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

[attachment=0:1gme2sdz]Rhombus.jpg[/attachment:1gme2sdz]
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October 23rd, 2013, 07:30 AM   #2
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Re: rhombus

a=5cm, f?6cm e?6cm

Using law of cosine and given inequalities on diagonals

e=a+a-2acos? ? 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 ~7344' )
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October 23rd, 2013, 12:08 PM   #3
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By Pythagoras, BD = 100 - AC.
(AC + BD) = AC + 2ACBD + 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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October 24th, 2013, 02:28 AM   #4
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Re:

Quote:
Originally Posted by skipjack
By Pythagoras, BD = 100 - AC.
(AC + BD) = AC + 2ACBD + 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.
you got it
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