Conical bottle

When a conical bottle rests on its flat base, the water in the bottle is 8 cm from it vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the height of the bottle?


h =  10.22 cm


h1=810.2195 cm h2=28.2195 cm  V=13πr2 h V3=13πr12 h1  V1=VV3 V1=13πr2 h13πr12 h1 V1=13π(r2 hr12 h1)  r1<r2<r r1:h1=r:h r1=r h1/h r2:(hh2)=r:h r2=r (hh2)/h  V2=13πr22(hh2)  V1=V2  13π(r2 hr12 h1)=13πr22(hh2) (r2 hr12 h1)=r22(hh2) (r2 h(r h1/h)2 h1)=(r(hh2)/h)2(hh2)  (h(h1/h)2 h1)=((hh2)/h)2(hh2)  (h(8/h)2 8)=((h2)/h)2(h2)  h22h84=0  a=1;b=2;c=84 D=b24ac=2241(84)=340 D>0  h1,2=b±D2a=2±3402=2±2852 h1,2=1±9.2195444572929 h1=10.219544457293 h2=8.2195444572929   Factored form of the equation:  (h10.219544457293)(h+8.2195444572929)=0  h>0  h=h1=10.219510.219510.22 cmh_{1}=8 \doteq 10.2195 \ \text{cm} \ \\ h_{2}=2 \doteq -8.2195 \ \text{cm} \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \pi r^2 \ h \ \\ V_{3}=\dfrac{ 1 }{ 3 } \pi r_{1}^2 \ h_{1} \ \\ \ \\ V_{1}=V-V_{3} \ \\ V_{1}=\dfrac{ 1 }{ 3 } \pi r^2 \ h-\dfrac{ 1 }{ 3 } \pi r_{1}^2 \ h_{1} \ \\ V_{1}=\dfrac{ 1 }{ 3 } \pi (r^2 \ h - r_{1}^2 \ h_{1}) \ \\ \ \\ r_{1}<r_{2}<r \ \\ r_{1}:h_{1}=r:h \ \\ r_{1}=r \cdot \ h_{1}/h \ \\ r_{2}:(h-h_{2})=r:h \ \\ r_{2}=r \cdot \ (h-h_{2})/h \ \\ \ \\ V_{2}=\dfrac{ 1 }{ 3 } \pi r_{2}^2 (h-h_{2}) \ \\ \ \\ V_{1}=V_{2} \ \\ \ \\ \dfrac{ 1 }{ 3 } \pi (r^2 \ h - r_{1}^2 \ h_{1})=\dfrac{ 1 }{ 3 } \pi r_{2}^2 (h-h_{2}) \ \\ (r^2 \ h - r_{1}^2 \ h_{1})=r_{2}^2 (h-h_{2}) \ \\ (r^2 \ h - (r \cdot \ h_{1}/h)^2 \ h_{1})=(r(h-h_{2})/h)^2 (h-h_{2}) \ \\ \ \\ (h - (h_{1}/h)^2 \cdot \ h_{1})=((h-h_{2})/h)^2 (h-h_{2}) \ \\ \ \\ (h - (8/h)^2 \cdot \ 8)=((h-2)/h)^2 (h-2) \ \\ \ \\ h^2 -2h -84=0 \ \\ \ \\ a=1; b=-2; c=-84 \ \\ D=b^2 - 4ac=2^2 - 4\cdot 1 \cdot (-84)=340 \ \\ D>0 \ \\ \ \\ h_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 2 \pm \sqrt{ 340 } }{ 2 }=\dfrac{ 2 \pm 2 \sqrt{ 85 } }{ 2 } \ \\ h_{1,2}=1 \pm 9.2195444572929 \ \\ h_{1}=10.219544457293 \ \\ h_{2}=-8.2195444572929 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (h -10.219544457293) (h +8.2195444572929)=0 \ \\ \ \\ h>0 \ \\ \ \\ h=h_{1}=10.2195 \doteq 10.2195 \doteq 10.22 \ \text{cm}

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