Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm, `angleA=90^(@)` and BC = CD= 52 cm.

Answer

To calculate the area of quadrilateral ABCD where \( AB = 32 \, \text{cm} \), \( AD = 24 \, \text{cm} \), \( \angle A = 90^\circ \), and \( BC = CD = 52 \, \text{cm} \), we can break the quadrilateral into two triangles: \( \triangle ABD \) and \( \triangle BCD \).

### Step-by-Step Solution:

1. **Draw the Quadrilateral**: Start by sketching quadrilateral ABCD with the given dimensions. Mark point A at the origin (0,0), point B at (32,0), point D at (0,24), and point C at an unknown position that maintains the lengths \( BC \) and \( CD \).

2. **Identify the Triangles**: The quadrilateral can be divided into two triangles: \( \triangle ABD \) and \( \triangle BCD \).

3. **Calculate the Area of Triangle ABD**:
- Since \( \angle A = 90^\circ \), triangle ABD is a right triangle.
- The area of a right triangle is given by:
\[ \text{Area}_{ABD} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, \( AB = 32 \, \text{cm} \) (base) and \( AD = 24 \, \text{cm} \) (height).
- Therefore,
\[ \text{Area}_{ABD} = \frac{1}{2} \times 32 \times 24 = \frac{768}{2} = 384 \, \text{cm}^2 \]

4. **Calculate the Length of BD**:
- Use the Pythagorean theorem in triangle ABD to find \( BD \):
\[ BD^2 = AB^2 + AD^2 = 32^2 + 24^2 = 1024 + 576 = 1600 \]
- Thus,
\[ BD = \sqrt{1600} = 40 \, \text{cm} \]

5. **Calculate the Height CE of Triangle BCD**:
- Since \( BC = CD = 52 \, \text{cm} \), triangle BCD is isosceles.
- Let \( E \) be the midpoint of \( BD \). Therefore, \( BE = DE = \frac{BD}{2} = 20 \, \text{cm} \).
- Using the Pythagorean theorem in triangle BCE:
\[ BC^2 = BE^2 + CE^2 \]
\[ 52^2 = 20^2 + CE^2 \implies 2704 = 400 + CE^2 \implies CE^2 = 2304 \]
- Thus,
\[ CE = \sqrt{2304} = 48 \, \text{cm} \]

6. **Calculate the Area of Triangle BCD**:
- The area of triangle BCD is given by:
\[ \text{Area}_{BCD} = \frac{1}{2} \times BD \times CE \]
- Substituting the values:
\[ \text{Area}_{BCD} = \frac{1}{2} \times 40 \times 48 = \frac{1920}{2} = 960 \, \text{cm}^2 \]

7. **Calculate the Total Area of Quadrilateral ABCD**:
- The total area of quadrilateral ABCD is the sum of the areas of triangles ABD and BCD:
\[ \text{Area}_{ABCD} = \text{Area}_{ABD} + \text{Area}_{BCD} = 384 + 960 = 1344 \, \text{cm}^2 \]

### Final Answer:
The area of quadrilateral ABCD is \( 1344 \, \text{cm}^2 \).