Opposite and Adjacent

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(C)Copyright 2018, C. Burke.

I have this conversation every term. Sometimes multiple times.




Come back often for more funny math and geeky comics.

Daily Regents: Right Triangle Altitude Theorem (June 2015)

Even though the title says "Daily", I won't be doing them daily any more, until the exams get closer. But they will be frequent.

Geometry, June 2015, Question 34

34. In the diagram below, the line of sight from the park ranger station, P, to the lifeguard chair, L, on the beach of a lake is perpendicular to the path joining the campground, C, and the first aid station, F. The campground is 0.25 mile from the lifeguard chair. The straight paths from both the campground and first aid station to the park ranger station are perpendicular.

If the path from the park ranger station to the campground is 0.55 mile, determine and state, to the nearest hundredth of a mile, the distance between the park ranger station and the lifeguard chair.

Gerald believes the distance from the first aid station to the campground is at least 1.5 miles. Is Gerald correct? Justify your answer.

First part: PLC is a right triangle. This means that you can find the missing side using the Pythagorean Theorem.
So (PL)² + (LC)² = (LC)²
x² + .25² = .55²
x² + .0625 = .3025
x² = .24
x = 0.48989794855 = 0.49 to the nearest hundredth mile.

The second part of the question wants to know, essentially, what is the length of FC. We already know the length of LC, so we need to find the length of FL. We can do this using the Right Triangle Altitude Theorem.
(PL)² = (FL) * (LC) Remember, we found that x² = .24. Do NOT square 0.49
.24 = FL * .25
.24 / .25 = FL
FL = .96
FC = FL + LC = .96 + .25 = 1.21
Gerald is incorrect. The distance is shorter than 1.5 miles.

Alternate Solution

All three right triangles in the diagram are similar triangles, so their corresponding sides are proportional.
Therefore, the proportion (base1) / (hypontenuse1) = (base2) / (hypontenuse2) must be a true statement.

Fill in the information we know: .25 / .55 = .55 / FC
Cross-multiply: .25 FC = (.55)²
.25 FC = .3025
FC = .3025 / .25 = 1.21, which is less than 1.5

Daily Regents: Right Triangle Altitude Theorem (January 2016)

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams next week. At least, that is the plan.

I had a question about this problem on the thread where I posted all the multiple-choice answers: http://mrburkemath.blogspot.com/2016/02/january-2016-new-york-geometry-common.html

Common Core Geometry, January 2016, Question 22

30. 2 In the diagram below, CD is the altitude drawn to the hypotenuse AB of right triangle ABC.
Which lengths would not produce an altitude that measures 6√2?

The Right Triangle Altitude Theorem tells us that (AD)(DB) = (CD)²
The square of the altitude is (6√2)(6√2) = (36)(2) = 72.
So which choice has lengths of AD and DB that have a product of something other than 72?

Careful! Two of the options give you AB instead of DB. Subtract AB - AD to get DB.

Look at the four choices:

1. - 2 * 36 = 72. Not the answer
2. - 3 * (24 - 3) = 3 * 21 = 63, not 72. This is the correct answer.
3. - 6 * 12 = 72. Not the answer
4. - 8 * (17 - 8) = 8 * 9 = 72. Not the answer.

The correct answer is choice (2).

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

(blog): 360, 180, 90, 2 and 1/2: I'm Talking Arcs and Inscribed Angles

Assisting today in a Geometry class. It was Day 2 of Arcs of a Circle. Yesterday, the teacher covered central angles with the class, which meant that today's lesson moved to the inscribed angles. She carefully and diligently explained what makes an inscribed angle, and how the line segments intercept an arc in a similar way that they saw in the previous lesson. And then we got into the relationship between the arcs and the two types of angles.

The measure of an arc of a circle is equal to the size of the central angle which intercepts it. The measure of the inscribed angle is half the size of the arc it intercepts. Pause. What does this make the relationship between the central and the inscribed angle. Pause. Wait. Rephrase? Response? Good -- but try again in a full sentence...

"It's half the size." sounds good. No, really, it does -- it means someone's paying attention and either getting it, or somewhat getting it. Following up: "What is half of what?" The inscribed angle is half of the central angle.

Okay. So if the inscribed angle is 60 degrees, how big is the central angle? Let them think about it. Did they come up with 120 degrees? Or 30 degrees? If the smaller angle is half the bigger angle, then the bigger angle is ... ? (Okay, it's a leading question, and I hate leading questions, but sometimes you do need to just pull that one number out of them so you can move on.)

