A circular disc of area `A_1` is given. With its radius as diameter, a circular disc of area `A_2` is cut out of it. The area of the remaining disc is denoted by `A_3`. Then,

Options

A. `A_1A_3 lt 16 A_2^2`
B. `A_1A_3 gt 16 A_2^2`
C. `A_1A_3 = 16A_2^2`
D. `A_1A_3 gt 2A_2^2`

Answer

To solve the problem step by step, we will follow the logical flow of the video transcript while ensuring clarity in each step.

### Step-by-Step Solution:

1. **Understanding the Areas**:
- Let the radius of the original circular disc be \( r \).
- The area of the original disc, denoted as \( A_1 \), is given by the formula:
\[ A_1 = \pi r^2 \]

2. **Finding the Radius of the Cut-Out Disc**:
- A circular disc of area \( A_2 \) is cut out from the original disc, and its radius is half of the original disc's radius. Therefore, the radius of the cut-out disc is:
\[ r_2 = \frac{r}{2} \]

3. **Calculating the Area of the Cut-Out Disc**:
- The area \( A_2 \) of the cut-out disc can be calculated using the formula for the area of a circle:
\[ A_2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]

4. **Finding the Area of the Remaining Disc**:
- The area of the remaining disc, denoted as \( A_3 \), is the area of the original disc minus the area of the cut-out disc:
\[ A_3 = A_1 - A_2 = \pi r^2 - \pi \frac{r^2}{4} \]
- Simplifying this expression:
\[ A_3 = \pi r^2 \left(1 - \frac{1}{4}\right) = \pi r^2 \left(\frac{3}{4}\right) = \frac{3\pi r^2}{4} \]

5. **Calculating the Product of Areas \( A_1 \) and \( A_3 \)**:
- Now, we need to find the product \( A_1 \times A_3 \):
\[ A_1 \times A_3 = \left(\pi r^2\right) \times \left(\frac{3\pi r^2}{4}\right) = \frac{3\pi^2 r^4}{4} \]

6. **Relating the Areas**:
- We can express \( A_2 \) in terms of \( A_1 \):
\[ A_2 = \frac{1}{4} A_1 \]
- Therefore, we can relate \( A_1 \), \( A_2 \), and \( A_3 \):
\[ A_1 \times A_3 = \frac{3}{4} A_1^2 \]
- Since \( A_2 = \frac{1}{4} A_1 \), we can also express \( A_1 \times A_3 \) in terms of \( A_2 \):
\[ A_1 \times A_3 = 3 A_2 \]

7. **Final Conclusion**:
- From the calculations, we conclude that:
\[ A_1 \times A_3 < 16 A_2^2 \]
- Thus, the relationship holds true as per the problem statement.