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October 23rd, 2013, 06:45 AM   #1
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rhombus

[attachment=0:1gme2sdz]Rhombus.jpg[/attachment:1gme2sdz]
Attached Images Rhombus.jpg (16.7 KB, 138 views) October 23rd, 2013, 07:30 AM #2 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: rhombus a=5cm, f?6cm e?6cm Using law of cosine and given inequalities on diagonals e�=a�+a�-2a�cos? ? 36 2a�(1-cos?) ? 36 1-cos? ? 18/25 7/25?cos? ( we get -7/25 ? cos? from other diagonal) e+f=?(e+f)�=?(e�+f�+2ef)=?(4a�+2ef)=?(4a�+4ah)=2?( a�+ah)= 2?(a�+a� sin?)=2a?(1+sin?) (e�+f�=4a�, 2ef=4ah , h=a sin? ) Maximum occurs when sin ? is max , that is when cos? is min, and that is for cos?=7/25 then sin?=24/25. e+f=2a?(1+sin?) e+f=14 (e=6, f=8, ?=arcsin 0.96 ~73�44' ) October 23rd, 2013, 12:08 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,635 Thanks: 2080 By Pythagoras, BD� = 100 - AC�. (AC + BD)� = AC� + 2AC�BD + BD� = 2?(AC�(100 - AC�)) + 100 = 2?(50� - (50 - AC�)�) + 100, which is maximized when AC has its maximum value of 6. For AC = 6, BD = 8 and so AC + BD = 14. October 24th, 2013, 02:28 AM   #4
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Quote:
 Originally Posted by skipjack By Pythagoras, BD� = 100 - AC�. (AC + BD)� = AC� + 2AC�BD + BD� = 2?(AC�(100 - AC�)) + 100 = 2?(50� - (50 - AC�)�) + 100, which is maximized when AC has its maximum value of 6. For AC = 6, BD = 8 and so AC + BD = 14.
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