help plz

To find the area of trapezoid ABCD, we can use the formula for the area of a trapezoid, which is given by:

$$\text{Area} = \frac{1}{2} \times (\text{sum of the lengths of the bases}) \times (\text{height})$$

In this case, the bases are AB and CD, and the height can be found using trigonometry.

Given:
- AB = 96
- CD = 44
- ∠D = 110°
- ∠B = 55°

First, let's find the height of the trapezoid, which is the perpendicular distance between AB and CD. We can use the law of sines to find the height.

$$\frac{\sin(55°)}{CD} = \frac{\sin(110°)}{AB}$$

$$\frac{\sin(55°)}{44} = \frac{\sin(110°)}{96}$$

Now, let's solve for the height:

$$\text{Height} = \frac{\sin(55°) \times 44}{\sin(110°)}$$

$$\text{Height} \approx \frac{0.8192 \times 44}{1}$$

$$\text{Height} \approx 36$$

Now, we have the height. Let's calculate the area using the formula for the area of a trapezoid:

$$\text{Area} = \frac{1}{2} \times (AB + CD) \times \text{Height}$$

$$\text{Area} = \frac{1}{2} \times (96 + 44) \times 36$$

$$\text{Area} = \frac{1}{2} \times 140 \times 36$$

$$\text{Area} = 70 \times 36$$

$$\text{Area} = 2520 \, \text{square units}$$

So, the area of trapezoid ABCD is 2520 square units.