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Diagonal in a rectangle

This topic has 2 expert replies and 2 member replies
isisalaska Community Manager Default Avatar
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Diagonal in a rectangle

Post Sun Jan 28, 2007 3:34 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Hi
    If you have the diagonal in a rectangle cand you get the lengh? Example

    (1) The perimeter of rectangle PQRS is 28 feet.
    (2) Each diagonal of rectangle PQRS is 10 feet long.

    I though it was only with a square. The correct answer is C here

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    Stacey Koprince GMAT Instructor
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    Post Wed Jan 31, 2007 2:08 pm
    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).

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    seshakesav Newbie | Next Rank: 10 Posts Default Avatar
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    Post Tue Feb 06, 2007 4:35 am
    hi

    For this problem as we have two equations and only two variables we can solve it.
    This is the Best way of dealing the Data sufficiency problems as answer is not required for this type of questions

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    Post Tue Feb 06, 2007 12:18 pm
    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).

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    Post Thu Feb 08, 2007 12:36 am
    Stacey is absolutely correct in pointing out that with quadratics you do have to be careful about the two possibilities. In the above data sufficiency question, if the question asked for what the area is then there is only one unique answer because if length = 8 and width =6, then area is 48, and it does not change if length = 6 and width =8.

    What if they asked what is the ratio of the length to width, now we have two possibilities: 8/6 or 6/8, so the answer would be E. However, my observations of questions that involve real physical quantities (length, money, people, etc.) is that I have never come across a question that plays with this angle. ETS/Pearson are mean, but not that mean. They have their certain boundaries and I believe this is one of them.

    On a side, a general rule for rectangles is that if you know any two of the three quantities: area, diagonal, and perimeter, then the third can be obtained from the other two.

    The best way to do this is by using the following algebraic identity:

    (l+w)^2 = l^2 + w^2 + 2lw

    Here (l+w) = P/2 (here P is the perimeter), l^2 + w^2 = d^2 (diagonal square), and lw = A (area).

    Therefore, the identity becomes:

    (P/2)^2 = d^2 + 2A

    The above equation summarizes the relationship between the diagonal, perimeter, and area of any rectangle.

    This approach avoids using the quadratic, which I believe is too time consuming. It is interesting that ETS in the official guide solves this problem via the quadratic approach, which involves solving for l and w, and then finding the area.

    In the given problem, P = 28 and d = 10, therefore:

    (28/2)^2 = 10^2 + 2A

    196 = 100 + 2A
    A = 96/2 = 48

    I have seen couple variants of this same idea on the exam, where the perimeter and diagonal were strange numbers and solving the quadratic was too involved.

    I hope this helps.
    Mark

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