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To find the area of this quadrilateral, we can break it down into two triangles using diagonal \( AC \) or \( BD \), but first, let's attempt a simpler approach using the formula for a general quadrilateral when one angle is known:

The formula to compute the area of a quadrilateral when two adjacent sides and the included angle are known is:

\[
\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin(\theta)
\]

Where:
- \( a = 4.8 \) yards (side \( AB \))
- \( b = 6.1 \) yards (side \( BC \))
- \( \theta = 108^\circ \) (angle \( \angle ABC \))

### Step 1: Calculate the area of triangle \( ABC \)
We can use the formula for the area of a triangle using two sides and the included angle:

\[
\text{Area of } \triangle ABC = \frac{1}{2} \cdot 4.8 \cdot 6.1 \cdot \sin(108^\circ)
\]

First, compute \( \sin(108^\circ) \). Using a calculator, we get:

\[
\sin(108^\circ) \approx 0.9511
\]

Now, calculate the area of triangle \( ABC \):

\[
\text{Area of } \triangle ABC = \frac{1}{2} \cdot 4.8 \cdot 6.1 \cdot 0.9511 = 13.897 \, \text{square yards}
\]

### Step 2: Find the area of triangle \( ACD \)
For triangle \( ACD \), we would use Heron's formula, as we know all three sides (\( AD = 12.3 \), \( CD = 9.2 \), and diagonal \( AC \) can be calculated using the Law of Cosines).

First, find \( AC \) using the Law of Cosines:

\[
AC^2 = 4.8^2 + 6.1^2 - 2 \cdot 4.8 \cdot 6.1 \cdot \cos(108^\circ)
\]

Substitute the values:

\[
AC^2 = 23.04 + 37.21 - 2 \cdot 4.8 \cdot 6.1 \cdot (-0.309)
\]
\[
AC^2 = 60.25 + 18.21 \approx 78.46
\]
\[
AC \approx \sqrt{78.46} = 8.86 \, \text{yards}
\]

Now, with sides \( AD = 12.3 \), \( CD = 9.2 \), and \( AC = 8.86 \), apply Heron’s formula for triangle \( ACD \).

First, compute the semi-perimeter \( s \):

\[
s = \frac{12.3 + 9.2 + 8.86}{2} = 15.18
\]

Now, the area \( A \) of triangle \( ACD \) is:

\[
A = \sqrt{s(s - 12.3)(s - 9.2)(s - 8.86)}
\]
\[
A = \sqrt{15.18(15.18 - 12.3)(15.18 - 9.2)(15.18 - 8.86)}
\]
\[
A = \sqrt{15.18 \cdot 2.88 \cdot 5.98 \cdot 6.32}
\]
\[
A \approx \sqrt{15.18 \cdot 108.69} = \sqrt{1650.85} \approx 40.63 \, \text{square yards}
\]

### Step 3: Add the areas of the two triangles
Now, sum the areas of triangles \( ABC \) and \( ACD \):

\[
\text{Total Area} = 13.897 + 40.63 = 54.53 \, \text{square yards}
\]

Thus, the area of the quadrilateral is approximately **54.53 square yards**.

Would you like any further details or clarifications?

### Related Questions:
1. What is the Law of Cosines, and how can it be applied to find unknown sides of triangles?
2. How does Heron's formula work for calculating the area of a triangle with known sides?
3. How would the result change if the given angle was less than \( 90^\circ \)?
4. How do sine and cosine rules help in solving non-right-angled triangles?
5. Can the same method be applied to other quadrilaterals, such as rectangles or parallelograms?

### Tip:
When dealing with irregular quadrilaterals, breaking them into triangles and applying known formulas simplifies the calculation of the area.

What is the area of the quadrilateral with sides 4.8 yards, 6.1 yards, 9.2 yards, and 12.3 yards, and an angle of 108° between two of the sides?

Learn how to find the area of a quadrilateral with sides 4.8, 6.1, 9.2, and 12.3 yards, and an included angle of 108°. This step-by-step solution uses trigonometry and Heron's formula for area calculation. Suitable for students in Grades 10-12.

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