA following Step 10, which is absurd. But since three lines (AB, CD, and the transversal BD) were proved coplanar in Step 12 (they all lie in plane Q), we complete the proof that AB | | CD.
David Joyce points out that Euclid omitted a step here. It could be the case that the two lines meet the plane at the same point (i.e., B and D are the same point). But Euclid proves that this is impossible in a subsequent proposition.
Here are the worksheets:
No comments:
Post a Comment