# Mar 2 ACT & SAT Question of the Day

http://sat.collegeboard.org/practice/sat-question-of-the-day?src=R&questionId=20130302 (This link takes you to today’s question. If you use my archive, you will see the question related to my SAT explanation for that date.)

After reading their explanation, I have to think the SAT folks have been reading my blog and have responded to my criticisms.  Their explanation this morning is formatted differently than previous ones and it gives a nice “math teacher” explanation.  No more cryptic explanation that is harder to understand than the question.  Good for them.  I’m looking forward to seeing whether this is the beginning of a new process for them or a one time event!

The answer is B and you can read their explanation to see how to to do the math.  Let’s see what you can learn from this question about the test so that I can raise your score.  First, you need to realize that the SAT (and ACT) folks are infatuated with right triangles.  They love to bury them in situations that require you to realize that there’s a right triangle without telling you that.  You will need to calculate an area, perimeter, or the length of a side.  This morning’s question is a simple example.

In addition, if you see a cone or pyramid question, you are very likely looking at a right triangle question in three dimensions!  Take a look at DVDs #7 and #8 or my course and look at the practice questions.

Second, there’s another thing they like to ask about the areas of triangles and it has to do with the concept of “area.”  Let’s use this question to understand it.  Look at the diagram and you’ll see the right triangle with sides of 3 and 5.  When you calculate the area, you get 7.5; that’s easy enough.  Now, here’s something you need to understand: if the right angle at (0,0) is changed (either increased or decreased) the area of the triangle is reduced if the lengths of the sides stay the same (3 and 5).  For example, imagine what would happen if the y-coordinate side of the triangle (3) were “pivoted” to the right (clockwise) until the angle at (0,0) is 45 degrees.  The height would be smaller.  (If you can’t imagine this, I want you to take a pencil or some other straight item and place one end at (0,0) then turn it clockwise and you’ll see how the top of the triangle gets lower.)  That reduces the height and, therefore, the area is smaller.   The principle is that when two sides of a triangle are at right angles to one another, that is the largest triangle (area) that you can get with those two lengths.

Third, you need to understand that if you take the original triangle and turn the height clockwise any amount (for example, 30 degrees) and you turn it counter-clockwise the same amount (30 degrees), the resulting triangles each have the same area.  That’s because the heights will be the same.  You should draw some pictures to see how this works.

Fourth, you should notice that the farther you lower the height, the less area you have.  This pattern will continue until the angle at (0,0) is zero and the triangle disappears because there is no longer a height.

Fifth, whenever you see a right triangle, be sure to use the two sides to calculate the area.  Let’s say you have a right triangle and its “base” is the hypotenuse.  That is, the hypotenuse is horizontal and the two sides are above it.  (Sorry.  I can’t draw but I hope you see what I mean.)  That means the highest point in the triangle is a right angle somewhere above the base/hypotenuse.  All too often, I see students using the hypotenuse for the base and trying to calculate a height from it up to the right angle.  Whoa!  Slam on those brakes!!  Don’t do that.  Use the two sides adjoining the right angle for the base and height to do the calculation.  You don’t care how a right triangle is tilted.  Since the two sides are at right angles, they form a “natural” base and height.

Finally, this concept also applies to circles but we’ll wait until we get an SAT or ACT question related to it to discuss it.  My students learned about this last week or tomorrow depending on the location.  I’ll give “Wizard Bonus Points” to the first student who raises this issue in each class this week.

Gosh, I hope the SAT folks keep explaining the math so I don’t have to bother.  I would much rather be blogging about how to think about math and teaching you the thinking skills that will be required on test day.  That will raise your score much more than learning how to do individual questions.

The College Board announced this week that there are dramatic changes coming in the SAT.  They didn’t give a timeline but the process usually takes 2-3 years.  My current 8th grade students will certainly be affected by these changes.  Probably not current sophomores and older.  Stay tuned.

http://www.act.org/qotd/ (The ACT staff does not put a date on their questions so if you click on an archived blog, you’ll get today’s question and the old explanation. Sorry. The SAT staff has dated their questions; so, the archive is helpful. The ACT folks simply don’t do that.)-

I’m sure they asked this question a few weeks ago.  As I said yesterday, “The ACT folks need to get some new material.” There’s not much incentive to do their questions when they keep repeating the same ones.

The answer is J.  Ms. Murphy, my ninth grade English teacher, loved to say, “shorter is better.”  If you can send the same message with fewer words, do it.  The ACT explanation for this question gives you the details why this is true; so, I don’t need to bother.  Here’s my strategy: pick the shortest answer that paraphrases the words that are underlined to avoid “wordiness” problems.

If the grammar problem is not wordiness, the shortest answer may not be correct.  For example, if the problem is non-parallel structures, you have to make them parallel and not worry about the length of the answers.  DVD #9 deals with these issues and many others related to the SAT and ACT grammar/composition tests.

Enjoy your weekend.  I’m glad you took a few minutes to read my blog.  It should be time well spent.

The Wizard