Find the area of quadrilateral ABCD in which AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and `angleABC = 90^(@)`.

Options

A. `300cm^(2)`
B. `180cm^(2)`
C. `126cm^(2)`
D. `306cm^(2)`

Answer

To find the area of quadrilateral ABCD, we can break it down into two triangles: triangle ABC and triangle ACD. Given that angle ABC is 90 degrees, we can easily calculate the area of triangle ABC using the formula for the area of a right triangle.

### Step-by-Step Solution:

1. **Identify the dimensions of triangle ABC:**
- AB = 9 cm (base)
- BC = 40 cm (height)

2. **Calculate the area of triangle ABC:**
\[ \text{Area}_{ABC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times 40 \]
\[ \text{Area}_{ABC} = \frac{1}{2} \times 360 = 180 \text{ cm}^2 \]

3. **Use the Pythagorean theorem to find AC:**
- AC is the hypotenuse of triangle ABC.
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \text{ cm} \]

4. **Identify the dimensions of triangle ACD:**
- AD = 15 cm
- CD = 28 cm
- AC = 41 cm

5. **Calculate the semi-perimeter (s) of triangle ACD:**
\[ s = \frac{AD + CD + AC}{2} = \frac{15 + 28 + 41}{2} = \frac{84}{2} = 42 \text{ cm} \]

6. **Use Heron's formula to find the area of triangle ACD:**
\[ \text{Area}_{ACD} = \sqrt{s(s - AD)(s - CD)(s - AC)} \]
\[ = \sqrt{42(42 - 15)(42 - 28)(42 - 41)} \]
\[ = \sqrt{42 \times 27 \times 14 \times 1} \]
\[ = \sqrt{42 \times 27 \times 14} \]

To simplify:
\[ 42 = 14 \times 3 \quad \text{and} \quad 27 = 9 \times 3 \]
\[ = \sqrt{(14 \times 14) \times (3 \times 3) \times 1} = \sqrt{196 \times 9} = 14 \times 3 = 42 \text{ cm}^2 \]

7. **Calculate the total area of quadrilateral ABCD:**
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = 180 + 42 = 222 \text{ cm}^2 \]

### Final Result:
The area of quadrilateral ABCD is **222 cm²**.