## Saturday, February 2, 2013

### The Super Bowl and the ratio 16 to 9

NOTE: The following classroom scenario requires a reimagining of how we normally present a math lesson and I already have heard all the negatives: not enough time for this, too much material to cover, this won't be on the test so why bother...
I hope you will be open-minded.

The 16 to 9 ratio of course refers to the aspect ratio of  high definition LCD or LED TV screens today.

Let's say you just purchased a 55 inch LED HDTV. We all know the 55 inches refers to the diagonal of the screen and, in fact it's slightly less than 55 inches. So what would the width of the screen actually be?

How does our method of presentation and the questions posed affect concept development?

We know how to solve this, no nonsense. Just apply the Pythagorean Theorem with some algebra and voila. "Tradition" as the song title goes from Fiddler on the Roof.

A Non-Traditional Classroom Scenario

Let's try some estimates, boys and girls...
54" 52" 50" 48"?? Hmm, most 'guesstimated' 50?

Can anyone guess why there are sheets of paper, scissors and rulers on the table? Right, we will first "construct" a solution! Oh, so a 16" width is too big for standard 8.5×11 paper. Any ideas? Oh, it's a ratio so we don't have to use 16" and 9". Ok, we'll use 8"×4.5". OK, go to it...

Each member of your team should measure the diagonal to the nearest 1/8". Oh, that's right we could have used cm instead to make measurements more precise...

Alright, so most of you got around 9 1/8" for the diagonal.
(Aside: This is definitely an imaginary scenario!).
Guess, metric measurement would have been better.  Let's verify this using the Pythagorean Thm. Ok, 9.18" to nearest hundredth. So how will we apply this to a 55" diagonal? Oh, make a proportion, and we obtain 55×8/9.18 ≈ 47.9"!

In your groups, solve the ratio problem algebraically and compare results...

So a 55" screen is less than 48" in width. Wonder why they use diagonal measurements in the ads...

Yes, Alex. You found another way to estimate this mentally?

The closest Pythagorean triple to 16 and 9 is 15 and 8. The hypotenuse would be 17 and 17×3 is 51, sorta' close to 55. So we can triple the dimensions to get 16×3=48" for the width.  Hmm...

What non-traditional approaches for this kind of problem have my colleagues used? Share!

This imaginary lesson would consume the entire period, yes? Do you think it's worth it?

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