[Risolto] Problemi rombo

45*5/3=75=D     A=75*45/2=1687,5cm2

d2 = 45 cm

d1 = 45*5/3 = 75 cm 

area A = 45*75/2 = 1.687,50 cm^2

d1 = 80 m

d2 = 80*3/4 = 60 m

area A = 60*40 = 2400 m^2

bonus :

lato L = 10√4^2+3^2 = 10*5 = 50 m 

d1 = 24 cm

d2 = 24*3/4 = 18 cm

area A = 12*18 = 216 cm^2

perimetro 2p = 160 cm

lato L = 2p/4 = 160/4 = 40 cm

area A = L*h = 40*38,4 = 1.536 cm^2

lato L = A/h = 1.176/33,6 = 35,0 cm

perimetro 2p = 4L = 35*4 = 140 cm 

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$\small\text{Diagonale maggiore: \(D= d÷\dfrac{3}{5} = \cancel{45}^{15}×\dfrac{5}{\cancel3_1} = 15×5 = 75\,cm;\)}$

$\small\text{area del rombo: \(A= \dfrac{D×d}{2} = \dfrac{75×45}{2} = \dfrac{3375}{2} = 1687,5\,cm^2.\)}$ 

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$\small\text{Diagonale minore: \(d= D÷\dfrac{4}{3} = \cancel{80}^{20}×\dfrac{3}{\cancel4_1} = 20×3 = 60\,m;\)}$

$\small\text{area del rombo: \(A= \dfrac{D×d}{2} = \dfrac{80×\cancel{60}^{30}}{\cancel2_1} = 80×30 = 2400\,m^2.\)}$ 

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$\small\text{Diagonale incognita: \(= 24÷\dfrac{3}{4} = \cancel{24}^8×\dfrac{4}{\cancel3_1} = 8×4 = 32\,cm;\)}$

$\small\text{area del rombo: \(A= \dfrac{D×d}{2} = \dfrac{32×\cancel{24}^{12}}{\cancel2_1} = 32×12 = 384\,cm^2.\)}$

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$\small\text{Perimetro del rombo: \(2p= 160\,cm;\)}$

$\small\text{altezza: \(h= 38,4\,cm;\)}$

$\small\text{lato: \(l= \dfrac{2p}{4} = \dfrac{160}{4} = 40\,cm;\)}$

$\small\text{area: \(A= l×h = 40×38,4 = 1536\,cm^2.\)}$

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$\small\text{Area del rombo: \(A= 1176\,cm^2;\)}$

$\small\text{altezza: \(h= 33,6\,cm;\)}$

$\small\text{lato: \(l= \dfrac{A}{l} = \dfrac{1176}{33,6} = 35\,cm;\)}$

$\small\text{perimetro: \(2p= 4×l = 4×35  = 140\,cm.\)}$