Дано:
Sбок = 12 Pi
h = 3
Найти: S - ?
r= \frac{S_b_o_k}{h*2Pi} \\ S_o_s_n = Pi*(\frac{S_b_o_k}{h*2Pi})^2=Pi*(\frac{12Pi}{3*2Pi})^2=4Pi \\ S = S_b_o_k+2*S_o_s_n = 12Pi+2*4Pi=20Pi" alt="S = S_b_o_k+2*S_o_s_n \\ S_o_s_n = Pi*r^2; \\ S_b_o_k= h*l=h*2Pi*r => r= \frac{S_b_o_k}{h*2Pi} \\ S_o_s_n = Pi*(\frac{S_b_o_k}{h*2Pi})^2=Pi*(\frac{12Pi}{3*2Pi})^2=4Pi \\ S = S_b_o_k+2*S_o_s_n = 12Pi+2*4Pi=20Pi" align="absmiddle" class="latex-formula">
Ответ: 20 Pi
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ДАНО:
Sсеч = 10 см^2
Sосн = 5 см^2
Найти: h = ? ; Sбок = ?
Sсеч= a*h, где а = 2r
r = \sqrt{S_o_c/Pi} \\ h=S_c_e_g /a=\frac{S_c_e_g}{2r}=\frac{S_c_e_g}{2\sqrt{S_o_c/Pi}}= 5 : \sqrt{5/Pi}=\sqrt{5Pi}" alt="S_o_c=Pi*r^2 => r = \sqrt{S_o_c/Pi} \\ h=S_c_e_g /a=\frac{S_c_e_g}{2r}=\frac{S_c_e_g}{2\sqrt{S_o_c/Pi}}= 5 : \sqrt{5/Pi}=\sqrt{5Pi}" align="absmiddle" class="latex-formula">
Sбок =
ОТВЕТ: