Se considera un tetraedru \( MNPQ \) cu \( MN=3cm, PQ=5cm, NP=4cm \) si \( m(\angle(MN,PQ))=30^\circ. \). Demonstrati ca volumul tetraedrului nu depaseste \( 5cm^3 \).
Argentina Dobrescu, R.M.T. 2/2007
Tetraedru de volum mai mic ca 5
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Claudiu Mindrila
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Tetraedru de volum mai mic ca 5
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Marius Mainea
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Se stie ca volumul unui unui tetraedru ABCD este : \( V=\frac{AB\cdot CD\cdot d(AB,CD)\cdot\sin\angle(AB,CD)}{6} \) , unde \( d(AB,CD) \) este distanta dintre dreptele AB si CD.
Asadar \( V[MNPQ]=\frac{MN\cdot PQ\cdot d(MN,PQ)\cdot \sin30^\circ}{6}\le\frac{3\cdot 5\cdot 4\cdot\frac{1}{2}}{6}=5cm^2 \)
Asadar \( V[MNPQ]=\frac{MN\cdot PQ\cdot d(MN,PQ)\cdot \sin30^\circ}{6}\le\frac{3\cdot 5\cdot 4\cdot\frac{1}{2}}{6}=5cm^2 \)