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determinati aria totala a unei prisme hexagonale regulate cu aria bazei de
[tex]54 \sqrt{3} [/tex]
si volumul de 324 cm


Răspuns :

[tex]Ab = 54 \sqrt{3} \: {cm}^{2} [/tex]

[tex]Ab = \frac{3 {l}^{2} \sqrt{3} }{2} [/tex]

[tex] \frac{3 {l}^{2} \sqrt{3} }{2} = 54 \sqrt{3} [/tex]

[tex]3 {l}^{2} \sqrt{3} = 2 \times 54 \sqrt{3} [/tex]

[tex]3 {l}^{2} \sqrt{3} = 108 \sqrt{3} \: | \div 3 \sqrt{3} [/tex]

[tex] {l}^{2} = 36[/tex]

[tex]l = \sqrt{36} [/tex]

[tex]l = 6 \: cm[/tex]

[tex]Pb = 6l[/tex]

[tex]Pb = 6 \times 6[/tex]

[tex]Pb = 36 \: cm[/tex]

[tex]V = 324 \: cm[/tex]

[tex]V = Ab \times h[/tex]

[tex]324 = 54 \sqrt{3} \times h[/tex]

[tex]h = \frac{324}{54 \sqrt{3} } [/tex]

[tex]h = \frac{6}{\sqrt{3}} [/tex]

[tex]h=\frac{6\sqrt{3}}{3}[/tex]

[tex]h=2\sqrt{3}\:cm[/tex]

[tex]Al = Pb \times h[/tex]

[tex]Al = 36 \times 2\sqrt{3} [/tex]

[tex]Al = 72 \sqrt{3} \: {cm}^{2} [/tex]

[tex]At = Al + 2Ab [/tex]

[tex]At = 72 \sqrt{3} + 2 \times 54 \sqrt{3} [/tex]

[tex]At = 72 \sqrt{3} + 108 \sqrt{3} [/tex]

[tex]At = 180 \sqrt{3} \: {cm}^{2} [/tex]