Question

VA ROG MULT CINE MA POATE AJUTA !!!! PLS

Answer 1

[tex]\it \ c)\ \ \mathcal{V} = \pi R^2G = 40\pi \Rightarrow\pi\cdot2^2G=40\pi \Rightarrow G= \dfrac{40\pi}{4\pi} = 10cm \\ \\ \\ \mathcal{A}_{\ell} = 2\pi RG = 2\pi\cdot2\cdot10 = 40\pi \ cm^2 \\ \\ \\ \mathcal{A}_{t} = 2\pi R(R+G) = 2\pi\cdot2(2+10) = 4\pi\cdot12= 48\pi\ cm^2[/tex]


[tex]\it d) \ \mathcal{A}_{b} = \dfrac{1}{2} ( \mathcal{A}_{t} - \ \mathcal{A}_{\ell}) = \dfrac{1}{2} (72\pi - 40\pi) = \dfrac{1}{2} \cdot32\pi = 16\pi \ \ \ \ (1) \\ \\ \\ \mathcal{A}_{b} = \pi R^2 \ \ \ \ (2) \\ \\ \\ (1), (2) \Rightarrow \pi R^2= 16\pi |_{:\pi} \Rightarrow R^2 = 16 \Rightarrow R = 4\ cm \\ \\ \\ \mathcal{A}_{\ell} = = 2\pi RG = 40\pi |_{:\pi} \Rightarrow 2\cdot4\cdot G = 40 \Rightarrow 8G = 40 \Rightarrow G = 5\ cm[/tex]

$\it \ \mathcal{V} = \pi R^2G = \pi\cdot4^2\cdot5 = 80\pi \ cm^3$


b) 

[tex]\it \mathcal{A}_{\ell} = 2\pi RG = 80\pi \Rightarrow 2\pi R\cdot10=80 \pi \Rightarrow R = 4\ cm \\ \\ \mathcal{A}_{b} = \pi R^2 = \pi \cdot 4^2 = 16\pi \ cm^2 \\ \\ \mathcal{A}_{t} = \mathcal{A}_{\ell} +2\cdot \mathcal{A}_{b} = 80\pi +2\cdot16\pi= 80\pi + 32\pi = 112\pi \ cm^2 \\ \\ \mathcal{V} = \pi R^2G = \pi \cdot4^2\cdot10 = 160\pi\ cm^3 [/tex]