The volume \( V \) of a right circular cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius, and \( h \) is the height of the cone. Suppose the original volume is \( V_1 \) and the new volume after the increase is \( V_2 \). We are given that the volume increased by a factor of 25:
\[ V_2 = 25 V_1 \]
Since the height \( h \) remains fixed, we focus on how the radius \( r \) changes. Let \( r_1 \) be the original radius and \( r_2 \) be the new radius. Initially, the volume \( V_1 \) of the cone can be expressed as:
\[ V_1 = \frac{1}{3} \pi r_1^2 h \]
After the change, the new volume \( V_2 \) is:
\[ V_2 = \frac{1}{3} \pi r_2^2 h \]
Given \( V_2 = 25 V_1 \), we substitute the expressions for \( V_1 \) and \( V_2 \):
\[ \frac{1}{3} \pi r_2^2 h = 25 \left( \frac{1}{3} \pi r_1^2 h \right) \]
Simplifying this, we get:
\[ r_2^2 h = 25 r_1^2 h \]
Since \( h \) is constant and nonzero, we can divide both sides by \( h \):
\[ r_2^2 = 25 r_1^2 \]
Taking the square root of both sides, we find:
\[ r_2 = \sqrt{25} r_1 \]
\[ r_2 = 5 r_1 \]
Thus, the radius was changed by a factor of 5. Therefore, the correct answer is:
\[ \boxed{5} \]
The volume of a right circular cone increased by a factor of 25. If the height remained fixed, by what factor was the radius changed?
A. 5
B. 25
C. 125
D. 225
1 answer