Problemi parallelogrammo

questo è il primo quesito 

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Parallelogramma verde:

$AB= 48\,cm;$

$BH= 2×AH= \dfrac{48}{2+1}×2 = 16×2 = 32\,cm;$

$AH= 48-32 = 16\,cm;$

$DH= 30\,cm;$

$AD= \sqrt{30^2+16^2} = 34\,cm$ (teorema di Pitagora);

perimetro $2p= 2(48+34) = 2×82 = 164\,cm;$

area $A= AB×DH = 48×30 = 1440\,cm^2.$

Parallelogramma giallo:

$BH= \dfrac{50}{3+7}×3 = 5×3 = 15\,cm;$

$AH= \dfrac{50}{3+7}×7 = 5×7 = 35\,cm;$

$CH= \dfrac{12}{\cancel5_1}×\cancel{15}^3 = 12×3 = 36\,cm;$

$BC= \sqrt{36^2+15^2} = 39\,cm;$

perimetro $2p= 2(50+39) = 2×89 = 178\,cm;$

area $A= 50×36 = 1800\,cm^2.$

Parallelogramma rosso:

$AD= \dfrac{40+10}{2} = 25\,cm;$

$DH= \dfrac{40-10}{2} = 15\,cm;$

$AH= \sqrt{25^2-15^2} = 20\,cm;$

$BH= \dfrac{3}{\cancel2_1}×\cancel{20}^{10} = 3×10 = 30\,cm;$

$AB= AH+BH = 20+30 = 50\,cm;$

perimetro $2p= 2(50+25)= 2×75 = 150\,cm;$

area $A= 50×15 = 750\,cm^2.$ 

prossima volta, foto dritta Grazie. 

AH = AB/3 = 48/3 = 16 cm

AD = √AH^2+DH^2 = √30^2+16^2 = 34,0 cm 

perimetro 2p = 2(AB+AD) = 2(48+34) = 164 cm

area A = AB*DH = 48*30 = 1440 cm^2

AB = AH+BH = AH+3AH/7 = 10AH/7 = 50

AH = 50*7/10 = 35 cm

BH = 50-35 = 15 cm

CH= 12BH/5 = 15/5*12 = 36 cm 

BC = √CH^2+BH^2 = √36^2+15^2 = 39,0 cm

perimetro 2p = 2(AB+BC) = 2(50+39) = 178 cm

area A = AB*CH =  50*36 = 1.800 cm^2

AD+DH = 40 cm

AD-DH = 10 cm

somma m. a m.

2AD = 50 cm

AD = 25 cm 

DH = 40-25 = 15 cm

AH = √AD^2-DH^2 = √25^2-15^2 = 20 cm 

BH = 3AH/2 = 20*3/2 = 30 cm 

AB = AH+BH = 20+30 = 50 cm 

perimetro 2p = 2(AD+AB) = 2(25+50) = 150 cm

area A = AB*DH = 50*15 = 750 cm^2

Io li risolverei bene, se solo riuscissi a leggere di traverso.

ora provo a fare gli altri