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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
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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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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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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GMAT/MBA Expert

Stacey Koprince GMAT Instructor
Joined
27 Dec 2006
Posted:
2228 messages
Followed by:
680 members
Upvotes:
639
GMAT Score:
780
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).

_________________
Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!

Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT

Contributor to Beat The GMAT!

Learn more about me

  • +1 Upvote Post
  • Quote
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