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)?

Comments

- Example of "Grid-In" or student-constructed response question on the SAT
- This question seems more difficult than it really is. Students often give up on questions near the end of a section. DON'T!!
- Hopefully you will view the utility of questions like this as I do:

Students become more competent problem solvers only when challenged with nonroutine problems which are not always to be found in the textbooks. Questions like these should become more routine in our texts and in our classes (not only honors!).

Answer: 12

Solution (no explanation, details omitted):

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

Discussion Points

- What are some problem-solving strategies we need to review with our students here? Draw a diagram for sure but what are some other general attack strategies students should employ in triangle area problems?
- Although advanced theorems could be used here, the actual solution given above is efficient and fairly basic. But what insights are needed to use that approach? What geometry or algebra standards are being tested here?
- I chose this problem because coordinate geometry problems connect many important ideas in geometry and algebra. Not to mention that they are becoming more common on standardized tests like SATs, ACTs and state assessments. Besides, I enjoyed writing the question! Sometimes I'll get the germ of an idea, re-work it many times and then the question takes on a life of its own.
- If you find an error in my work or want to share your thoughts, please add a comment!

## 9 comments:

My kids usually scare me when I ask them to do Geometry problems, but I had a good alg II kid come in after school and I popped this one on him... He looks, never uncrossed his arms, and said "12 square units".. Elapsed time, 30 seconds......

HOW?? so he explains... let the base be the two points on y=x, translate them down to the origin so that they are at (0,0) and (4,4) and then you can do 1/2 (6)4 = 12....

I decided NOT to show him how I had done it with stuff involving square roots of 2...

Why not Pat? There's nothing wrong with saying that you went another direction. We all approach problems differently depending on what frame of mind we're in at the time. We find a solution and then move on.

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.

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.

On another note, I could see this as an SAT question, but without a diagram it would be rated as hard and placed late in the section.

With a diagram, it would be in the medium section, middle.

Very interesting.

I wonder whether any student would blow the diagram up by a factor of 2 and use Pick's Theorem.

Enclosing the triangle in a rectangle is also useful:

www.mathopenref.com/coordtriangleareabox.htmlThen just calculate the area of the rectangle and subtract the areas of the triangular pieces.

I've been holding off reacting to the insightful comments made thus far to this seemingly routine triangle area problem. I gave this problem to a small group of students (fairly strong sophs, juniors) yesterday and all but 2 struggled with it. Those who got it used Maria's method. When asked where they had seen that approach before, one replied, "Somewhere" and the other said "It just came to me." This should be a standard method shown to all geometry students IMO.

Pat, a few students considered the expression 4√2 but got stuck from that point. The method your student used (translation to origin I believe) is fascinating but I would like more clarification of his thinking. He is replacing the original triangle by a triangle with base 6 and height 4, or vice versa or what? Curmudgeon seemed to get this transformation idea, but it doesn't seem transparent to me.

I used a straightforward subtraction of 2 areas as you can see from the computation. Interestingly, the computation leads to (1/2)(6)(4) which suggests a single triangle with base 6 and height 4, but...

Eric, I love Pick's Theorem as you know from my previous posts on that topic. All students should be exposed to it with some indication for why it works at least for right triangles or triangles in general, but I don't believe too many students, even your top "mathletes" would consider it. I went through the details of it just to confirm hte result. That's always fun!

Curmudgeon, I deeply appreciate your thoughts and suggestions. This would be a fairly challenging SAT problem even with the diagram, IMO, I drew the diagram for the students and they still struggled, many going off in various distance formula or Pythagorean directions. No one seemed to recognize that I chose two of the vertices on the line y = x for a reason. They may have noticed the coordinates were the same but they didn't make the connections. Not unusual in my experience. That's the main reason I wrote this question. I didn't pick the vertices randomly!!

"He is replacing the original triangle by a triangle with base 6 and height 4, or vice versa or what? Curmudgeon seemed to get this transformation idea, but it doesn't seem transparent to me."

This is why it's so amazing to me that that kid DID get it. This idea requires a visual flexibility that isn't common in students. If he truly KNEW that the two triangles were identical instead of just guessing correctly, this kid will go far.

I put a few diagrams here:

http://mathcurmudgeon.blogspot.com/2009/05/in-discussing-following-problem-over-on.html

Dave,

He actually held his two hands up spreading thumb and finger in the air to show the base on the line y=x and the pointer of his left hand to show the vertex of the altitude as he viewed it at (6,0), then he translated the base along the line y=x so the area remained unchanged,(thumb and finger slide down and to his left) until one end was at the origin.. then cleverly (or at least I thought it was a stoke of brillance) reversed his choice of base to the line from the origin to (6,0) and made the point at (4,4) the apex of the altitude... I was, as the Brits say, Gobsmacked...

Thanks, Pat...

Curmudgeon,

First of all, you are dead on in your comments about the importance of encouraging different solution paths and then discussing more efficient methods.

The diagrams you posted on your site are awesome. Talk about " one picture is worth..."

As soon as I saw your diagram of what the student did it was perfectly clear to my decrepit and spatially challenged brain. Over the years I 've had a few students who were capable of that kind of visualization, but Pat's student may be in a class by himself if in fact he translated and then replaced the triangle with another of equal area by switching the base and height.

The solution actually reminds me of the famous "median divides a triangle into 2 equal areas" argument. Nice!

I can recall a student I had in geometry honors back in the 90's. We finished discussing the "what is the maximum number of pennies can you place around a central penny so that each of the additional pennies are tangent to the central one and to two others." This proved to be fairly straightforward leading to a nice discussion of hexagons and then I posed the 3-dimensional version: How many basketballs of equal size can I place around a central basketball of the same size? I knew this was a famous sphere-packing problem although I wasn't sure of the answer. I asked the students to make educated guess without a model. I recall guesses like 6,9 12, 15, 16 or 18. One young man named Ken (if you're reading this, how are you!) felt pretty certain it would be 12 and I asked him how he could "prove it?" He said he would build a model and bring it in for us. The next day he brought in a carefully engineered collection of 12 tennis balls surrounding a central ball! How did he connect these? He punctured small holes in each and ran thread attaching the balls in a tightly knit fashion. It was so well done he bounced the arrangement on the table without loosening the tight packing of the balls. The class could easily see all 12 surrounding ball and gave him a standing ovation. Why am I bringing this up here? Because Ken visualized there would be 12 almost immediately. When I asked him what he saw in his mind's eye, he said he could see two groups of six but the rest of his explanation was beyond my "ken!" Oh, how I envy that kind of spatial reasoning. I'm just a mere algebra-symbolic person, and clearly I'm not afraid to risk exposing my lack of aptitude!

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