Површина и запремина купе. Решени задаци.
Текст задатака објашњених у видео лекцији.
Пр.1) Угао при врху осног пресека купе је прав, а површина пресека је $18c{m^2}$. Наћи површину и запремину купе.
Пр.2) Дужина висине и изводнице купе се односе 4:5. Запремина купе је $96\pi c{m^3}$. Израчунати површину ове купе.
Пр.1)

\[\begin{array}{*{20}{c}}
\begin{gathered}
{P_{op}} = \frac{{{s^2}}}{2} \hfill \\
18 = \frac{{{s^2}}}{2} \hfill \\
{s^2} = 36 \hfill \\
s = 6 \hfill \\
\end{gathered} &{}&\begin{gathered}
r = H = \frac{d}{2} = \frac{{3\sqrt 2 }}{2} \hfill \\
r = \frac{{6\sqrt 2 }}{2} \hfill \\
r = 3\sqrt 2 \hfill \\
H = 3\sqrt 2 \hfill \\
\end{gathered}
\end{array}\]
\[\begin{gathered}
P = B + M \hfill \\
P = {r^2}\pi + rs\pi \hfill \\
P = {\left( {3\sqrt 2 } \right)^2}\pi + 3\sqrt 2 \cdot 6\pi \hfill \\
P = 9 \cdot 2\pi + 18\sqrt 2 \pi \hfill \\
P = 18\pi + 18\sqrt 2 \pi \hfill \\
P = 18\pi \left( {1 + \sqrt 2 } \right)c{m^2} \hfill \\
\end{gathered} \]
Пр.2)

\[\begin{gathered}
H:s = 4:5 \Rightarrow 49 = sH \Rightarrow s = \frac{5}{4}H \hfill \\
{s^2} = {H^2} + {r^2} \hfill \\
{\left( {\frac{5}{4}H} \right)^2} = {H^2} + {r^2} \hfill \\
{r^2} = {\left( {\frac{5}{4}H} \right)^2} - {H^2} \hfill \\
{r^2} = \frac{{9{H^2}}}{{16}} \hfill \\
r = \frac{{3H}}{4} \hfill \\
\hfill \\
V = \frac{1}{3}B \cdot H \hfill \\
V = \frac{1}{3}{r^2}\pi \cdot H \hfill \\
96\pi = \frac{1}{3}{\left( {\frac{{3H}}{4}} \right)^2}\pi \cdot H \hfill \\
96 = \frac{1}{3} \cdot \frac{{9{H^2}}}{{16}}H \hfill \\
96 = \frac{{3H3}}{{16}} \hfill \\
3{H^3} = 1536 \hfill \\
{H^3} = 512 \hfill \\
H = 8cm \hfill \\
\hfill \\
s = \frac{5}{4} \cdot 8 = 10cm \hfill \\
r = \frac{3}{4} \cdot 8 = 6cm \hfill \\
\hfill \\
P = B + M \hfill \\
P = {r^2}\pi + rs\pi \hfill \\
P = {6^2}\pi + 6 \cdot 10\pi \hfill \\
P = 36\pi + 60\pi \hfill \\
P = 96\pi c{m^2} \hfill \\
\end{gathered} \]