[Risolto] Geometria

@gggggggh 

$A_ABCD=b*h=45*24=1080~cm^2$

unità frazionaria=(AE+EB)/(3+2)=45/5=9 cm

$AE=u.f.*3=9*3=27~cm$

$EB=u.f.*2=9*2=18~cm$

$EC=\sqrt{EB^2+BC^2}=\sqrt{18^2+24^2}=\sqrt{324+576}=\sqrt{900}=30~cm$

$AC=\sqrt{AD^2+AC^2}=\sqrt{24^2+45^2}=\sqrt{576+2025}=\sqrt{2601}=51~cm$

$p_AEC=AC+EC+AE=51+30+27=108~cm$

$A_ACD=\frac{AD*DC}{2}=\frac{24*45}{2}=540~cm^2$

$A_EBC=\frac{EB*BC}{2}=\frac{18*24}{2}=216~cm^2$

$A_AEC=A_ABCD-A_ACD-A_EBC=1080-540-216=324~cm^2$

EB+3EB/2 = 5EB/2 = 45 cm

EB = 45/5*2 = 18 cm

AE = 45-18 = 27 cm

AC = √45^2+24^2 = 51,0 cm

EC = √18^2+24^2 = 30,0 cm

area AEC = AE*BC/2 = 27*12 = 324 cm^2

perimetro AEC = 51+30+27 = 108 cm 

@gggggggh 

$AC= \sqrt{45^2+24^2} = 51~cm$ (teorema di Pitagora);

$AB= AE+EB =45~cm$;

rapporto tra $AE ~e~ EB = \frac{3}{2}$, quindi:

$AE= \frac{45}{3+2}×3 = 27~cm$;

$EB= \frac{45}{3+2}×2 = 18~cm$;

$EC= \sqrt{24^2+18^2}= 30~cm$ (teorema di Pitagora);

perimetro triangolo AEC $2p= AE+EC+AC = 27+30+51 = 108~cm$;

area triangolo AEC $A= \frac{AE×BC}{2} = \frac{27×24}{2}= 324~cm^2$.