I finally found Fawn Nguyen’s current blog! SUH-WEET. I feel like I’ve struck MTBoS gold…*again*. I’ve decided to read it from her first post forward.

This is not going to be about what an freakishly amazing and intelligent teacher she is, or about how reading her blog makes me waver between feeling incredibly inspired and feeling incredibly inferior. Nope, none of that.

On 11/6/14, she shared this problem that she had given her students to wrestle with. The problem intrigued me, so I thought I’d try it; I even averted my eyes from the photos of student work so that I would not be influenced. (Want to do the problem, too? Spoiler alert! More photos ahead! Avert eyes!) In the few minutes I had before leaving for an appointment, I made a sketch and threw down some variables for the missing lengths.

I noticed that the two right triangles were similar and the requested area was a trapezoid. I wondered if it would be possible to express the variables in terms of one of them; I can easily express “a” as “10 – b”. Could I do the same for “x”?

I looked at the similar triangles first.

Nah. Still two variables after I simplify. What about working with the triangle with the known area?

Also a bust. My time was almost up, anyway, so I quickly fiddled with the area formula for the trapezoid, which does not use x, but does still have a second variable, A.

*As I copy my work neatly for this post, I am VERY AWARE that by this point*** I already had everything I needed***. Between my time interruption and my laser-like focus on and blind attachment to my original strategy, I missed that tidbit*.

When I again had a few minutes the next day, I started over with a fresh sketch, this time keeping all three variables as I worked. I revisited the similar triangles and the known area to record everything I knew to be true. I felt these were key. Using the proportions and the ax = 48 fact, I guessed that 6 and 8 were reasonable options to try for “a” since the options for a + b = 10 were limited. Confident, I tested them out….

…and arrived at a lovely WTF moment. Maybe I need to let go of my assumption that the measurements are whole numbers? Apparently 5(9.6) = 48, but that’s not helpful in my proportion.

I looked at my proportion set up again, this time circling both cross products, and…AHA!

(Finally, right?) BOTH products = 48. Duh. So b has to be 3. (Remember, I missed this previously… If 16b = 10x – bx and 48 = 10x – bx, then 16b = 48. So what. Now I have two ways so solve this problem!) This is probably the first time I have ever put cross products to good use. (I refrain from introducing this memorized-process-to-get-the-right-answer to students because I rather they’d focus on what’s-the-relationship? instead.)

Figured out what x was (just for kicks..and to validate my proportion) and subbed in 3 for b in my trapezoid area equation to find the solution. Ta DA!

To celebrate, I watched the video Fawn linked at the end of her post, in which Mike Lawler also uses similarity to find the solution, but via a different path, which is cool to see. Check out also Mike’s student’s area solution, as well as a completely different area solution by Mike that employs a system. Holy Multiple Pathways, Batman!

This task is such a poster child for rich problems. I shared it with my entrepreneurial daughter. She doodled around a bit, dusting off had some long-idle knowledge, determined the exact question she needed to ask me, solved it, cheered, and then demanded another one. EXACTLY what I want for all my students.