They moved along with the notes and did a couple of practice problems before moving on to the next step. What if two inscribed angles intercepted the same arc? What could we conclude about the two angles? The teacher waited to see if they could reach the statement before she gave it. It took a moment to realize that A was half the arc and B was half the same arc, so angles A and B had to be congruent even if we didn't know how big the arc was. We didn't need to know. But if we did, we could work things out.

Then things started getting complicated because when you start putting in two many line segments and too many inscribed angles, triangles start forming. Wait! What are we supposed to do with those?! Treat them like three inscribed angles, of course, but don't forgot those properties of triangles, either. Particularly, the one about the sum of the angles!

So if we had a problem that looked like this:

... we have enough information to fill in both angles BAC and BCA as well as arcs AC and BC. We just might not know that we know yet. Not unless we remember some other facts about circles and triangles. The total measure of the central angles in a circle is 360 degrees, so the total of the arcs of the circle is also 360 degrees. The sum of the angles of a triangle is 180 degrees. Notice that if each angle of the triangle is inscribed that make each part of the circle twice the size of inscribed angle -- and 360 degrees is twice as big as 180!

Excellent discovery, if they can make it on there own. One student was hovering about it while he was talking. If he made the connection, he didn't share it with me, but he was close to it.

Finally, why is 90 so important that I included it in the title?

Because many of these problems use diameters as one of the line segments. A diameter cuts the circle in half, into two semicircles, each 180 degrees. The central angle formed by the two radii joining into a diameter is a straight angle, measuring 180 degrees. Any inscribed triangle using the diameter as one of its sides would, by necessity, have an angle that measures half of 180 degrees, which is 90 degrees.

Wait a minute!

So any inscribed triangle using the diameter of a circle is a right triangle? And any inscribed right triangle has to include the diameter?

It's almost as if someone planned it that way. Maybe not, but that's how we planned the lesson.

Perimeter, Right Triangles and Radicals

In my Algebra class today, we were reviewing the process for adding and subtracting radical numbers. Previously, we simplified irrational numbers, such as the square of 80, which becomes 4 times the square root of 5.

To give them a more thoughtful question than just what is the sum of SQRT(75) + SQRT(48), where the only "thought" is to get past thinking that it's SQRT(123), I decided to pull out Right Triangles and that Old Favorite, the Pythagorean Theorem. Now, I didn't want them to just simplify the irrational number, I wanted some kind of addition in the problem. That brings us to Perimeter of a Right Triangle.

There are basically two types of problems you can offer up for consideration: problems with one radical number, and problems with two radical numbers. Three is just being mean, and overly complicates things -- on the other hand, it could make it interesting. Give the students the length of the legs (the base and the height), establish that there is a right angle between them, and let them go to work.

In the first kind of problem, you can pick any two numbers and you'll most likely have an irrational hypotenuse. Okay, but boring. It's more interesting if you make one of the legs irrational in such a way to make the hypotenuse rational. Surprise them.

It keeps it interesting when you consider these two problems, which look very similar, but are very different.

Consider the square root of 13 triangle first. If we square 6, we get 36. If we take the square of the square root of 13, we get 13. 36 + 13 = 49, which is the square of the hypotenuse. Therefore, the hypotenuse is 7.

Finding the perimeter is as simple as adding the three sides, which in this case means combining the like terms, which would be the integers 6 and 7. The perimeter is 13 + root 13.

In the square root 12 problem, the numbers had to be carefully selected. In this case, there will be two irrational numbers. If anything is to be combined, then the radicals have to simplify to the same radicand. Otherwise, the exercise is pointless.

If we square 6, we get 36 again. If we take the square of the square root of 12, we get 12, of course. 36 + 12 = 48, which is the square of the hypotenuse. Therefore, the hypotenuse is root 48.

We can't add any of the numbers as they are written, but we can simplify both of the radicals, as shown:

Both of the radicals have the square root of 3 in simplest form, and their coefficients can be added. The perimeter is 6 + 6 square root 3.

Interestingly, in both examples I wrote for lessons today, a double number appeared in the answer. This is actually something else to be careful about. Some students might see that as a co-incidence. Others might see it as a pattern and come to expect it.

There is an easy solution to that: give them some more problems to work on!

Happy Fourth of July 2014

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(C)Copyright 2014, C. Burke.

