Equilateral cone

We pour so much water into a container that has 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 result:

h2 =  3.4641 cm


r=6 cm s=2 r=2 6=12 cm h=s2r2=12262=6 3 cm10.3923 cm  V=13 π r2 h=13 3.1416 62 10.3923391.7807 cm3 V1=13 V=13 391.7807130.5936 cm3 V2=VV1=391.7807130.5936261.1871 cm3  V1=13 π r2 h2  h2=3 V1π r2=3 130.59363.1416 62=2 3=3.4641 cm

We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!

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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