Cone and the ratio

The rotational cone has a height of 59 cm, and the ratio of the base surface to the lateral surface is 10: 12. Calculate the surface of the base and the lateral surface.

Final Answer:

S₁ =  4811.79 cm²
S₂ =  5774.15 cm²

Step-by-step explanation:

$$\begin{gathered} v=59\text{cm} \\ {S}_1=\pi r^2 \\ {S}_2=\pi rs \\ q=12/10=\frac{6}{5}=1.2 \\ {S}_1:S_2=10:12=10/12=q \\ \frac{\pi r^2}{\pi rs}=10/12 \\ r/s=10/12 \\ s=r\cdot 12/10=r\cdot q \\ {s}^2=v^2+r^2 \\ {r}^2\cdot q^2=59^2+r^2 \\ r=v\cdot \sqrt{q^2-1}=59\cdot \sqrt{1.2^2-1}\doteq 39.1362\text{cm} \\ s=r\cdot q=39.1362\cdot 1.2\doteq 46.9634\text{cm} \\ {S}_1=\pi \cdot r^2=3.1416\cdot 39.1362^2=4811.79{\text{cm}}^2 \end{gathered}$$

$S_2=\pi \cdot r\cdot s=3.1416\cdot 39.1362\cdot 46.9634=5774.15{\text{cm}}^2$

Try another example