As far as I know, there aren't rules for displaying a flag at 45 degrees. It probably shouldn't. You probably will never have a radical-two length flag, either.

June 2014 Geometry Regents, Parts 2

If you find this page helpful, please Like or Share or Leave a Comment. Thank you!

I really can't do Parts 3 and 4 until I take the time to create the necessary illustrations. Sorry. So here's Part 2. If you missed Part 1, it's here.

Part 2

29. The coordinates of the endpoints of BC are B(5, 1) and C(-3, -2). Under the transformation R₉₀, the image of BC is B'C'. State the coordinates of points B' and C'.

Remember, if they don't state which direction you are rotating, the default is counterclockwise. That might seem counter-intuitive, but you are moving from quadrant I to II, from II to III, etc.

Under R90, a point P(x, y) becomes P'(-y, x). So these points will become B'(-1, 5) and C'(2, -3). Interestingly enough, according to the key, you didn't need to say which was which, but you did need to include the parentheses.

30. As shown in the diagram below [COMING SOON], AS is a diagonal of trapezoid STAR, RA || ST, m<ATS = 48, m<RSA = 47, and m<ARS = 68.
Determine and state the longest side of triangle SAT.

You needed to calculate that m<AST = 65 degrees. You can get this using triangle RAS or knowing that the same-side interior angles are supplementary.
Once you know that, you can use 65 + 48 + x = 180 to find m<SAT, which is 67.

Side ST is the longest side because it is opposite the largest angle. You needed to give the side AND show some kind of work that indicates that you didn't go eeny-meeny... Seriously.
ST without any work -- even just angle sizes after you did the subtraction in your head or on your calculator -- is worth 0 points.

31. In right triangle ABC shown below [I'LL GET TO IT!], altitude BD is drawn to hypotenuse AC. If AD = 8 and DC = 10, determine and state the length of AB.

There's a short way and a long way to do this. Both are okay for full credit.

First, remember that there are three right triangles in this diagram. Both of the smaller ones are similar (and therefore proportional) to the larger one and to each other.

You could write a proportional comparing ( base / hypotenuse) = ( base / hypotenuse) of triangle BAD and triangle ABC.
You would have ( 8 / x ) = ( x / 18 ). (It's 18 because it is 10+8.)
Cross-multiply: x² = 144.
x = 12, which is the length of AB.

The longer method: If you forgot about that proportionality, but remembered the Altitude Rule, you would have found that altitude (BD)² = (AD)(DC), so y² = (8)(10) or 80. So the altitude is the square root of 80. Okay, that's good for partial credit, but we aren't done.

Now you can use Pythagorean Theorem: 8² + 80 = x². Remember: (the square root of 80) squared is just 80.
64 + 80 = x²
144 = x²
x = 12, again.

Either method is valid and will get full credit. I saw quite a few of each type. The only problem with the longer method was that some students started estimating an answer in the middle and that introduced an error into the rest of the calculations.

32. Two prisms with equal altitudes have equal volumes. The base of one prism is a square with a side length of 5 inches. The base of the second prism is a rectangle with a side length of 10 inches. Determine and state, in inches, the measure of the width of the rectangle.

I graded a lot of these. Two things: First, I can't believe how many Geometry students can do so well and still not know the Area of a rectangle or the Volume of a rectangular prism.
Second: This is a good case of Less is More. What do I mean by that? A lot of students talked or worked themselves out of a 1 or 2 points. Seriously.

Despite the fact that the word prism is used repeatedly, students draw pyramids, which by themselves, might not have been work and could've been overlooked. But then they used formulas for Volume that included 1/3 in them. That's a conceptual error. What makes this worse is that the (1/3)s cancel out and don't affect the outcome -- just the final score. I had to call over an arbitrator on this one because I didn't want to make that call.

Another error that was made repeatedly was that students assumed that the first prism was a cube even though there is nothing supporting this. They substituted 5 for the height! There is nothing about 5 being the height. This led to a discussion about whether that could be a case of "guess and check" and not an actual "conceptual error". Again, the mistake factors out.

But the worst part of it all is that because the heights were the same, the formula for Volume wasn't even needed! You only needed to compare the Area of the two bases.

This became confusing as well. People did perimeter of one and Area of the other. Or they assume that both bases were squares or other crazy stuff.

If was a very simple problem. How simple?

5² = 10x.

That simple: x = 2.5.

