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 11-14-2012, 07:34 PM #1 New Member   Join Date: Nov 2012 Posts: 8 Gruber's Math Workbook Help Does anyone have Gruber's Math Workbook that can help me understand the problem on page 268. ((1/2)(AB)h)/((1/2)(AB)h)=1/3 Why is the second AB not AD since its the base of the whole triangle? And where did the 1/3 come from? Reply
 11-14-2012, 07:36 PM #2 New Member   Join Date: Nov 2012 Posts: 8 By the way, I understand why it's 1/3 because the lines trisect the whole triangle in three equal parts which mean the area of one part is a third of the whole, but I want to understand it the way they did it in case I encounter more complex problems. Reply
 11-14-2012, 08:06 PM #3 Member   Join Date: Jun 2012 Posts: 340 The second.. what? Reply
 11-14-2012, 08:13 PM #4 New Member   Join Date: Nov 2012 Posts: 8 I mean't the bottom of the fraction. Reply
 11-14-2012, 08:21 PM #5 Member   Join Date: Jun 2012 Posts: 340 Oh no you're right; it should be AD. Otherwise the answer would be 1. Which would be wrong. x/x=1 Grubers has tons of mistakes. Reply
 11-14-2012, 08:25 PM #6 New Member   Join Date: Nov 2012 Posts: 8 Thanks! I thought I was missing something, but apparently not. This would've saved me much more time if I knew that earlier. Reply
 11-15-2012, 04:04 PM #7 New Member   Join Date: Nov 2012 Posts: 8 Wait, I don't think it was a mistake because he talks about how the AB cancels out. But, I just don't understand where the equation came from. Anyone else have Gruber's Math Workbook that can help? Reply

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