potreste aiutarmiii,grazjeee

l1=V 38,88*3/4=5,4   l2=5,4*4/3=7,2    d=V 7,2^2+5,4^2=9

2p=2*(7,2+5,4)=25,2    l rombo=25,2/4=6,3

=========================================================

$\small\text{Rettangolo:}$

$\small\text{dimensione minore: } l_1= \sqrt{A÷\dfrac{3}{4}} = \sqrt{38,88×\dfrac{4}{3}}=\sqrt{\dfrac{1296}{25}} = \dfrac{36}{5} = 7,2\,cm;$

$\small\text{dimensione maggiore: } l_2= \dfrac{A}{l_1} = \dfrac{38,88}{7,2} = 5,4\,$

$\small\text{perimetro: } 2p= 2(l_1+l_2) = 2(7,2+5,4) = 2×12,6 = 25,2\,cm;$

a) $\small\text{diagonale: } d= \sqrt{(l_1)^2+(l_2)^2} = \sqrt{7,2^2+5,4^2} = 9\,cm\quad\text{(teorema di Pitagora)}.$

$\small\text{Rombo isoperimetrico:}$

$\small\text{perimetro: } 2p= 25,2\,cm;$

b) $\small\text{lato: } l= \dfrac{2p}{4} = \dfrac{25,2}{4} = 6,3\,cm.$

h = 3b/4 

area A = 38,88 = b*3b/4 = 3b^2/4

base b = √38,88*4/3 = 7,20 cm

altezza h = 7,20*3/4 = 5,40 cm 

perimetro 2p = 2(7,20+5,40) = 25,20 cm 

diagonale d = √7,20^2+5,40^2 = 9,0 cm

spigolo s = 25,20/4 = 6,30 cm