Funniest answer: Someone wrote 2.5in = w, with a little bit of cursive and I thought it said "2 Sin w". Okay, it was funny to me.

33. As shown in the diagram below, BO and tangents BA and BC are drawn from external point B to circle O. Radii OA and OC are drawn. If OA = 7 and DB = 18, determine and state the length of AB.

A very similar question was asked in August 2011, but that was multiple choice.

Again, there are TWO methods to solve this: Pythagorean Theorem, and Tangent-Secant.

First, realize that OD is also a radius, with length 7. That means that OB is 25. OA is perpendicular to the tangent line BA. Therefore, OAB is a right triangle with OB as its hypotenuse and OA, a radius, as one of its legs. So 7² + (AB)² = 25².
Solving the equation, we find that AB has a length of 24.

The other way: Draw diameter DOE (make point E on the circle opposite from D). BE is now a secant line with a length of 18 + 7 + 7 = 32.
Tangent-secant rule: (BA)² = (BD)(BE) = 18 X 32 = 576. AB = 24.

Both methods are valid.

34. Triangle RST is similar to triangle XYZ with RS = 3 inches and XY = 2 inches. If the area of triangle RST is 27 inches, determine and state the area of triangle XYZ, in square inches.

Short way: The ratio of the areas of two similar triangles is the square of the ratio of their sides. The scale factor of the sides is 3:2, so the scale factor for the areas is 9:4. If the bigger one has area 27, then ( 9 / 4 ) = ( 27 / x ).
Cross-multiply and 9x = 108, and x = 12.

LONG WAY: If you forgot about that, you could have done the following: if the large triangle has Area 27 and base 3, and A = 1/2 b h, then 27 = (1/2)(3)(h), and h = 18.
Then to find the height of the smaller triangle, ( 3 / 2 ) = ( 18 / h ). Then 3h = 36 and h = 12 inches. But that isn't the answer -- that's just the height.
A = (1/2) b h = (1/2) (2) (12) = 12 sq inches. That's the answer. Even though both numbers are 12, if you didn't complete it, you lost a point.

35.

Almost Right

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(C)Copyright 2011, C. Burke. All rights reserved.


If Bob is a isosceles triangle and his legs have a length of two, then you could call him Bob Twos Triangle.

Purple Mountains

(Click on the cartoon to see the entire image.) (C)Copyright 2009, C. Burke. All rights reserved.


This discovery was a watershed moment in history.

Braves

(Click on the cartoon to see the entire image.) (C)Copyright 2009, C. Burke. All rights reserved.


I wanted to work in Geronimo, but it was getting crowded.

Pythagorean Triples: Introduction

Sometimes I get a bug in my head about something and it won't go away until I write it down and work it out. This time I recorded it in a journal, so I decided to upload it here on a non-comic days. Comments are welcome, especially from my students.

A few years back, I got tired of every right triangle problem involving a 3-4-5, 6-8-10, 30-40-50 or 5-12-13 triangle. I started creating a list of other triples that I could use when avoiding irrational numbers.

First discovery: Triples are either E² + O² = O² or E² + E² = E²
(where E is an even number, and O is an odd number), and the latter form could always be reduced to the first form. For need of a term, let's say that the triples that are reduced are in the simplest form, which means that the numbers are relatively prime.

Second discovery: Many (a, b, c) triples could be reduced to (a, b, b + 1), when a was odd, or (a, b, b + 2), when a was even. (There are some that don't fit either, but those will have to wait.)

Squares can be made by summing consecutive odd numbers. When that addend is a perfect square, the result is a Pythagorean Triple.
9 = 3² ==> 3² + 4² = 5²
25 = 5² ==> 5² + 12² = 13²

Was there a quicker way to find the triples for 7, 9, 11, etc?
I'll get to that.

Also notice that 31 + 33 = 64, which is 8², so 8² + 15² = 17².
And it had been under my nose all the time that 4² + 3² = 5².
The simplest Pythagorean Triple of them all, fits both these models.

(to be continued)

Questions for my students

Extra credit for answering or for participating in the discussion.

1. Can you explain in your own words what I meant by "relatively prime"? What do you know about prime numbers that might give you a hint?
   
   
2. Can you find Pythagorean Triples where the smallest side is 9, 11, and 13?
   Remember: 9-12-15 doesn't count. It can be reduced to 3-4-5.
   
   
3. Why can't there be any Triples of the form: Even² + Even² = Odd²?