Two vases

Michaela has two vases in her collection. The first vase has the shape of a cone with a base diameter of d = 20 cm; the second vase has the shape of a truncated cone with a lower base of d1 = 25 cm and a diameter of the upper base d2 = 15 cm. Which vase can hold more water if the height of both vases is 0.5 m?

Correct answer:

V₁ =  5235.9878 cm³
V₂ =  5104.1667 cm³

Step-by-step explanation:

$$\begin{gathered} d=20\text{cm} \\ {d}_1=25\text{cm} \\ {d}_2=15\text{cm} \\ h=0.5\text{m}\to \text{cm}=0.5\cdot 100\text{cm}=50\text{cm} \\ r=d/2=20/2=10\text{cm} \\ {r}_1=d_1/2=25/2=\frac{25}{2}=12.5\text{cm} \\ {r}_2=d_2/2=15/2=\frac{15}{2}=7.5\text{cm} \\ {V}_1=\frac{1}{3}\cdot \pi \cdot r^2\cdot h=\frac{1}{3}\cdot 3.1416\cdot 10^2\cdot 50=5235.9878{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} S_1=\pi \cdot r_1^2=3.1416\cdot 12.5^2\doteq 490.8739{\text{cm}}^2 \\ {S}_2=\pi \cdot r_2^2=3.1416\cdot 7.5^2\doteq 176.7146{\text{cm}}^2 \\ {V}_2=\frac{1}{3}\cdot h\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=\frac{1}{3}\cdot 50\cdot (490.8739+\sqrt{490.8739\cdot 176.7146}+176.7146)=\frac{30625}{6}\doteq 5104.1667{\text{cm}}^3 \\ {V}_2<V_1 \\ \text{ Verifying Solution: } \\ {V}_2=\frac{1}{3}\cdot h\cdot (r_1^2+r_1\cdot r_2+r_2^2)=\frac{1}{3}\cdot 50\cdot ({\frac{25}{2}}^2+\frac{25}{2}\cdot \frac{15}{2}+{\frac{15}{2}}^2){\text{cm}}^3=\frac{30625}{6}{\text{cm}}^3=5104.1667{\text{cm}}^3 \end{gathered}$$

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