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{l = √((d/2)^2 + (Δ/2)^2) = lato rombo

d= diagonale minore  (in dm)

Δ = diagonale maggiore  (in dm)

{l/d = 3·√6/4     

{1/2·d·Δ = 80·√2 (in dm^2)

d = 160·√2/Δ

l = √((160·√2/Δ/2)^2 + (Δ/2)^2)

l/(160·√2/Δ) = 3·√6/4

l = 240·√3/Δ

240·√3/Δ = √((160·√2/Δ/2)^2 + (Δ/2)^2)

240·√3/Δ = √(12800/Δ^2 + Δ^2/4)

Risolvo:

Δ = 20·√2 dm

l = 240·√3/(20·√2)-----> l = 6·√6 dm

perimetro rombo=4·6·√6 = 24·√6 dm

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Visto il rapporto poni il lato e la diagonale minore come segue:

lato $\small l= (3\sqrt6)x;$

diagonale minore $\small d= 4x;$

quindi:

diagonale maggiore utilizzando il teorema di Pitagora:

$\small D= 2×\sqrt{\left(3\sqrt6·x\right)^2-\left(\dfrac{4x}{2}\right)^2}$

$\small D= 2×\sqrt{9·6x^2-(2x)^2}$

$\small D=2×\sqrt{54x^2-4x^2}$

$\small D=2×\sqrt{50x^2}$

$\small D= 2×5\sqrt2x = 10\sqrt2x;$

conoscendo l'area del rombo, calcola:

$\small  \dfrac{D×d}{2} = A$

$\small \dfrac{10\sqrt2x×4x}{2} = 80\sqrt2$

$\small 10\sqrt2x×4x = 160\sqrt2$

$\small 40\cancel{\sqrt2}x^2 = 160\cancel{\sqrt2}$

$\small 40x^2 = 160$

$\small \dfrac{\cancel{40}x^2}{\cancel{40}} = \dfrac{\cancel{160}^4}{\cancel{40}_1}$

$\small x^2 = 4$

$\small \sqrt{x^2} = \sqrt4$

$\small x= 2$

per cui:

lato $\small l= (3\sqrt6)x = 3\sqrt6×2 = 6\sqrt6\,dm;$

perimetro $\small 4×l = 4×6\sqrt6 = 24\sqrt6\,dm.$

L/d = 3√6 /4

4L = 3d√6 

d = 4L/(3√6)

d/2 = 2L/(3√6)

D^2/4 = L^2-(d/2)^2 = L^2- 4L^2/54 

D^2 = 4L^2(1- 4/54) 

D = 2L√(1- 4/54) = 2L√50/54

2A = d*D 

160√2 = (8L^2/3)√50/(54*6)

60√2 = L^2√50/(54*6)

3600*2 = L^4*50/324

144*324 = L^4

L^2 = 12√(9*6^2) = 12*18 = 4*3*9*2 dm^2

L = 6√6 dm 

perimetro 2p = 4*6√6 = 24√6 dm