It's the same school as Tuesday, so we know that today is even blocks. Second period is Spanish III, fourth period is Spanish II Honors (for freshmen, as non-honors ninth graders would be in Spanish I), and sixth period is sports -- the female regular teacher is a coach of both the boys basketball and boys baseball teams! As basketball is a winter sport and baseball a spring sport, the former ends just as the latter's about to start, unless the hoops team makes the high school playoffs -- which it did. Indeed, the second-round playoff game is tonight.
Since this is a Spanish class, the song reverts to Square One TV's "Sign of the Times." I also include "Mathematics of Love" for Valentine's Day, except I sing Spanish (instead of Roman) numerals.
Today is Tenday on the Eleven Calendar:
Resolution #10: We are not truly done until we have achieved excellence.
The Spanish III students have several tasks to complete today (including a short vocab quiz), but I don't give the top students the holiday pencils and candy until they finish the fifth task (a worksheet on the subjunctive mood). The Spanish II kids have only two tasks, but the second is a project to create on Chromebooks, and so I give the reward after the first task. Unfortunately, it's difficult to tell whether the students have achieved excellence on the computer project. (Of course, the basketball team needs to follow this resolution in tonight's game.)
Today on her Daily Epsilon of Math 2020, Rebecca Rapoport writes:
What is the area of LOVES?
[Here is the given info from the diagram: LO = 6.5, OV = 5, VE = sqrt(29), ES = 1.5, SL = 6, and also EOV and OES are right angles.]
Hey, that's a Valentine's Day problem for you -- find the area of pentagon LOVES. Well, we notice that Triangle OVE is a right triangle with base 5 and hypotenuse sqrt(29). Thus we can use the Pythagorean Theorem to find the height:
a^2 + b^2 = c^2
a^2 + 5^2 = sqrt(29)^2
a^2 + 25 = 29
a^2 = 4
a = 2
So the legs of right Triangle OVE are 5 and 2, allowing us to find the area:
Area(OVE) = (1/2)(2)(5) = 5
We can draw in
Area(EOS) = (1/2)(2)(1.5) = 1.5
Let's use the Pythagorean Theorem to find the hypotenuse:
1.5^2 + 2^2 = OS^2
2.25 + 4 = OS^2
6.25 = OS^2
OS = 2.5
Why did we find OS, anyway? Well, it's because the last part of the area we need is Triangle LOS, and now we know all three of its sides -- 2.5, 6, and 6.5. We could use Heron's Formula to find the area -- or if we're lucky, it's a right triangle, so we can just take half the product of the legs again. So how can we tell whether it's a right triangle? That's correct -- we use Converse Pythagorean Theorem:
2.5^2 + 6^2 = 6.5^2
6.25 + 36 = 42.25
42.25 = 42.25
Thus it's indeed a right triangle. So it's easy to find the area:
Area(LOS) = (1/2)(6)(2.5) = 3(2.5) = 7.5
We finally add up the sum of all three right triangles:
Area(LOVES) = 7.5 + 1.5 + 5 = 9 + 5 = 14
Therefore the desired area is 14 square units -- and of course, today's date is the fourteenth of February, Valentine's Day.
We used the Pythagorean Theorem thrice today (technically, the forward theorem twice and the converse once). And we'll use the theorem even more, since it's Distance Formula day today.
This is what I wrote last year about today's lesson:
Lesson 11-2 of the U of Chicago text is called "The Distance Formula." In the modern Third Edition of the text, the Distance Formula appears in Lesson 11-5.
Let's get to today's lesson. Many students have trouble with graphing throughout Chapter 11, and furthermore, today we learn the Distance Formula, which of course will be difficult for some students.
In the past, I combined Lesson 11-2 with Lesson 8-7, on the Pythagorean Theorem (and indeed, this lesson in the Third Edition is titled "The Pythagorean Distance Formula").
David Joyce has more to say about the Distance Formula:
Also in chapter 1 there is an introduction to plane coordinate geometry. Unfortunately, there is no connection made with plane synthetic geometry. Here in chapter 1, a distance formula is asserted with neither logical nor intuitive justification. Of course, the justification is the Pythagorean theorem, and that's not discussed until chapter 5. In that chapter there is an exercise to prove the distance formula from the Pythagorean theorem. The Pythagorean theorem itself gets proved in yet a later chapter.
Fortunately, the U of Chicago text avoids this problem. Our text makes it clear that the Distance Formula is derived from the Pythagorean Theorem.
Today I post an old worksheet from a few years ago. It introduces the Distance Formula -- but of course, it teaches (or reviews) the Pythagorean Theorem as well -- including its similarity proof, which is mentioned in the Common Core Standards.
Today is also an activity day, so let's add an activity to this old worksheet. The one Exploration Question given in this lesson is:
22. The distance from point X to (2, 8) is 17.
a. Show that X could be (10, 23).
b. Name five other possible locations of point X. (Hint: Draw a picture.)
Of course, there are infinitely many possible locations for X. The assumption is that students will look for lattice points (points whose coordinates are both integers). These are based on the 8-15-17 Pythagorean triple. Even though we haven't covered vectors yet, it might be helpful for teachers to think in terms of vectors when correcting the student work. Basically, the solutions are to add all versions of the vector <8, 15> to the original point (2, 8). This includes changing one of both signs of <8, 15> as well as switching the x- and y-components. There are seven possible lattice points in addition to the given (10, 23), and students only need to find five of them.
President's Day is on Monday, and so my next post will be Tuesday.
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