Problema

168/(19-12)=24    A1=24*19=456cm2   A2=24*12=288cm2     l1=456/8=57    l2=288/12=24

2p1=2(57+8)=130cm     2p2=2(24+12)=72cm

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Differenza (168 cm²) e rapporto tra le aree (12/19) dei due rettangoli, quindi:

area del 1° rettangolo $\small A_1= \dfrac{168}{19-12}×12 = \dfrac{\cancel{168}^{24}}{\cancel7_1}×12 = 24×12 = 288\,cm^2;$

area del 2° rettangolo $\small A_2= \dfrac{168}{19-12}×19 = \dfrac{\cancel{168}^{24}}{\cancel7_1}×19 = 24×19 = 456\,cm^2;$

oppure direttamente $\small A_2= 288+168 = 456\,cm^2;$

per cui:

1° rettangolo:

base $\small b_1= \dfrac{A_1}{h_1} = \dfrac{288}{12} = 24\,cm;$

perimetro $\small 2p_1= 2(b_1+h_1) = 2(24+12) = 2×36 = 72\,cm;$

2° rettangolo:

base $\small b_2= \dfrac{A_2}{h_2} = \dfrac{456}{8} = 57\,cm;$

perimetro $\small 2p_2= 2(b_2+h_2) = 2(57+8) = 2×65 = 130\,cm.$

A2/A1 = 19/12 

A2 = 19A1/12

A2-A1 = 19A1/12-A1 = 7A1/12 = 168

A1 = 168*12/7 = 288 cm^2 

b1 = A1/h1 = 288/12 = 24 cm

2p1 = 2(24+12) = 72 cm

A2 = 288*19/12 = 456 cm^2

b2 = A2/h2 = 456/8 = 57 cm

2p2 = 2(57+8) = 130 cm

oppure

8b2/12b1 = 19/12

96b2 = 228b1

b2 = 228b1/96 = 57b1/24 

57b1/24*8 -12b1 = 168

b1(57/3-12) = 168

7b1 = 168

b1 = 24 cm

b2 = 24*57/24 = 57 cm

2p1 = 2*(24+12) =72 cm

2p2 = 2*(57+8) = 130 cm