12/8/09

Test 4 Section 6 - #8 (page 595)


I see variables, so I'm thinking PLUG-IN (yes, even in a triangle problem)!  Plug-in works really well in triangles, provided you follow 2 very important rules:
  1. Don't break a triangle!  A triangle's angles must add up to 180°.
  2. Don't break a straight line!  A straight line ALSO must be 180°.
So we can't plug-in willy-nilly, but with practice we can get very good and very quick at using plug-in, even in geometry questions.  I'm choosing to make a = 80, and b = 60.  Let's leave c alone for now.

It's super important from here to be careful filling in the rest of the angles.  (This is why I haven't plugged-in for c yet.)  I'll use a black pen to fill in what I can from here, based on the numbers I chose.  I've gotta make sure my triangles that I've already given two angles to add up to 180°, and that the straight lines on the left and bottom of the big triangle also end up being 180°.  Here goes:


Now it's clear what c has to be: c = 20.  Let's fill that in (in red again -- I ran out of colored pens), and fill in the rest of the angles.


Ok.  Do a quick check and make sure all the triangles (including the big one, four in all) add up to 180°, and make sure all the lines do as well.  We good?  Good.  That was the hard part, but with practice you'll find that you can do this very quickly.  And as always, the beauty of plug-in is that once you've done the work, you can be super sure of your answer.

From here, all we do is check the answer choices with our numbers plugged in.  Remember, we said a = 80 and b = 60.  We're looking for the answer that gives us c = 20.

(A) 80 + 3(60) - 180 = 80
(B) 2(80) + 2(60) - 180 = 100
(C) 180 - 80 - 60 = 40
(D) 360 - 80 - 60 = 220
(E) 360 - 2(80) - 3(60) = 20

Bingo.

There are a few other ways to do this problem, too.  Read on if you're interested.

Notice that the labeled angles actually outline a quadrilateral.  The angles in a quadrilateral always add up to 360°, just like a triangle's angles always total 180°.  So c is going to be what's left over when we start at 360° and take away everything else.

c = 360 - 2a - 3b.

You might also have noticed that there's only one answer choice that actually acknowledges the fact that there are 3 b's and 2 a's in the diagram.  Sometimes the SAT will do that, and it's often a giveaway.  On the hardest multiple choice question in the section I don't know if I'd feel comfortable just picking (E) and moving on based on that, but there you go.

2 comments:

  1. Thanks mike. This problem really confused me, and the explanation from the college board website didn't help much either. You did a great job here. Thanks again!

    ReplyDelete