To determine the area of the right triangle with sides [tex]\(2 \sqrt{2}\)[/tex], [tex]\(6 \sqrt{2}\)[/tex], and [tex]\(\sqrt{80}\)[/tex] units, we can follow these steps:
1. Identify which sides are the legs and the hypotenuse:
- Because this is a right triangle, one of the sides must be the hypotenuse. Typically, the hypotenuse is the longest side. We compare [tex]\(2 \sqrt{2}\)[/tex], [tex]\(6 \sqrt{2}\)[/tex], and [tex]\(\sqrt{80}\)[/tex].
- [tex]\(2 \sqrt{2} \approx 2.828\)[/tex]
- [tex]\(6 \sqrt{2} \approx 8.485\)[/tex]
- [tex]\(\sqrt{80} \approx 8.944\)[/tex]
- Thus, [tex]\(\sqrt{80}\)[/tex] (approximately 8.944) is the hypotenuse, and therefore, [tex]\(2 \sqrt{2}\)[/tex] and [tex]\(6 \sqrt{2}\)[/tex] must be the legs of the triangle.
2. Calculate the area of the right triangle:
- For a right triangle, the area can be calculated using the formula:
[tex]\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\][/tex]
- Here, the base and height correspond to the legs of the triangle, which are [tex]\(2 \sqrt{2}\)[/tex] and [tex]\(6 \sqrt{2}\)[/tex].
3. Substitute the lengths of the legs into the area formula and compute:
[tex]\[
\text{Area} = \frac{1}{2} \times 2 \sqrt{2} \times 6 \sqrt{2}
\][/tex]
- First, calculate the product of the legs:
[tex]\[
2 \sqrt{2} \times 6 \sqrt{2} = (2 \times 6) \times (\sqrt{2} \times \sqrt{2}) = 12 \times 2 = 24
\][/tex]
- Then, take half of this product:
[tex]\[
\frac{1}{2} \times 24 = 12
\][/tex]
So, the area of the right triangle is [tex]\(\boxed{12}\)[/tex].
