In a rectangle ABCD, its diagonal AC = 15 cm and `angleACD = alpha` If `cot alpha = 3/2` , find the perimeter and the area of the rectangle.

Answer

To solve the problem, we need to find the perimeter and area of rectangle ABCD given that the diagonal AC = 15 cm and cot(α) = 3/2, where α is the angle ∠ACD.

### Step-by-step Solution:

1. **Identify the sides using cotangent**:
Given that cot(α) = 3/2, we can express the sides of the rectangle in terms of a variable \( k \).
- Let \( CD = 3k \) (base) and \( AD = 2k \) (perpendicular).

2. **Use the Pythagorean theorem**:
In triangle ACD, we can apply the Pythagorean theorem:
\[ AC^2 = AD^2 + CD^2 \]
Substituting the known values:
\[ 15^2 = (2k)^2 + (3k)^2 \]
This simplifies to:
\[ 225 = 4k^2 + 9k^2 \]
\[ 225 = 13k^2 \]

3. **Solve for \( k^2 \)**:
Rearranging gives:
\[ k^2 = \frac{225}{13} \]

4. **Calculate \( k \)**:
Taking the square root:
\[ k = \frac{15}{\sqrt{13}} \]

5. **Find the lengths of the sides**:
Now we can find the lengths of the sides:
- Length \( AD = 2k = 2 \times \frac{15}{\sqrt{13}} = \frac{30}{\sqrt{13}} \)
- Width \( CD = 3k = 3 \times \frac{15}{\sqrt{13}} = \frac{45}{\sqrt{13}} \)

6. **Calculate the perimeter**:
The perimeter \( P \) of the rectangle is given by:
\[ P = 2 \times (AD + CD) = 2 \times \left(\frac{30}{\sqrt{13}} + \frac{45}{\sqrt{13}}\right) \]
Simplifying this:
\[ P = 2 \times \frac{75}{\sqrt{13}} = \frac{150}{\sqrt{13}} \]

7. **Calculate the area**:
The area \( A \) of the rectangle is given by:
\[ A = AD \times CD = \left(\frac{30}{\sqrt{13}}\right) \times \left(\frac{45}{\sqrt{13}}\right) \]
This simplifies to:
\[ A = \frac{1350}{13} \text{ cm}^2 \]

### Final Answers:
- **Perimeter**: \( \frac{150}{\sqrt{13}} \) cm
- **Area**: \( \frac{1350}{13} \) cm²