If the area of the base of a cone is increased by 100%, then th volume increases by

Options

A. `200%`
B. `182%`
C. `141%`
D. `100%`

Answer

To solve the problem of how the volume of a cone changes when the area of its base is increased by 100%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Volume Formula:**
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius of the base and \( h \) is the height of the cone.

2. **Initial Area of the Base:**
The area \( A \) of the base of the cone is:
\[ A = \pi r^2 \]

3. **Increase in Area:**
If the area of the base is increased by 100%, the new area \( A' \) becomes:
\[ A' = A + 100\% \text{ of } A = A + A = 2A \]
Therefore,
\[ A' = 2 \pi r^2 \]

4. **New Radius Calculation:**
Since the area of the base is now \( 2 \pi r^2 \), we can find the new radius \( r' \) using the area formula:
\[ A' = \pi (r')^2 \]
Setting the two expressions for area equal gives:
\[ 2 \pi r^2 = \pi (r')^2 \]
Dividing both sides by \( \pi \):
\[ 2 r^2 = (r')^2 \]
Taking the square root:
\[ r' = r \sqrt{2} \]

5. **Volume with New Radius:**
The new volume \( V' \) of the cone with the new radius \( r' \) and the same height \( h \) is:
\[ V' = \frac{1}{3} \pi (r')^2 h = \frac{1}{3} \pi (r \sqrt{2})^2 h = \frac{1}{3} \pi (2r^2) h = \frac{2}{3} \pi r^2 h \]

6. **Percentage Increase in Volume:**
The original volume \( V \) was:
\[ V = \frac{1}{3} \pi r^2 h \]
The increase in volume is:
\[ V' - V = \frac{2}{3} \pi r^2 h - \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 h \]
The percentage increase in volume is calculated as:
\[ \text{Percentage Increase} = \left( \frac{V' - V}{V} \right) \times 100\% = \left( \frac{\frac{1}{3} \pi r^2 h}{\frac{1}{3} \pi r^2 h} \right) \times 100\% = 100\% \]

### Conclusion:
Thus, the percentage increase in the volume of the cone when the area of the base is increased by 100% is **100%**.