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[Есра]

h и l се отнасят както 1:3, V=72П см3.Търси се S1=?

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h и l се отнасят както 1:3, V=72П см3.Търси се S1=?

[tex]CH = h ,BC = l ,HB = r ,\angle HBC = \alpha[/tex]
[tex]\displaystyle \frac{h}{l} = \displaystyle \frac{1}{3} \Rightarrow h = \displaystyle \frac{l}{3}[/tex]
[tex]\displaystyle \frac{h}{l} = sin\alpha = \displaystyle \frac{1}{3} \Rightarrow cos\alpha = \displaystyle \frac{2\sqrt{2}}{3}[/tex] , [tex]cos\alpha = \displaystyle \frac{r}{l} \Rightarrow \displaystyle \frac{r}{l} = \displaystyle \frac{2\sqrt{2}}{3} \Rightarrow r = \displaystyle \frac{2\sqrt{2}}{3}l[/tex]
[tex]V = \displaystyle \frac{\pi r^{2} h}{3} = 72\pi \Rightarrow \displaystyle \frac{1}{3} \pi \displaystyle(\frac{2\sqrt{2}l}{3})^{2}\displaystyle \frac{l}{3} = 72 \pi \Leftrightarrow \displaystyle \frac{8l^{3}\pi}{81} = 72\pi \Leftrightarrow l^{3} = 9^{3} \Rightarrow l = 9[/tex]
[tex]r = \displaystyle \frac{2\sqrt{2}}{3}.9 \Rightarrow r = 6\sqrt{2}[/tex]
[tex]S = \pi.r.l = 54\sqrt{2}\pi , S_{1 } = S + \pi.r^{2} = 54\sqrt{2}\pi + 72\pi = (54\sqrt{2} + 72)\pi = (6\sqrt{2} + 8).9\pi[/tex]