{"id":259,"date":"2014-06-10T12:00:13","date_gmt":"2014-06-10T19:00:13","guid":{"rendered":"http:\/\/mostlymental.com\/?p=259"},"modified":"2014-06-10T00:04:39","modified_gmt":"2014-06-10T07:04:39","slug":"pythagoras-week-day-3-2","status":"publish","type":"post","link":"http:\/\/mostlymental.com\/?p=259","title":{"rendered":"Pythagoras Week: Day 3"},"content":{"rendered":"

Hello everyone!<\/p>\n

Welcome back to Pythagoras Week.\u00a0 If you missed it, be sure to check it out from the beginning<\/a>.\u00a0 Today's proof uses an interesting \"sliding\" property of parallelograms.\u00a0 Essentially, we can take a parallelogram and slide opposite sides parallel to each other without changing the area, as below.\u00a0 It works because the area for a parallelogram is the base times the height, and we aren't changing either by sliding them.\"parallelogram\"<\/a><\/p>\n

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Now that we have that fact, what can we do with it?\u00a0 Well, let's start with our right triangle, as below.\u00a0 We can place squares on either of the legs, as shown.\"pyth_proof_3_1\"<\/a>Using the sliding property, we can slide the two squares until they meet, as shown below.\"pyth_proof_3_2\"<\/a>Then we can just slide the two parallelograms down until they fill the square at the bottom.\"pyth_proof_3_3\"<\/a>The original squares had area \"\"<\/span>