837/VB03. zadatak / Matematika - Zbirka rešenih zadataka

MR-837 / 1200. zadatak

Zapremina prave pravilne osmostrane prizem je 8 m³, a njena visina je $(\sqrt{2} + 1)$ m. Izračunati površinu omotača.

$B = 8 \cdot \frac{a \cdot h}{2}$

$t g 22, 5 ° = \frac{\frac{a}{2}}{h}$

$h = \frac{\frac{a}{2}}{t g 22, 5 °} = \frac{a}{2 \cdot t g 22, 5 °} = \frac{a}{2} \cdot c t g 22, 5 °$

$c t g^{2} 22, 5 ° = \frac{1 + \cos 45 °}{\sin 45 °}$

$c t g^{2} 22, 5 ° = \frac{1 + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$c t g^{2} 22, 5 ° = \frac{\frac{2 + \sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$c t g^{2} 22, 5 ° = \frac{2 + \sqrt{2}}{\sqrt{2}} \cdot | \frac{\sqrt{2}}{\sqrt{2}}$

$c t g^{2} 22, 5 ° = \frac{(2 + \sqrt{2}) \cdot \sqrt{2}}{2}$

$c t g^{2} 22, 5 ° = \frac{2 \sqrt{2} + 2}{2}$

$c t g^{2} 22, 5 ° = \frac{2 (\sqrt{2} + 1)}{2}$

$c t g^{2} 22, 5 ° = \sqrt{2} + 1$

$h = \frac{a}{2} (\sqrt{2} + 1)$

$B = 8 \cdot \frac{a \cdot h}{2} = 8 \cdot \frac{a \cdot a}{2 \cdot 2} (\sqrt{2} + 1)$

$B = 2 a^{2} (\sqrt{2} + 1)$

$V = B \cdot H = 2 a^{2} (\sqrt{2} + 1) \cdot (\sqrt{2} + 1)$

$8 = 2 a^{2} {(\sqrt{2} + 1)}^{2}$

$4 = a^{2} {(\sqrt{2} + 1)}^{2}$

$a^{2} = \frac{{(\sqrt{2} + 1)}^{2}}{4}$

$a = \frac{2}{\sqrt{2} + 1}$

$M = 8 \cdot a \cdot H$

$M = 8 \cdot \frac{2}{\sqrt{2} + 1} \cdot (1 + \sqrt{2})$

$M = 16$

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