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# help plz

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In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.

Apr 28, 2024

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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.

Apr 28, 2024