Find the Area of a Triangle

In discussing the following problem over on Math Notations, there's been an interesting mix of methodology.

In the coordinate plane, what is the area of ΔPQR given the coordinates P(4.5,4.5), Q(8.5,8.5), and R(6,0)?

The first level of difficulty lies, of course, in getting the image correct. Leaving this up to the kids makes for a tougher problem. Here's the diagram.

So the question fairly begs for a solution and it was interesting to me how many different ideas came up. Before you scroll down and see what the others did, try this problem yourself.

Dee-dee-dee-dee, dee-dee-dee, dee-dee-dee-dee-DUM-da-dee-dee
(Badly transcribed Jeopardy theme tune)

Here we go!

All of them are valid but operate under different frames of reference - your starting point, I think, depends a lot on what you're working on in math or the most recent similar problem you've solved.

Here's probably the simplest version. Big triangle minus the yellow one:

(1/2)(6)(8.5) - (1/2)(6)(4.5) = 12

I did the problem mentally and didn't twig on the two triangles. Instead I found a base and an altitude. 1/2* 4(root2) * 3(root2) like so:

The most interesting one was done by a student:
Translate the points P and Q down the line y=x. The orange and red triangles are still equal areas because they have the same height and equal bases.

Now find the area of the orange triange with a height of 4 and a base of 6.

Ain't that cool?

To repeat my comment on MathNotations:
We all approach problems differently depending on what frame of mind we're in at the time.

A basic math student would probably use the 2triangles method. An algebra student who just got through pythagorean theorem, distance formula and simplifying radicals might gravitate to the other. The algebra student who's been translating stuff might think of that - though the idea that sliding the two points down y=x doesn't change the area is NOT something that many students can do in their heads! Someone else might go to the trouble of flipping out Heron's formula or 1/2ab*sinC if the points were set up differently. Still others use Pick's formula since they can see the points in the diagram.

If your student weren't able to immediately follow your using PQ as a base and finding the altitude to it using perpendicular slopes (the problem, it seemed to me, was set up to push the solver in that direction) then he isn't really comfortable with algebra.

Don't knock your initial instincts until they get in the way of solutions. Students do like to see patterns that flow through many courses and like to see that old ideas are springboards to new solutions. We tell them all the time to "use critical thinking" and "Choose the best method," so we should show them different methods when the opportunity arises.

Hindsight is 20-20 in mathematics, too. For the record, my first instinct was 1/2 PQ * altitude. Had P and Q not been on y=x, I might have changed course but the solution was easy enough.

Here's some variations on this theme which may push the solver in different directions depending on what he's just been working on:

P(4.5,4.5), Q(8.5,8.5), R(5,1)
or
P(4.5,5.5), Q(8.5,9.5), R(6,1)
or
P(1.5,6), Q(4.5,4.5), R(6.5,8.5)

Just a few thoughts on a Saturday morning early Sunday afternoon.