In rectangle ABCD, diagonal BD = 26 cm and cotangent of angle ABD = 1.5. Find the area and the perimeter of the rectangle ABCD.

Answer

To solve the problem step by step, we will use the information given about rectangle ABCD, the diagonal BD, and the cotangent of angle ABD.

### Step 1: Understand the Given Information
We have a rectangle ABCD with diagonal BD = 26 cm and cotangent of angle ABD = 1.5. We need to find the area and perimeter of the rectangle.

### Step 2: Draw the Rectangle
Draw rectangle ABCD and label the corners A, B, C, and D. The diagonal BD divides the rectangle into two right triangles, ABD and BCD.

### Step 3: Define the Angles and Sides
Let:
- \( L \) = length of rectangle (AB)
- \( B \) = breadth of rectangle (AD)
- \( \theta \) = angle ABD

Given:
- \( \cot \theta = 1.5 \)

### Step 4: Use the Cotangent Definition
The cotangent of an angle is defined as:
\[ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{L}{B} \]
From the given cotangent value:
\[ \frac{L}{B} = 1.5 \quad \text{or} \quad L = 1.5B \]

### Step 5: Use the Diagonal Length
In triangle ABD, we can apply the Pythagorean theorem:
\[ BD^2 = AB^2 + AD^2 \]
Substituting the known values:
\[ 26^2 = L^2 + B^2 \]
\[ 676 = L^2 + B^2 \]

### Step 6: Substitute for Length
Substituting \( L = 1.5B \) into the Pythagorean theorem:
\[ 676 = (1.5B)^2 + B^2 \]
\[ 676 = 2.25B^2 + B^2 \]
\[ 676 = 3.25B^2 \]

### Step 7: Solve for Breadth
Now, solve for \( B^2 \):
\[ B^2 = \frac{676}{3.25} \]
Calculating this gives:
\[ B^2 = 208 \]
Taking the square root:
\[ B = \sqrt{208} = 4\sqrt{13} \, \text{cm} \]

### Step 8: Find Length
Now substitute back to find \( L \):
\[ L = 1.5B = 1.5 \times 4\sqrt{13} = 6\sqrt{13} \, \text{cm} \]

### Step 9: Calculate Area
The area \( A \) of the rectangle is given by:
\[ A = L \times B = (6\sqrt{13}) \times (4\sqrt{13}) = 24 \times 13 = 312 \, \text{cm}^2 \]

### Step 10: Calculate Perimeter
The perimeter \( P \) of the rectangle is given by:
\[ P = 2(L + B) = 2(6\sqrt{13} + 4\sqrt{13}) = 2(10\sqrt{13}) = 20\sqrt{13} \, \text{cm} \]

### Final Answers
- Area = \( 312 \, \text{cm}^2 \)
- Perimeter = \( 20\sqrt{13} \, \text{cm} \)