Find the area of quadrilateral ABCD in which `angleB=90^(@)`, BC = 32 cm, AB = 24 cm and CD=DA = 25 cm.

Answer

To find the area of quadrilateral ABCD where angle B = 90°, BC = 32 cm, AB = 24 cm, and CD = DA = 25 cm, we can break the quadrilateral into two triangles: triangle ABC and triangle ADC.

### Step-by-Step Solution:

1. **Draw the Quadrilateral**:
- Draw quadrilateral ABCD with points A, B, C, and D. Mark angle B as 90°.
- Label the sides: AB = 24 cm, BC = 32 cm, CD = 25 cm, and DA = 25 cm.

2. **Calculate AC using Pythagorean Theorem**:
- Since triangle ABC is a right triangle (angle B = 90°), we can use the Pythagorean theorem to find AC.
- \( AC = \sqrt{AB^2 + BC^2} = \sqrt{24^2 + 32^2} \)
- \( AC = \sqrt{576 + 1024} = \sqrt{1600} = 40 \) cm.

3. **Calculate the Area of Triangle ABC using Heron's Formula**:
- First, find the semi-perimeter (s) of triangle ABC:
\[ s = \frac{AB + BC + AC}{2} = \frac{24 + 32 + 40}{2} = \frac{96}{2} = 48 \text{ cm} \]
- Now apply Heron's formula:
\[ \text{Area}_{ABC} = \sqrt{s(s - AB)(s - BC)(s - AC)} = \sqrt{48(48 - 24)(48 - 32)(48 - 40)} \]
\[ = \sqrt{48 \times 24 \times 16 \times 8} \]
- Calculate:
\[ = \sqrt{48 \times 24 \times 128} = 384 \text{ cm}^2 \]

4. **Calculate the Area of Triangle ADC using Heron's Formula**:
- Find the semi-perimeter (s) of triangle ADC:
\[ s = \frac{AD + CD + AC}{2} = \frac{25 + 25 + 40}{2} = \frac{90}{2} = 45 \text{ cm} \]
- Now apply Heron's formula:
\[ \text{Area}_{ADC} = \sqrt{s(s - AD)(s - CD)(s - AC)} = \sqrt{45(45 - 25)(45 - 25)(45 - 40)} \]
\[ = \sqrt{45 \times 20 \times 20 \times 5} \]
- Calculate:
\[ = \sqrt{45000} = 300 \text{ cm}^2 \]

5. **Calculate the Total Area of Quadrilateral ABCD**:
- The area of quadrilateral ABCD is the sum of the areas of triangles ABC and ADC:
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ADC} = 384 + 300 = 684 \text{ cm}^2 \]

### Final Answer:
The area of quadrilateral ABCD is **684 cm²**.