Challenge problem

Posted on May 29, 2014


The area, A, in square inches, of a parallelogram with a height of 4 inches is given by the equation  A = 4L, where L is the length, in inches, of the base of the parallelogram.  The table shows the area, M, in square inches, of a triangle with a height of 4 inches and a base of p inches.

p (inches) M (square inches)
1 2
3 6
36 72
52 104

The area of the triangle is ___________ the area of the parallelogram.

Fill in the blank.

Happy Discussing – I will post the solution next week!


I am *almost* hoping that … I’m missing somethign about this problem and it will all become clear to me. 

However, I cannot find, anywhere in that problem, an indicator of what triangle they’re talking about.   

I did some utterly contrived algebra and came up with a “solution,” with two variables in it — but in my opinion that is *not* “mathematical reasoning” but a perverse exercise in symbol manipulation.  Yet it is, I am told, the kind of questions people are getting in the new GED test.   

Looking back at this, oh, six months later… gee, if a student didn’t bother to read the problem at all they’d prob’ly have a 50/50 chance of getting the right answer since the relationship of the two columns (with completely different labels than what the problem asks for, but — details! details!) is 1:2. And if one of the choices wasn’t twice as much instead of half as much, or they figgered triangles would be smaller than parallelograms, then — home free!

With, of course, the lesson reinforced: don’t read the problem! Then you get one of those math teachers or texts where (like ALEKS and our Algebra course materials) actually, it matters a *ton* to read closely and traps are set … well… oops… but hey, it’s not as if we want students to understand this stuff, is it? We just want to weed out the dross who aren’t worthy of higher degrees.

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