MHB Can the Maximum Sum of Diagonals in a Rhombus Be Proven to Be 14?

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The discussion focuses on proving that the maximum sum of the diagonals AC and BD in a rhombus with a side length of 5 is 14, given the constraints that diagonal BD is at least 6 and diagonal AC is at most 6. By setting BD as x and AC as y, the equation x² + y² = 100 is established. Through algebraic manipulation, it is shown that the maximum value of the expression AC + BD can be achieved when specific conditions for a and b are met. Ultimately, the proof confirms that the maximum sum of the diagonals is indeed 14. The conclusion reinforces the validity of the maximum diagonal sum in the specified rhombus configuration.
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A rhombus with side length 5

if diagonal BD $\geq 6$

and diagonal AC $\leq 6$

Prove : max (AC+BD)=14
 
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Re: Prove max(AC+BD)=14

Albert said:
A rhombus with side length 5

if diagonal BD $\geq 6$

and diagonal AC $\leq 6$

Prove : max (AC+BD)=14

let $BD=x=6+a,AC=y=6-b$
for $x^2+y^2=100$
$we\,\, have\,\, (a>0 ,b\leq0$)
$\therefore (6+a)^2+(6-b)^2=100$
$(a-b)^2+12(a-b)+2ab-28=0$
$a-b=-6\pm\sqrt{64-2ab}$
$\therefore max(a-b)=2 \,\, (here \,\, b=0,a=2)$
$and\,\,we\,\,get :max(y+x)=max(AC+BD)=(6-0)+(6+2)=14$
 
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