Triangle and Cone

A right triangle has legs 3 cm and 4 cm long. Cone A was created by rotating this triangle around the longer leg, and cone B by rotating it around the shorter leg. Which cone has:
a) a larger volume?
b) a smaller lateral surface area?
c) a larger total surface area?

Final 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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