Diagonal in a rectangle

You haven't actually typed the question stem itself, so I'll show you what you can calculate and you'll have to figure out whether that answers the question.

Please also remember to give the source of any questions you post. Thanks!

let's call the length "x" and the width "y" for the rectangle.

Statement 1 lets you write: 2x + 2y = 28, or x + y = 14

Statement 2 lets you say that the diagonal is 10

1 AND 2 gives you:
x + y = 14
x^2 + y^2 = 10^2

You have two equations and two variables, so you can try to solve.
x = 14 - y
(14 - y)^2 + y^2 = 10^2 (substitute in for x)
196 - 28y + y^2 + y^2 = 100
2y^2 - 28y + 96 = 0
y^2 - 14y + 48 = 0
(y-6)(y-8) = 0
so, y = 6 or 8

If y is 6, x is 8. If y is 8, x is 6. If the question stem tells you one is longer, or tells you which one is the length (longer) or width (shorter), then you can tell the length of each individual side. Or, if you're not asked to specify which is which, you can tell that the lengths of the two sides are 6 and 8 - whichever way you arrange it.

I also recommend that you memorize the common right triangles. 6-8-10 is the second most common and if you say that the diagonal was 10, you could have just tried 6 and 8 in your two formulas to see if those were in fact the answers (rather than having to solve the quadratic).

On this test, you can't just assume the "2 equations, 2 variables" rule we learned in high school. You do have to check a couple of things because they try to trick you on harder questions. If one (or both) of the equations contains a quadratic, you probably have multiple answers, which obviously means you don't have one definitive answer.

They will also write an equation to disguise the fact that it's either (a) identical to one they've already given you (and therefore useless) or (b) a quadratic that doesn't look like a quadratic at first glance (in which case, see above).