Equilateral cone

We pour so much water into a container with the shape of an equilateral cone, the base of which has a radius r = 6 cm, that one-third of the volume of the cone is filled. How high will the water reach if we turn the cone upside down?

Correct answer:

x =  1.3138 cm

Step-by-step explanation:

r=6 cm s=d=2r s=2 r=2 6=12 cm  s2 = r2 + h2 h=s2r2=12262=6 3 cm10.3923 cm  V=31 π r2 h=31 3.1416 62 10.3923391.7807 cm3 V1=31 V=31 391.7807130.5936 cm3 V2=VV1=391.7807130.5936261.1871 cm3  V2 = 31  π r22   h2 r2:h2 = r:h  V2 = 31  π (r/h h2)2   h2 3 V2/π = r2/h2 h22   h2 3 V2/π = r2/h2 h23  h2=3π r23 V2 h2=33.1416 623 261.1871 10.392329.0785 cm  x=hh2=10.39239.07851.3138 cm  Verifying Solution:  r2=h2 r/h=9.0785 6/10.39235.2415 cm V22=31 π r22 h2=31 3.1416 5.24152 9.0785261.1871 cm3 V22=V2  r1:h1 = r:h V1 = 31 π r12 h1 V1 = 31 π r12 (r1 h/r) V1 = 31 π r13 (h/r)  r1=33 V1 r/(π h)=33 130.5936 6/(3.1416 10.3923)4.1602 cm h1=r1 h/r=4.1602 10.3923/67.2056 cm V11=31 π r12 h1=31 3.1416 4.16022 7.2056130.5936 cm3

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Showing 1 comment:
Dear Sirs! Considering English is not my mother tongue, I hope I understood the word problem as well as the solution you have provided (since there's no image in solution I hope that I understood the labels of the variables that you've uses in the solution) . .. Anyway, I think there is an error in the given solution. .. When you turn the cone upside down, the shape of the cone part filled with water is a truncated cone with the height h2, so it's volume can not be calculated using formula that you have used, that is V1=1/3 * pi * r2 * h2 . .. This isn't the formula for the volume of a truncated cone. ..

Best regards

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