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A right triangle has sides of length $2\\sqrt{2}$, $6\\sqrt{2}$, and $\\sqrt{80}$ units. What is the area of the triangle, in square units?

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A right triangle has sides of length $2\\sqrt{2}$, $6\\sqrt{2}$, and $\\sqrt{80}$ units. What is the area of the triangle, in square units?

","comment":{"@type":"Comment","text":"

Choice B is correct. The area, $A$, of a triangle can be found using the formula $A=\\frac{1}{2}bh$, where $b$ is the length of the base of the triangle and $h$ is the height of the triangle. It's given that the triangle is a right triangle. Therefore, its base and height can be represented by the two legs. It's also given that the triangle has sides of length $2\\sqrt{2}$, $6\\sqrt{2}$, and $\\sqrt{80}$ units. Since $\\sqrt{80}$ units are the greatest of these lengths, it's the length of the hypotenuse. Therefore, the two legs have lengths $2\\sqrt{2}$ and $6\\sqrt{2}$ units. Substituting these values for $b$ and $h$ in the formula $A=\\frac{1}{2}bh$ gives $A=\\frac{1}{2}\\left(2\\sqrt{2}\\right)\\left(6\\sqrt{2}\\right)$, which is equivalent to $A=6\\sqrt{4}$ square units, or $A=12$ square units.

Choice A is incorrect. This expression represents the perimeter, rather than the area, of the triangle.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

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A right triangle has sides of length $2\\sqrt{2}$, $6\\sqrt{2}$, and $\\sqrt{80}$ units. What is the area of the triangle, in square units?

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

$8\\sqrt{2}+\\sqrt{80}$

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$24\\sqrt{80}$

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"}],"acceptedAnswer":{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

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A right triangle has sides of length $2\sqrt{2}$, $6\sqrt{2}$, and $\sqrt{80}$ units. What is the area of the triangle, in square units?