October 23rd, 2013, 07:45 AM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  rhombus
[attachment=0:1gme2sdz]Rhombus.jpg[/attachment:1gme2sdz]

October 23rd, 2013, 08:30 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 e²=a²+a²2a²cos? ? 36 2a²(1cos?) ? 36 1cos? ? 18/25 7/25?cos? ( we get 7/25 ? cos? from other diagonal) e+f=?(e+f)²=?(e²+f²+2ef)=?(4a²+2ef)=?(4a²+4ah)=2?( a²+ah)= 2?(a²+a² sin?)=2a?(1+sin?) (e²+f²=4a², 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, 01:08 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
By Pythagoras, BD² = 100  AC². (AC + BD)² = AC² + 2AC·BD + BD² = 2?(AC²(100  AC²)) + 100 = 2?(50²  (50  AC²)²) + 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, 03:28 AM  #4  
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: Quote:
 

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