A park is in the shape of quadrilateral ABCD in which AB = 9 cm, BC = 12 cm, CD = 5 cm, AD = 8 cm and `angle C = 90^(@)`. Find the area of the park.

Answer

To find the area of the quadrilateral ABCD, we can break it down into two triangles: triangle BCD and triangle ABD. Given the dimensions and the right angle at C, we can calculate the area step by step.

### Step 1: Identify the dimensions and structure
We have:
- AB = 9 cm
- BC = 12 cm
- CD = 5 cm
- AD = 8 cm
- Angle C = 90°

### Step 2: Calculate the length of diagonal BD
Since triangle BCD is a right triangle (angle C = 90°), we can use the Pythagorean theorem to find the length of diagonal BD.

Using the formula:
\[ BD = \sqrt{BC^2 + CD^2} \]

Substituting the values:
\[ BD = \sqrt{12^2 + 5^2} \]
\[ BD = \sqrt{144 + 25} \]
\[ BD = \sqrt{169} \]
\[ BD = 13 \, \text{cm} \]

### Step 3: Calculate the area of triangle BCD
The area of triangle BCD can be calculated using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, we can take:
- Base = CD = 5 cm
- Height = BC = 12 cm

So,
\[ \text{Area}_{BCD} = \frac{1}{2} \times 5 \times 12 \]
\[ \text{Area}_{BCD} = \frac{1}{2} \times 60 \]
\[ \text{Area}_{BCD} = 30 \, \text{cm}^2 \]

### Step 4: Calculate the area of triangle ABD using Heron's Formula
First, we need to calculate the semi-perimeter (s) of triangle ABD:
\[ s = \frac{AB + BD + AD}{2} \]
\[ s = \frac{9 + 13 + 8}{2} \]
\[ s = \frac{30}{2} \]
\[ s = 15 \, \text{cm} \]

Now we can apply Heron's formula:
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
Where:
- a = AB = 9 cm
- b = BD = 13 cm
- c = AD = 8 cm

Substituting the values:
\[ \text{Area}_{ABD} = \sqrt{15(15-9)(15-13)(15-8)} \]
\[ \text{Area}_{ABD} = \sqrt{15 \times 6 \times 2 \times 7} \]
\[ \text{Area}_{ABD} = \sqrt{15 \times 84} \]
\[ \text{Area}_{ABD} = \sqrt{1260} \]
\[ \text{Area}_{ABD} \approx 35.5 \, \text{cm}^2 \]

### Step 5: Calculate the total area of quadrilateral ABCD
Now, we can find the total area of quadrilateral ABCD by adding the areas of triangles BCD and ABD:
\[ \text{Area}_{ABCD} = \text{Area}_{BCD} + \text{Area}_{ABD} \]
\[ \text{Area}_{ABCD} = 30 + 35.5 \]
\[ \text{Area}_{ABCD} \approx 65.5 \, \text{cm}^2 \]

### Final Answer
The area of the park (quadrilateral ABCD) is approximately **65.5 cm²**.