Лопта и полиедри – примери 2

Задаци

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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)

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

Пр.4)

$$\begin{aligned}&{P_L} = \pi {m^2} = 100\pi d{m^2} \\ &\underline {{P_{op}} = 48d{m^2}} \\ &{P_V} = ?,{V_V} = ? \\ & \\ &{P_L} = 4{R^2}\pi \\ &100\pi = 4{R^2}\pi \\ &{R^2} = 25 \\ &R = 5cm \\ &\begin{array}{cccccccccccccccccccc} \begin{aligned}&{P_V} = 2B + M \\ &{P_V} = 2{r^2}\pi + 2r\pi H\end{aligned} &\begin{aligned}&{V_V} = BH \\ &{V_V} = {r^2}\pi H\end{aligned} \end{array} \\ & \\ &2rH = 48 \\ &rH = 24 \\ &H = \frac{{24}}{r}\end{aligned}$$

У правоуглом троуглу:

$$\begin{aligned}&{\left( {2R} \right)^2} = {\left( {2r} \right)^2} + {H^2} \\ &100 = 4{r^2} + {H^2} \\ &100 = 4{r^2} + {\left( {\frac{{24}}{r}} \right)^2} \\ &100 = 4{r^2} + \frac{{576}}{{{r^2}}} \\ &100{r^2} = 4{r^4} + 576\end{aligned}$$

Смена: ${r^2} = t$

$$\begin{aligned}&4{t^2} - 100t + 576 = 0 \\ &{t^2} - 25t + 144 = 0 \\ &{t_{1,2}} = \frac{{25 \pm \sqrt {625 - 576} }}{2} = \frac{{25 \pm 7}}{2} \\ &\begin{array}{cccccccccccccccccccc} \begin{aligned}&{t_1} = 16 \\ &{r^2} = 16 \\ &r = 4dm \\ &H = 6dm\end{aligned} &\begin{aligned}&{t_2} = 9 \\ &{r^2} = 9 \\ &r = 3dm \\ &H = 8dm\end{aligned} \end{array}\end{aligned}$$

Имамо два случаја:

1)

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

2)

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