Rectangles Diagonal Calculation I was having a problem with the following question, and could use some help:

If a rectangle with a perimeter of 48 inches is equal in area to a right triangle with legs of 12 inches and 24 inches, what is the rectangle's diagonal?

The answer to the above question is $12\sqrt{2}$. Frankly I am a bit confused by a part of the question stating

in area to a right triangle with legs of 12 inches and 24 inches

Is the area of rectangle equal to the area of the triangle? Frankly, the phrase "equal to triangle" doesn't say much. And 12 and 24 the lengths of which sides of a triangle?
Am I missing something here or are my concerns valid?
Edit:
After reading the suggestions posted here, here is what I did, and I am getting a square instead of a rectangle:
$$2x + 2y = 48 \qquad (A)$$
$$xy = 144 \qquad (B)$$ so $x=144/y$, inserting in $(A)$ I get $x=12$. So is this actually a square?
 A: There is a specific rectangle that the question is referring to, but we don't know all of its properties. The question tells us that this rectangle has a perimeter of 48 inches. When the question says that this rectangle is 

equal in area to a right triangle with legs of 12 inches and 24 inches 

it just means that the area of this rectangle is equal to the area of such a triangle.

The area of a shape doesn't depend on how you orient it. If you look at a right triangle that has legs of lengths $A$ and $B$, and draw it like this:

then the "height" is $B$, the "base" is $A$, and the area is $$\frac{1}{2}\times \text{base}\times\text{height}=\frac{1}{2} AB.$$
If you take the same right triangle and draw it like this:

then the "height" is $A$, the "base" is $B$, and the area is $$\frac{1}{2}\times \text{base}\times\text{height}=\frac{1}{2} BA=\frac{1}{2}AB.$$
So the area of a right triangle that has legs of length $A$ and $B$ is always $\frac{1}{2}AB$.

Your calculations are correct; the rectangle is a square. Note that a square is a quadrilateral with four right angles, and therefore is a rectangle (all squares are rectangles, though of course not all rectangles are squares).
Now, it should be fairly straightforward to show that, for a square with side lengths of $12$ inches, the length of the diagonal is $12\sqrt{2}$ inches (hint: use the Pythagorean theorem).
