Лопта и полиедри – примери 2

Текст задатака објашњених у видео лекцији.

Пр.3)   Израчунати однос површина и  запремина ваљка и лопте уписане у ваљак.

Пр.4)   Израчунати површину и запремину ваљка уписаног у лопту чија је површина $\pi {m^2}$, ако знамо да је површина осног пресека ваљка $48d{m^2}$.

Пр.3)

\[\begin{gathered}
  \underline {\frac{{{P_V}}}{{{P_L}}} = ?,\frac{{{V_V}}}{{{V_L}}} = ?}  \hfill \\
   \hfill \\
  {P_V} = 2B + M \hfill \\
  {P_V} = 2{r^2}\pi  + 2r\pi H \hfill \\
  {P_V} = 2{R^2}\pi  + 2R\pi 2R \hfill \\
  {P_V} = 2{R^2}\pi  + 4{R^2}\pi  \hfill \\
  {P_V} = 6{R^2}\pi  \hfill \\
   \hfill \\
  {P_L} = 4{R^2}\pi  \hfill \\
   \hfill \\
  \frac{{{P_V}}}{{{P_L}}} = \frac{{6{R^2}\pi }}{{4{R^2}\pi }} = \frac{3}{2} \hfill \\
   \hfill \\
  {V_V} = BH \hfill \\
  {V_V} = {r^2}\pi H \hfill \\
  {V_V} = {R^2}\pi  \cdot 2R \hfill \\
  {V_V} = 2{R^3}\pi  \hfill \\
   \hfill \\
  {V_L} = \frac{4}{3}{R^3}\pi  \hfill \\
   \hfill \\
  \frac{{{V_V}}}{{{V_L}}} = \frac{{2{R^3}\pi }}{{\frac{4}{3}{R^3}\pi }} = \frac{3}{2} \hfill \\
\end{gathered} \]

Пр.4)

\[\begin{gathered}
  {P_L} = \pi {m^2} = 100\pi d{m^2} \hfill \\
  \underline {{P_{op}} = 48d{m^2}}  \hfill \\
  {P_V} = ?,{V_V} = ? \hfill \\
   \hfill \\
  {P_L} = 4{R^2}\pi  \hfill \\
  100\pi  = 4{R^2}\pi  \hfill \\
  {R^2} = 25 \hfill \\
  R = 5cm \hfill \\
  \begin{array}{*{20}{c}}
  \begin{gathered}
  {P_V} = 2B + M \hfill \\
  {P_V} = 2{r^2}\pi  + 2r\pi H \hfill \\
\end{gathered} &\begin{gathered}
  {V_V} = BH \hfill \\
  {V_V} = {r^2}\pi H \hfill \\
\end{gathered}  
\end{array} \hfill \\
   \hfill \\
  2rH = 48 \hfill \\
  rH = 24 \hfill \\
  H = \frac{{24}}{r} \hfill \\
\end{gathered} \]

У правоуглом троуглу:

\[\begin{gathered}
  {\left( {2R} \right)^2} = {\left( {2r} \right)^2} + {H^2} \hfill \\
  100 = 4{r^2} + {H^2} \hfill \\
  100 = 4{r^2} + {\left( {\frac{{24}}{r}} \right)^2} \hfill \\
  100 = 4{r^2} + \frac{{576}}{{{r^2}}} \hfill \\
  100{r^2} = 4{r^4} + 576 \hfill \\
\end{gathered} \]

Смена: ${r^2} = t$

\[\begin{gathered}
  4{t^2} - 100t + 576 = 0 \hfill \\
  {t^2} - 25t + 144 = 0 \hfill \\
  {t_{1,2}} = \frac{{25 \pm \sqrt {625 - 576} }}{2} = \frac{{25 \pm 7}}{2} \hfill \\
  \begin{array}{*{20}{c}}
  \begin{gathered}
  {t_1} = 16 \hfill \\
  {r^2} = 16 \hfill \\
  r = 4dm \hfill \\
  H = 6dm \hfill \\
\end{gathered} &\begin{gathered}
  {t_2} = 9 \hfill \\
  {r^2} = 9 \hfill \\
  r = 3dm \hfill \\
  H = 8dm \hfill \\
\end{gathered}  
\end{array} \hfill \\
\end{gathered} \]

Имамо два случаја:

1)

\[\begin{gathered}
  r = 4dm;H = 6dm \hfill \\
  {P_V} = 2B + M \hfill \\
  {P_V} = 2{r^2}\pi  + 2r\pi H \hfill \\
  {P_V} = 2 \cdot 16\pi  + 2 \cdot 4\pi  \cdot 6 \hfill \\
  {P_V} = 32\pi  + 48\pi  \hfill \\
  {P_V} = 80\pi d{m^2} \hfill \\
   \hfill \\
  {V_V} = BH \hfill \\
  {V_V} = 16\pi  \cdot 6 \hfill \\
  {V_V} = 96\pi d{m^3} \hfill \\
\end{gathered} \]

2)

\[\begin{gathered}
  r = 3dm;H = 8dm \hfill \\
  {P_V} = 2B + M \hfill \\
  {P_V} = 2{r^2}\pi  + 2r\pi H \hfill \\
  {P_V} = 2 \cdot 9\pi  + 2 \cdot 3\pi  \cdot 8 \hfill \\
  {P_V} = 8\pi  + 48\pi  \hfill \\
  {P_V} = 66\pi d{m^2} \hfill \\
   \hfill \\
  {V_V} = BH \hfill \\
  {V_V} = 9\pi  \cdot 8 \hfill \\
  {V_V} = 72\pi d{m^3} \hfill \\
\end{gathered} \]