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.

Options

A. `65.4 cm^(2)`
B. `69.4 cm^(2)`
C. `66.4 cm^(2)`
D. `68.4 cm^(2)`

Answer

To find the area of the park shaped as quadrilateral ABCD, we can divide it into two triangles: triangle BCD and triangle ABD. Given the dimensions and angle, we can use the following steps:

### Step 1: Identify the triangles
We have quadrilateral ABCD with:
- AB = 9 cm
- BC = 12 cm
- CD = 5 cm
- AD = 8 cm
- Angle C = 90°

We can divide the quadrilateral into two triangles: triangle BCD and triangle ABD.

### Step 2: Calculate the area of triangle BCD
Triangle BCD is a right triangle (since angle C is 90°). The area of a right triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take CD as the base and BC as the height:
\[ \text{Area of } \triangle BCD = \frac{1}{2} \times CD \times BC = \frac{1}{2} \times 5 \times 12 = \frac{60}{2} = 30 \text{ cm}^2 \]

### Step 3: Calculate the length of diagonal BD using Pythagoras theorem
In triangle BCD:
\[ BD^2 = BC^2 + CD^2 \]
Substituting the values:
\[ BD^2 = 12^2 + 5^2 = 144 + 25 = 169 \]
Taking the square root:
\[ BD = \sqrt{169} = 13 \text{ cm} \]

### Step 4: Calculate the area of triangle ABD using Heron's formula
First, we need to find the semi-perimeter (s) of triangle ABD:
\[ s = \frac{AB + AD + BD}{2} = \frac{9 + 8 + 13}{2} = \frac{30}{2} = 15 \text{ cm} \]
Now, we can use Heron's formula to find the area:
\[ \text{Area} = \sqrt{s(s - AB)(s - AD)(s - BD)} \]
Substituting the values:
\[ \text{Area} = \sqrt{15(15 - 9)(15 - 8)(15 - 13)} = \sqrt{15 \times 6 \times 7 \times 2} \]
Calculating inside the square root:
\[ = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{1260} \]
Breaking it down:
\[ = \sqrt{15 \times 6 \times 14} = \sqrt{15 \times 84} = \sqrt{1260} \]
Calculating further:
\[ = 6\sqrt{35} \text{ cm}^2 \]

### Step 5: Calculate the total area of the park
Now, we add the areas of triangles BCD and ABD:
\[ \text{Total Area} = \text{Area of } \triangle BCD + \text{Area of } \triangle ABD = 30 + 6\sqrt{35} \text{ cm}^2 \]

### Step 6: Approximate the area
Using the approximate value of \(\sqrt{35} \approx 5.91\):
\[ 6\sqrt{35} \approx 6 \times 5.91 \approx 35.46 \]
Thus,
\[ \text{Total Area} \approx 30 + 35.46 = 65.46 \text{ cm}^2 \]

### Final Answer
The area of the park is approximately \(65.46 \text{ cm}^2\).

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