This square is circumscribed [sic]. Find the circle's area to the nearest whole number.
(A side of the square is given as length 4.)
Pappas makes an error here in stating the question. The square is inscribed, not circumscribed. At any rate, our students can solve this problem using the formulas of Chapter 8.
The side of the square is 4, and so its hypotenuse -- the diameter of the circle -- is 4sqrt(2). In the modern Third Edition, students learn about 45-45-90 right triangles in Chapter 8. Even though Second Edition students won't reach Special Right Triangles until Chapter 14, they can still use the Pythagorean Theorem to find the hypotenuse.
Then the radius of the circle is 2sqrt(2), and so its area is pi(2sqrt(2))^2 = 8pi square units. If we use an approximation such as 3.14 for pi, then the area becomes 25.12 square units. To the nearest whole number, this is 25 square units -- and of course, today's date is the 25th.
By the way, notice that 25/8 is the ancient Babylonian approximation to pi. Hmmm, perhaps August 25th should be declared Babylonian Pi Approximation Day? Hmm -- never mind! (By the way, I sneaked today's Pappas problem in on the review side of today's worksheet.)
Lesson 9-6 of the U of Chicago text is called "Views of Solids and Surfaces." In the modern Third Edition of the text, views of solids and surfaces appear in Lesson 9-5. (Recall that Lesson 9-4 of the new edition is "Drawing in Perspective," which is Lesson 1-5 of the old edition.)
This lesson is perfect for an architect. We've discussed architecture in many previous posts, most recently in my September 18th post as we read about buildings in Pappas.
Only two words are defined in this lesson -- views and elevations:
"Underneath the picture of the house are views of the front and right sides. These views are called elevations."
There's not much more for me to say about this lesson, except to add that the modern Third Edition includes some examples using isometric graph paper. This reminds me a little of the (mis)adventures my middle school students had with isometric graph paper last year.
And so we return to Euclid. Of course, none of his definitions or propositions have anything to do with views of solids and surfaces, so we just look at the next proposition in order:
This is the three-dimensional analog of yet another theorem, previously studied in Lesson 3-4:
Transitivity of Parallelism Theorem:
In a plane, if l | | m and m | | n, then l | | n.
Some texts omit "in a plane," (after all, as Euclid is about to prove, it holds in three dimensions as well) but only prove it for a plane. The U of Chicago text is more honest and admits that it is only proving it for the plane case. Indeed, the text gives an informal argument in terms of slope -- and of course, we only define slope for lines in a (coordinate) plane.
Euclid's proof is interesting in that unlike those of the previous propositions, he does not use plane Transitivity of Parallelism to prove the spatial case.
Given:
Prove:
Statements Reasons
1. bla, bla, bla 1. Given
2. Choose G on
H on
K on
3.
(call it plane P)
4.
5.
But in this case, Euclid's original proof is simple enough to show to high school students as it is:
Let each of the straight lines AB and CD be parallel to EF, but not in the same plane with it.
I say that AB is parallel to CD.
Now, since EF is at right angles to each of the straight lines GH and GK, therefore EF is also at right angles to the plane through GH and GK.
And EF is parallel to AB, therefore AB is also at right angles to the plane through HG and GK.
For the same reason CD is also at right angles to the plane through HG and GK. Therefore each of the straight lines AB and CD is at right angles to the plane through HG and GK.
But if two straight lines are at right angles to the same plane, then the straight lines are parallel. Therefore AB is parallel to CD.
Therefore, straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other.
Q.E.D.
Perhaps the only change we might make is replace "plane through HG and GK" with "plane P" in order to avoid repeating that cumbersome phrase.
David Joyce mentions another proof at the above link -- midpoint quadrilaterals, which are also known as Varignon quadrilaterals. The U of Chicago text only mentions midpoint quadrilaterals in an Exploration exercise (likewise in the new Third Edition, the only difference being that the new text actually mentions Varignon's name). But it's easy to prove that they are parallelograms -- just divide the quadrilateral into two triangles and apply the Midsegment Theorem of Lesson 11-5 to each.
But now Joyce points out that the theorem is true even for "space quadrilaterals." Our definition of polygon (and hence quadrilateral) states that the vertices must be coplanar. But even if we relax this requirement and allow for space quadrilaterals, Varignon's Theorem still holds. After all, even space quadrilaterals can be divided into two triangles (and each triangle lies in a plane), and so we can still apply the Midsegment Theorem to each one. The difference is that today's Proposition 9 is used to prove that opposite midsegments are parallel -- each is parallel to the the diagonal of the quadrilateral but the three lines aren't coplanar.
Notice that the final parallelogram always lies in a single plane -- this follows from Tuesday's Proposition 7 that two parallel lines and a transversal are always coplanar.
Meanwhile, I've been continuing to think about a proof of Proposition 4 that is based on rotations, since this might be simpler than Euclid's proof that requires seven pairs of congruent triangles. Let's recall what we're supposed to prove -- given that line l is perpendicular to both lines m and n, prove that l must be perpendicular to the entire plane containing m and n.
Euclid begins by defining points A, B on m, C, D on n, and E, F on l. (Actually, E is where all three lines intersect.) These points have the additional property that AE = BE = CE = DE. My goal was to demonstrate that a rotation of 180 degrees about axis EF maps A and B to each other, as well as C and D to each other. Of course, E and F are fixed points of this rotation.
Let's first find A', the rotation image of A. We first notice that rotations, like all isometries, preserve angle measure, and so Angle A'EF = AEF, which is known to be 90. Also, since rotations preserve distance, AE = A'E. Finally, since the magnitude of the rotation is 180, Angle AEA' = 180. The only point satisfying all these requirements is B, so A' = B. Similarly, we have B' = A, C' = D, and D' = C.
Of course, this requires a more rigorous definition of rotation about an axis. For plane rotations about a center O, we expect A' to be a point such that AO = A'O and Angle AOA' equals the magnitude. But for rotations about an axis, we can't use the same definition unless we know where O is along the entire axis. Otherwise we won't know where to measure AO, A'O, or Angle AOA'.
The correct answer is that O is chosen along the axis such that AO is perpendicular to the axis. Since AE is perpendicular to the axis EF, E is the correct point to choose. So that's why AE = A'E and Angle AEA' must be 180. If AE weren't perpendicular to the axis, choose O to be a different point instead.
Once this is complete, then we choose line o with G and H the points of intersection. It then follows that H = G' since rotations preserve collinearity, and the proof is complete.
Or is it? I'm wondering whether I slipped and accidentally assumed what we're trying to prove. For example, can we be sure that the rotation maps o to itself unless we already know that o is perpendicular to l? Also, if we define a rotation as the composite of two reflections in intersecting planes (similar to the definition for plane rotations), then we might not be able to prove that rotations work the way I said they do unless we've already assumed that Proposition 4 is true! (Why, for example, do we choose O such that AO is perpendicular to the axis?)
Making the proof work requires much more thought than I'm willing to take now, considering that we're not actually trying to teach Proposition 4 in a classroom.
Here is the worksheet for today: Students can fill out the front of the worksheet with examples of both a 3D figure (such as a pyramid) and a block-building (similar to #2 on the back) with the respective front, right, and top views of each.
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