![\"post5pic1\"](\"http:\/\/advancedmathyoungstudents.com\/blog\/wp-content\/uploads\/2014\/03\/post5pic1.jpg\")
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Here is the square with length (a + b<\/i>):<\/p>\n
(When we looked at binomial squares, I used x<\/i> and y<\/i>.\u00a0 This time, I\u2019m using a<\/i> and b<\/i>.\u00a0 You\u2019ll see why soon.)<\/p>\n
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Below, I have broken up the square two different ways.\u00a0 On the left, you see the familiar shapes that we used to derive the formula for the square of a binomial.\u00a0 And on the right, you see a square surrounded by four identical right triangles.<\/p>\n
This gives us two different ways to express the same area.\u00a0 On the left, we have:<\/p>\n Area = (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup><\/i><\/p>\n On the right, we have one tilted square in the middle with side c<\/i>, and 4 right triangles, each with a<\/i> and b<\/i> as their base and height.\u00a0 We can add up their areas to find:<\/p>\n Area = (a + b)2<\/sup>\u00a0<\/i>= c2<\/sup> + 4 \u00d7 <\/i>\u00bd \u00d7<\/i> a <\/i>\u00d7 b = c2<\/sup> + 2ab<\/i><\/p>\n But those\u00a0two areas are equal.\u00a0 They are two different ways of chopping up the same square.\u00a0\u00a0That means\u2026<\/p>\n a2<\/sup> + 2ab + b2<\/sup> = c2<\/sup> + 2ab<\/i><\/p>\n And if we subtract 2ab<\/i> from both sides, we get a pretty well known result.<\/p>\n \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a2<\/sup> + b2<\/sup> = c2<\/sup> <\/i><\/strong><\/p>\n where a and b are the legs and c is the hypotenuse \u00a0of a right triangle.<\/i><\/p>\n We use that<\/i> theorem all the time.\u00a0 Again, better with pictures.<\/p>\n All of these posts have been about using pictures to get an improved understanding of what the algebra is telling us.\u00a0 These are ideas that I have been kicking around for a while so I thought I would share.\u00a0 But they are leading up to something I’ll write about in the next post.\u00a0 And I’ll reveal what got me started down this path again this year.<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" In my last post, we saw how taking apart an area in different ways can lead to interesting results.\u00a0 Let\u2019s do that again. Here is the square with length (a + b): (When we looked at binomial squares, I … Continue reading <\/a><\/p>\n