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Question Number 47243 by MrW3 last updated on 06/Nov/18

Answered by MJS last updated on 07/Nov/18

left picture  (8/r)=(H/R) ⇒ r=((8R)/H)  v_(blue) =((πR^2 (H^2 +8H+64)(H−8))/(3H^2 ))  right picture  ((H−2)/r)=(H/R) ⇒ r=R(1−(2/H))  v_(blue) =((π(H−2)r^2 )/3)=((πR^2 (H−2)^3 )/(3H^2 ))    (H−2)^3 =(H^2 +8H+64)(H−8)  H=1+(√(85)) ≈ 10.22cm

$$\mathrm{left}\:\mathrm{picture} \\ $$$$\frac{\mathrm{8}}{{r}}=\frac{{H}}{{R}}\:\Rightarrow\:{r}=\frac{\mathrm{8}{R}}{{H}} \\ $$$${v}_{{blue}} =\frac{\pi{R}^{\mathrm{2}} \left({H}^{\mathrm{2}} +\mathrm{8}{H}+\mathrm{64}\right)\left({H}−\mathrm{8}\right)}{\mathrm{3}{H}^{\mathrm{2}} } \\ $$$$\mathrm{right}\:\mathrm{picture} \\ $$$$\frac{{H}−\mathrm{2}}{{r}}=\frac{{H}}{{R}}\:\Rightarrow\:{r}={R}\left(\mathrm{1}−\frac{\mathrm{2}}{{H}}\right) \\ $$$${v}_{{blue}} =\frac{\pi\left({H}−\mathrm{2}\right){r}^{\mathrm{2}} }{\mathrm{3}}=\frac{\pi{R}^{\mathrm{2}} \left({H}−\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{3}{H}^{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\mathrm{H}−\mathrm{2}\right)^{\mathrm{3}} =\left({H}^{\mathrm{2}} +\mathrm{8}{H}+\mathrm{64}\right)\left({H}−\mathrm{8}\right) \\ $$$${H}=\mathrm{1}+\sqrt{\mathrm{85}}\:\approx\:\mathrm{10}.\mathrm{22cm} \\ $$

Commented by MrW3 last updated on 07/Nov/18

thank you sir!

$${thank}\:{you}\:{sir}! \\ $$

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