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	<title>Comments on: Making them think</title>
	<atom:link href="http://blog.mathsage.com/?feed=rss2&#038;p=538" rel="self" type="application/rss+xml" />
	<link>http://blog.mathsage.com/?p=538</link>
	<description>Where Mr. K shares about learning about learning.</description>
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		<title>By: Mr. K</title>
		<link>http://blog.mathsage.com/?p=538#comment-447</link>
		<dc:creator>Mr. K</dc:creator>
		<pubDate>Wed, 11 Feb 2009 01:33:16 +0000</pubDate>
		<guid isPermaLink="false">http://blog.mathsage.com/?p=538#comment-447</guid>
		<description>That&#039;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 &amp; right parts of the hypotenuse as __a__ &amp; __b__ respectively, and you get x/a = b/x = 20/21, and x/21 = 20/(a+b).

I&#039;m not expecting the kids to be able to use that last one. I did, however, expect them to keep looking after they&#039;d already found the answer one way.

As far as I can tell, Kate&#039;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.</description>
		<content:encoded><![CDATA[<p>That's one approach.</p>
<p>The other is to use similar triangles.</p>
<p>You still find the hypotenuse=29 using the pythagorean formula, and then use x/21 = 20/29.</p>
<p>Or, (and this leads to the proof I alluded to earlier), you label the left &amp; right parts of the hypotenuse as <i>a</i> &amp; <i>b</i> respectively, and you get x/a = b/x = 20/21, and x/21 = 20/(a+b).</p>
<p>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.</p>
<p>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.</p>
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		<title>By: Druin</title>
		<link>http://blog.mathsage.com/?p=538#comment-446</link>
		<dc:creator>Druin</dc:creator>
		<pubDate>Wed, 11 Feb 2009 00:47:22 +0000</pubDate>
		<guid isPermaLink="false">http://blog.mathsage.com/?p=538#comment-446</guid>
		<description>Okay, I wouldn&#039;t use trig at all - isn&#039;t that funny how different people solve it differently!  I would use the area of the triangle and Pythagorean Theorem.

A = 0.5(21x20) = 210

Use Pythagorean Theorem to find the hypotense:  29

And the area formula again..
210 = 0.5(29)(x)</description>
		<content:encoded><![CDATA[<p>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.</p>
<p>A = 0.5(21&#215;20) = 210</p>
<p>Use Pythagorean Theorem to find the hypotense:  29</p>
<p>And the area formula again..<br />
210 = 0.5(29)(x)</p>
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		<title>By: Kate</title>
		<link>http://blog.mathsage.com/?p=538#comment-436</link>
		<dc:creator>Kate</dc:creator>
		<pubDate>Sun, 08 Feb 2009 15:00:18 +0000</pubDate>
		<guid isPermaLink="false">http://blog.mathsage.com/?p=538#comment-436</guid>
		<description>Geometry kids wouldn&#039;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.</description>
		<content:encoded><![CDATA[<p>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.</p>
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	<item>
		<title>By: Mr. K</title>
		<link>http://blog.mathsage.com/?p=538#comment-435</link>
		<dc:creator>Mr. K</dc:creator>
		<pubDate>Sun, 08 Feb 2009 14:43:05 +0000</pubDate>
		<guid isPermaLink="false">http://blog.mathsage.com/?p=538#comment-435</guid>
		<description>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&#039;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&#039;m pretty sure my kids&#039; algebra skills aren&#039;t strong enough for that, though.</description>
		<content:encoded><![CDATA[<p>I actually found 3 ways to solve this (so far).</p>
<p>The trig kids may jump straight to similar triangles with the pythagorean theorem, which is not actually the easiest way to solve this.</p>
<p>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.</p>
<p>I'm pretty sure my kids' algebra skills aren't strong enough for that, though.</p>
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		<title>By: Dan Stucke</title>
		<link>http://blog.mathsage.com/?p=538#comment-434</link>
		<dc:creator>Dan Stucke</dc:creator>
		<pubDate>Sun, 08 Feb 2009 10:22:42 +0000</pubDate>
		<guid isPermaLink="false">http://blog.mathsage.com/?p=538#comment-434</guid>
		<description>Is it 14.48?

Nice little problem, I have just finished teaching trig &amp; pythagoras with a class of mine and will set them this to warm them up on Monday morning.</description>
		<content:encoded><![CDATA[<p>Is it 14.48?</p>
<p>Nice little problem, I have just finished teaching trig &amp; pythagoras with a class of mine and will set them this to warm them up on Monday morning.</p>
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