Трећи разред средње школе

Лопта и полиедри - примери 3

Лопта уписана у трострaну призму. Сложенији пример.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.5)   У праву призму чија је основа правоугли троугао, оштрог угла ${30^ \circ }$ и хипитенузе 10cm, уписана је лопта. Израчунати однос површина лопте и призме.

Пр.5)

388

\[\begin{gathered}
  \alpha  = {30^ \circ } \hfill \\
  \underline {c = 10cm}  \hfill \\
  \frac{{{P_L}}}{{{P_V}}} = ? \hfill \\
   \hfill \\
  {P_\vartriangle } = \left( {\frac{{ab}}{2}} \right) = s \cdot {r_u} \hfill \\
   \hfill \\
  {P_L} = 4{R^2}\pi  \hfill \\
  s = \frac{{a + b + c}}{2} \hfill \\
  \sin {30^ \circ } = \frac{a}{c} = \frac{a}{{10}} \hfill \\
  \frac{1}{2} = \frac{a}{{10}} \hfill \\
  a = 5cm \hfill \\
   \hfill \\
  \cos {30^ \circ } = \frac{b}{c} \hfill \\
  \frac{{\sqrt 3 }}{2} = \frac{b}{{10}} \hfill \\
  2b = 10\sqrt 3  \hfill \\
  b = 5\sqrt 3 cm \hfill \\
\end{gathered} \]

\[\begin{gathered}
{P_\Delta } = \frac{{ab}}{2} \hfill \\
{P_\Delta } = \frac{{25\sqrt 3 }}{2} \hfill \\
{P_\Delta } = s{r_u};s = \frac{{a + b + c}}{2} = \frac{{15 + 5\sqrt 3 }}{2} \hfill \\
\frac{{25\sqrt 3 }}{2} = \frac{{15 + 5\sqrt 3 }}{2}{r_u} \hfill \\
{r_u} = \frac{{\frac{{25\sqrt 3 }}{2}}}{{\frac{{15 + 5\sqrt 3 }}{2}}} = \frac{{5\sqrt 3 }}{{3 + \sqrt 3 }} \cdot \frac{{3 - \sqrt 3 }}{{3 - \sqrt 3 }} \hfill \\
{r_u} = \frac{{5\sqrt 3 \cdot \left( {3 - \sqrt 3 } \right)}}{6} \hfill \\
R = \frac{{5\sqrt 3 \cdot \left( {3 - \sqrt 3 } \right)}}{6} \hfill \\
H = \frac{{5\sqrt 3 \cdot \left( {3 - \sqrt 3 } \right)}}{3} \hfill \\
\hfill \\
{P_L} = 4{R^2}\pi \hfill \\
{P_L} = 4{\left( {\frac{{5\sqrt 3 \cdot \left( {3 - \sqrt 3 } \right)}}{6}} \right)^2}\pi \hfill \\
{P_L} = 4\frac{{25 \cdot 3 \cdot \left( {9 - 2 \cdot 3 \cdot \sqrt 3 + 3} \right)}}{{36}}\pi \hfill \\
{P_L} = \frac{{25 \cdot \left( {12 - 6\sqrt 3 } \right)}}{3}\pi \hfill \\
{P_L} = \frac{{25 \cdot 6 \cdot \left( {2 - \sqrt 3 } \right)}}{3}\pi \hfill \\
{P_L} = 50 \cdot \left( {2 - \sqrt 3 } \right)\pi c{m^2} \hfill \\
\hfill \\
{P_P} = 2B + M \hfill \\
{P_P} = 2 \cdot \frac{{25\sqrt 3 }}{2} + \left( {a + b + c} \right) \cdot H \hfill \\
{P_P} = 25\sqrt 3 + \left( {15 + 5\sqrt 3 } \right) \cdot \frac{{5\sqrt 3 \cdot \left( {3 - \sqrt 3 } \right)}}{3} \hfill \\
{P_P} = 25\sqrt 3 + 5\left( {3 + \sqrt 3 } \right) \cdot \frac{{5\sqrt 3 \cdot \left( {3 - \sqrt 3 } \right)}}{3} \hfill \\
{P_P} = 25\sqrt 3 + \frac{{25\sqrt 3 \cdot \left( {9 - 3} \right)}}{3} \hfill \\
{P_P} = 75\sqrt 3 c{m^2} \hfill \\
\hfill \\
\frac{{{P_L}}}{{{P_P}}} = \frac{{50 \cdot \left( {2 - \sqrt 3 } \right)\pi }}{{75\sqrt 3 }} = \frac{{2 \cdot \left( {2 - \sqrt 3 } \right)\pi }}{{3\sqrt 3 }} \hfill \\
\end{gathered} \]