## Making them think

My kids are very answer oriented. I can tell them to focus on the process, but as any decent teacher knows, telling isn’t learning.

This is probably old news to some of you out there, but I’ve finally found a way to get my Geometry kids to focus on the process more than the answer:

Make them solve the problem two different ways.

Even if they only figure out one way, they’re spending time thinking about alternative approaches.

## { 5 } Comments

Is it 14.48?

Nice little problem, I have just finished teaching trig & pythagoras with a class of mine and will set them this to warm them up on Monday morning.

I actually found 3 ways to solve this (so far).

The trig kids may jump straight to similar triangles with the pythagorean theorem, which is not actually the easiest way to solve this.

I’m thinking that he fact that it can be solved using only similar triangles without the pythagorean theorem leads to a potential proof once you generalize the problem.

I’m pretty sure my kids’ algebra skills aren’t strong enough for that, though.

Geometry kids wouldn’t do it, but after a little right triangle trig, you could find the area with .5absinC, and then use the area to find the altitude.

Okay, I wouldn’t use trig at all – isn’t that funny how different people solve it differently! I would use the area of the triangle and Pythagorean Theorem.

A = 0.5(21×20) = 210

Use Pythagorean Theorem to find the hypotense: 29

And the area formula again..

210 = 0.5(29)(x)

That’s one approach.

The other is to use similar triangles.

You still find the hypotenuse=29 using the pythagorean formula, and then use x/21 = 20/29.

Or, (and this leads to the proof I alluded to earlier), you label the left & right parts of the hypotenuse as

a&brespectively, and you get x/a = b/x = 20/21, and x/21 = 20/(a+b).I’m not expecting the kids to be able to use that last one. I did, however, expect them to keep looking after they’d already found the answer one way.

As far as I can tell, Kate’s Trig method reduces to the area solution, since the only given angles (including the included angle of the two given sides) are right angles, and the sine of a right angle is 1.