PROBLEMI GEOMETRIA 3 MEDIA

Li svolgo uno per volta e me la prendo con calma

Sb=(1764-1176)/2=294cm2     c1*4/3=294*2=28     c2=21   ip=V 28^2+21^2=35

h=1176/84=14    V=14*294=4116cm3

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Area di base $Ab={\displaystyle \frac{At-Al}{2}}={\displaystyle \frac{1764-1176}{2}}={\displaystyle \frac{588}{2}}=294c{m}^2;$

calcola i lati del triangolo rettangolo di base:

cateto minore $c=\sqrt{2\times 294:{\displaystyle \frac{4}{3}}}=\sqrt{588\times {\displaystyle \frac{3}{4}}}=21cm;$

cateto maggiore $C={\displaystyle \frac{2A}{c}}={\displaystyle \frac{2\times {\cancel{294}}^{14}}{{\cancel{21}}_1}}=2\times 14=28cm;$

ipotenusa $i=\sqrt{C^2+c^2}=\sqrt{{28}^2+{21}^2}=35cm$ (teorema di Pitagora);

quindi:

perimetro di base del prisma $2p=C+c+i=28+21+35=84cm;$

altezza $h={\displaystyle \frac{Al}{2p}}={\displaystyle \frac{1176}{84}}=14cm;$

volume $V=Ab\times h=294\times 14=4116c{m}^3.$

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Rombo di base del prisma:

Diagonale maggiore $D=\sqrt{2\times 96:{\displaystyle \frac{3}{4}}}=\sqrt{192\times {\displaystyle \frac{4}{3}}}=\sqrt{256}=16cm;$ 

diagonale minore $d={\displaystyle \frac{2A}{D}}={\displaystyle \frac{2\times {\cancel{96}}^6}{{\cancel{16}}_1}}=2\times 6=12cm;$

lato $l=\sqrt{{\left({\displaystyle \frac{D}{2}}\right)}^2+{\left({\displaystyle \frac{d}{2}}\right)}^2}=\sqrt{{\left({\displaystyle \frac{16}{2}}\right)}^2+{\left({\displaystyle \frac{12}{2}}\right)}^2}=\sqrt{8^2+6^2}=10cm$ (teorema di Pitagora);

perimetro $2p=4l=4\times 10=40cm.$

Prisma:

area laterale $Al=At-2Ab=1152-2\times 96=1152-192=960c{m}^2;$

altezza $h={\displaystyle \frac{Al}{2p}}={\displaystyle \frac{960}{40}}=24cm;$

volume $V=Ab\times h=96\times 24=2304c{m}^3.$