MATHEMATIQUES – LES SOLIDES ET CHANGEMENT D’UNITES

Vtot = \(2000 m^3\)
P = 120 m

=> Vtot = \(π \times r^2 \times P\) => \(r^2\) = \(\frac{Vtot}{π \times P}\) => r = \(\sqrt\frac{Vtot}{π \times P}\) = \(\sqrt\frac{2000}{π \times 120}\) = 2,3 m

Longueur de l’échelle = \(\sqrt{d^2 + P^2}\) = \(\sqrt{(2 \times r)^2 + P^2}\)

Longueur échelle = \(\sqrt{(2 \times 2,3)^2 + 120^2}\) = 276 m

Atot = \(2 \times (ab + ac + bc)\) = \(2 \times (10 \times 5 + 10 \times 2 + 5 \times 2)\) = \(160 cm^2\)

Vtot = \(π \times r^2 \times P\) = \(π\times 36^2 \times 180\) = \(732870,73 m^3\)

Vdemi = (Vtot \div 2\) = \(732870,37 \div 2\) = \(366435,37 m^3\)

Changement d’unité car 1l = \(1 dm^3\)

=> \(366435,37 m^3\) = \(366435370 dm^3\) = \(366435370 l\)

Nb de barils = \(366435370 \div 2\) = 36643557

Vparall = \(a \times b \times c\) = \(50 \times 30 \times 20\) = \(30000 cm^3\)

r = \(\frac{d}{2}\) = \(\frac{10}{2}\) = 5 cm

Vcyl = \(r^2 \times π \times P\) = \(5^2 \times π \times 40\) = \(3141,60 cm^3\)

Veau = Vparall – Vcyl = 30000 – 3141,60 = \(26858,40 cm^3\)

d = diagonale de la base

d = \(\sqrt{c^2 + b^2}\) = \(\sqrt{5^2 + 10^5}\) = 11,18 cm

Aire = \(a \times d\) = \(25 \times 11,18\) = \(279,5 cm^2\)

Aire de la base = Ac = \(π \times r^2\)
r = \(d \div 2\) = \(4 \div 2\)
Ac = \(π \times 2^2\) =12,57 \(cm^2\)

Aire du cylindre = \(Ac \times P\) = \(12,5664 \times 48\) = \(603,19 cm^3\)

Atrapèze = \(\frac{(11+19) \times 30}{2}\)

Vtot = \(Atrapèze \times 30\) = \(\frac{(11+19) \times 30}{2} \times 30\) = \(13500 m^3\)