The height of a right circular cylinder is $6 \text{ m}$. If three times the sum of the areas of its two circular faces is twice the area of the curved surface, then the radius of its base is

### Concept & Formula This problem links the area of a cylinder's circular bases with its curved surface area. - Area of two circular faces = $2\pi r^2$ - Curved Surface Area (CSA) = $2\pi rh$ ### Step-by-Step Solution 1. **Set up the given equation:** - The problem states: $3 \times (\text{sum of areas of two circular faces}) = 2 \times (\text{area of curved surface})$ - $3 \times (2\pi r^2) = 2 \times (2\pi rh)$ 2. **Simplify the equation:** - $6\pi r^2 = 4\pi rh$ - Divide both sides by $2\pi r$ (since $r > 0$): - $3r = 2h$ 3. **Solve for the radius ($r$):** - We are given the height $h = 6 \text{ m}$. - $3r = 2(6)$ - $3r = 12$ - $r = 4 \text{ m}$ ### Exam Strategy & Shortcut Translate the word problem directly into a simplified algebraic form. Recognizing that the "sum of the areas of its two circular faces" is $2\pi r^2$ prevents the common mistake of only considering one face ($\pi r^2$). ### Common Pitfall Students often mistake "sum of the areas of its two circular faces" for just the top base ($\pi r^2$) or the total surface area ($2\pi r^2 + 2\pi rh$). Carefully parsing the text is crucial. ### Final Answer Therefore, the correct answer is **$4 \text{ m}$**.