# 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?

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The phrase "with a perimeter of 48 inches" describes the rectangle; the phrase "with legs of 12 inches and 24 inches" describes the right triangle. If you hide these phrases, then you can read the question like this: "If this rectangle is equal in area to that right triangle, then what is the rectangle's diagonal?" Does that help? – Blue Jun 17 '12 at 17:42
The legs part is throwing me off too .. are they perpendicular,base or the hypotenuse ? – MistyD Jun 17 '12 at 17:46

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).

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Okay what about the legs of the triangle ? How do we know are they the base or height etc ? – MistyD Jun 17 '12 at 17:42
Draw a right triangle - do you see why you it doesn't matter which leg you take be the height, and which to be the base? The area of a right triangle with legs of length $a$ and $b$ is $$\frac{1}{2}ab$$ so the area of the triangle mentioned in the question is $$\frac{1}{2}\times (12\text{ inches})\times(24\text{ inches})=144\text{ sq. inches}.$$ – Zev Chonoles Jun 17 '12 at 17:45
After solving this problem I get the answer but it turns out its a square with sides 12 and not a rectangle. Am i wrong ? – MistyD Jun 17 '12 at 17:56
Regarding base vs. height: There's an old joke about a guy standing out in a field, trying and trying to push a tape measure up a pole, but failing with each attempt. After watching the guy struggle for a while, his buddy asks, "Why don't you just knock the pole over so that you can easily measure it along the ground?" The guy says, "Because I want to know how tall it is, not how long it is!" – Blue Jun 17 '12 at 17:56
@MistyD: Nope, you did it exactly right! I've added some more to my answer in response. – Zev Chonoles Jun 17 '12 at 18:12