The area of a rectangle lies between 40 cm and 45 cm . If one of the sides is 5 cm, then its diagonal lies between

Options

A. 8 cm and 10 cm
B. 9 cm and 10 cm
C. 10 cm and 12 cm
D. 11 cm and 13 cm

Answer

To solve the problem step by step, we need to determine the range of the diagonal of a rectangle given that one side is 5 cm and the area lies between 40 cm² and 45 cm².

### Step 1: Understand the area of a rectangle
The area \( A \) of a rectangle is given by the formula:
\[ A = \text{length} \times \text{width} \]
In this case, we know one side (let's call it the width) is 5 cm.

### Step 2: Set up the inequality for the area
Given that the area lies between 40 cm² and 45 cm², we can express this as:
\[ 40 < 5 \times \text{length} < 45 \]

### Step 3: Solve for the length
To find the range of the length, we can divide the entire inequality by 5:
\[ \frac{40}{5} < \text{length} < \frac{45}{5} \]
This simplifies to:
\[ 8 < \text{length} < 9 \]
So, the length of the rectangle lies between 8 cm and 9 cm.

### Step 4: Calculate the diagonal
The diagonal \( d \) of a rectangle can be calculated using the Pythagorean theorem:
\[ d = \sqrt{(\text{length})^2 + (\text{width})^2} \]
Substituting the known width (5 cm) and the range for the length (between 8 cm and 9 cm), we have:
\[ d = \sqrt{(\text{length})^2 + 5^2} \]

### Step 5: Find the minimum diagonal
To find the minimum diagonal, use the minimum length (8 cm):
\[ d_{\text{min}} = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89} \]

### Step 6: Find the maximum diagonal
To find the maximum diagonal, use the maximum length (9 cm):
\[ d_{\text{max}} = \sqrt{9^2 + 5^2} = \sqrt{81 + 25} = \sqrt{106} \]

### Step 7: Approximate the values
Now we can approximate the values of \( \sqrt{89} \) and \( \sqrt{106} \):
- \( \sqrt{89} \approx 9.43 \)
- \( \sqrt{106} \approx 10.30 \)

### Final Conclusion
Thus, the diagonal of the rectangle lies between approximately 9.43 cm and 10.30 cm.

### Summary
The diagonal lies between:
\[ 9.43 \, \text{cm} < d < 10.30 \, \text{cm} \]