If the diameter of a circle is increased by 8%. then its area is increased by—

Options

A. 16-46%
B. 16·64%
C. 6-64%
D. 0.16

Answer

To solve the problem of how much the area of a circle increases when the diameter is increased by 8%, we can follow these steps:

### Step 1: Understand the relationship between diameter, radius, and area
The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle. The radius is half of the diameter \( d \):
\[ r = \frac{d}{2} \]

### Step 2: Determine the initial diameter and radius
Let the initial diameter of the circle be \( d \). Therefore, the initial radius \( r \) is:
\[ r = \frac{d}{2} \]

### Step 3: Calculate the new diameter and radius after an 8% increase
An 8% increase in the diameter means the new diameter \( d' \) is:
\[ d' = d + 0.08d = 1.08d \]
The new radius \( r' \) will then be:
\[ r' = \frac{d'}{2} = \frac{1.08d}{2} = 0.54d \]

### Step 4: Calculate the initial and new areas
Now, we can calculate the initial area \( A \) and the new area \( A' \):
\[ A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4} \]
\[ A' = \pi (r')^2 = \pi \left(0.54d\right)^2 = \pi (0.2916d^2) = 0.2916\pi d^2 \]

### Step 5: Find the increase in area
The increase in area \( \Delta A \) is given by:
\[ \Delta A = A' - A = 0.2916\pi d^2 - \frac{\pi d^2}{4} \]
To simplify this, we need a common denominator:
\[ \Delta A = 0.2916\pi d^2 - 0.25\pi d^2 = (0.2916 - 0.25)\pi d^2 = 0.0416\pi d^2 \]

### Step 6: Calculate the percentage increase in area
The percentage increase in area is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta A}{A}\right) \times 100 = \left(\frac{0.0416\pi d^2}{\frac{\pi d^2}{4}}\right) \times 100 \]
This simplifies to:
\[ = \left(\frac{0.0416 \times 4}{1}\right) \times 100 = 0.1664 \times 100 = 16.64\% \]

### Conclusion
Thus, the area of the circle increases by **16.64%** when the diameter is increased by 8%.

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