Triangle and Cone

A right triangle has legs 3 cm and 4 cm long. One cone (let's call it A) was created by rotating this triangle around the long leg, and the other (let's call it B) by rotating it around the shorter leg. Which cone has:
a) a larger volume
b) a smaller surface area
c) a larger total surface area?

Correct answer:

a = B
b = A
c = B

Step-by-step explanation:

$$\begin{gathered} x=3\text{cm} \\ y=4\text{cm} \\ {r}_1=x=3\text{cm} \\ {r}_2=y=4\text{cm} \\ {h}_1=y=4\text{cm} \\ {h}_2=x=3\text{cm} \\ {S}_1=\pi \cdot r_1^2=3.1416\cdot 3^2\doteq 28.2743{\text{cm}}^2 \\ {S}_2=\pi \cdot r_2^2=3.1416\cdot 4^2\doteq 50.2655{\text{cm}}^2 \\ {V}_1=\frac{1}{3}\cdot S_1\cdot h_1=\frac{1}{3}\cdot 28.2743\cdot 4\doteq 37.6991{\text{cm}}^3 \\ {V}_2=\frac{1}{3}\cdot S_2\cdot h_2=\frac{1}{3}\cdot 50.2655\cdot 3\doteq 50.2655{\text{cm}}^3 \\ {V}_2>V_1 \\ a=B \end{gathered}$$

$$\begin{gathered} s=\sqrt{x^2+y^2}=\sqrt{3^2+4^2}=5\text{cm} \\ {S}_{11}=\pi \cdot r_1\cdot s=3.1416\cdot 3\cdot 5\doteq 47.1239{\text{cm}}^2 \\ {S}_{22}=\pi \cdot r_2\cdot s=3.1416\cdot 4\cdot 5\doteq 62.8319{\text{cm}}^2 \\ {S}_{22}>S_{11} \\ b=A \end{gathered}$$

$$\begin{gathered} P_1=S_1+S_{11}=28.2743+47.1239\doteq 75.3982{\text{cm}}^2 \\ {P}_2=S_2+S_{22}=50.2655+62.8319\doteq 113.0973{\text{cm}}^2 \\ {P}_2>P_1 \\ c=B \end{gathered}$$

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