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 = 2 r s = 2 \cdot r = 2 \cdot 6 = 12 cm s^{2} = r^{2} + h^{2} h = s^{2} - r^{2} = 1 2^{2} - 6^{2} = 63 cm ≐ 10.3923 cm V = \frac{1}{3} \cdot π \cdot r^{2} \cdot h = \frac{1}{3} \cdot 3.1416 \cdot 6^{2} \cdot 10.3923 ≐ 391.7807 cm^{3} V_{1} = \frac{1}{3} \cdot V = \frac{1}{3} \cdot 391.7807 ≐ 130.5936 cm^{3} V_{2} = V - V_{1} = 391.7807 - 130.5936 ≐ 261.1871 cm^{3} V_{2} = \frac{1}{3} \cdot π \cdot r_{2}^{2} \cdot h_{2} r_{2} : h_{2} = r : h V_{2} = \frac{1}{3} \cdot π \cdot (r / h \cdot h_{2})^{2} \cdot h_{2} 3 \cdot V_{2} / π = r^{2} / h^{2} \cdot h_{2}^{2} \cdot h_{2} 3 \cdot V_{2} / π = r^{2} / h^{2} \cdot h_{2}^{3} h_{2} = 3 \frac{3 \cdot V _{2} \cdot h ^{2}}{π \cdot r ^{2}} = 3 \frac{3 \cdot 2 6 1 . 1 8 7 1 \cdot 1 0 . 3 9 2 3 ^{2}}{3 . 1 4 1 6 \cdot 6 ^{2}} ≐ 9.0785 cm x = h - h_{2} = 10.3923 - 9.0785 ≐ 1.3138 cm Verifying Solution: r_{2} = h_{2} \cdot r / h = 9.0785 \cdot 6 / 10.3923 ≐ 5.2415 cm V_{22} = \frac{1}{3} \cdot π \cdot r_{2}^{2} \cdot h_{2} = \frac{1}{3} \cdot 3.1416 \cdot 5.241 5^{2} \cdot 9.0785 ≐ 261.1871 cm^{3} V_{22} = V_{2} r_{1} : h_{1} = r : h V_{1} = \frac{1}{3} \cdot π \cdot r_{1}^{2} \cdot h_{1} V_{1} = \frac{1}{3} \cdot π \cdot r_{1}^{2} \cdot (r_{1} \cdot h / r) V_{1} = \frac{1}{3} \cdot π \cdot r_{1}^{3} \cdot (h / r) r_{1} = 3 3 \cdot V_{1} \cdot r / (π \cdot h) = 3 3 \cdot 130.5936 \cdot 6 / (3.1416 \cdot 10.3923) ≐ 4.1602 cm h_{1} = r_{1} \cdot h / r = 4.1602 \cdot 10.3923 / 6 ≐ 7.2056 cm V_{11} = \frac{1}{3} \cdot π \cdot r_{1}^{2} \cdot h_{1} = \frac{1}{3} \cdot 3.1416 \cdot 4.160 2^{2} \cdot 7.2056 ≐ 130.5936 cm^{3}

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Showing 1 comment:

Aleksandra-maria

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

2 years ago Like

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