What is the area of rectangular region R?

This is a common trap.

Knowing that the hypotenuse equals to 5 does NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: \(x^2+y^2=5^2\) does NOT mean that \(x=3\) and \(y=4\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=5^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=3\) and \(y=5\).

For example: \(x=1\) and \(y=\sqrt{24}\) or \(x=2\) and \(y=\sqrt{21}\)...

I collected the questions which use this trap:
https://gmatclub.com/forum/the-circular- ... 67645.html
https://gmatclub.com/forum/figure-abcd-i ... 48899.html
https://gmatclub.com/forum/in-right-tria ... 63591.html
https://gmatclub.com/forum/m22-73309-20.html
https://gmatclub.com/forum/points-a-b-an ... 84423.html
https://gmatclub.com/forum/if-vertices-o ... 59-20.html
https://gmatclub.com/forum/if-p-is-the-p ... 35832.html
https://gmatclub.com/forum/if-the-diagon ... 04205.html
https://gmatclub.com/forum/what-is-the-a ... 05414.html
https://gmatclub.com/forum/what-is-the-p ... 96381.html
https://gmatclub.com/forum/pythagorean-t ... 31161.html
https://gmatclub.com/forum/given-that-ab ... 27051.html
https://gmatclub.com/forum/m13-q5-69732-20.html#p1176059
https://gmatclub.com/forum/m20-07-triang ... 71559.html
https://gmatclub.com/forum/what-is-the-p ... 96381.html

Hope this helps.
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SOLUTION

What is the area of rectangular region R?

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we cannot get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Answer: C.
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Bunuel,

isn't this a special Pythagorean triangle? with a,b,c 3,4,5? Hence it would be A not C...

Is there a quick rule of thumb to know that we can't get the value of \(xy\) from \(x^2+y^2=5^2\)? (I find myself trying to think of multiple combinations of x and y that would fulfill the equation, but that tends to take quite a bit of time.)

Hi swaggerer,

The rule that you're describing is actually an Algebra rule. If you have just one equation (with no other restrictions) and two variables, then there will be more than one set of values that will "fit" the variables.

Here's a simple example:

X + Y = 10

What COULD (X)(Y) equal? Come up with a few examples.

Note: there are no other restrictions here, so X and Y could be non-integers, 0, negatives, square roots, etc.

You'll come to find that there are LOTS of possible values for (X)(Y). That same concept exists in this prompt.

GMAT assassins aren't born, they're made,
Rich
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We're asked to find the area of a rectangle, which means we need to know its exact length and width.

Fact 1: The diagonal is 5.

You are correct that the Pythagorean Theorem is appropriate here:

L^2 + W^2 = 5^2

HOWEVER, we were NOT told that the two dimensions have to be integers.

The two dimensions COULD be 3 and 4, in which case the answer to the question is (3)(4) = 12
The two dimensions COULD also be (Root1) and (Root24), in which case the answer to the question is (Root24).
Fact 1 is INSUFFICIENT

Fact 2: The perimeter of the rectangle is 14

Perimeter = 2L + 2W = 14

This tells us that L + W = 7

The two dimensions COULD be 3 and 4, in which case the answer to the question is (3)(4) = 12
The two dimensions COULD be 2 and 5, in which case the answer to the question is (2)(5) = 10
Fact 2 is INSUFFICIENT

Combined, we know....
L^2 + W^2 = 25
L + W = 7

Since we cannot have "negative side lengths", we now have a "system" of equations (2 variables and 2 unique equations), so we CAN solve for the two dimensions. With these rules, the dimensions can only be 3 and 4.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Can the triangle made by the diagonal be considered as a 30-60-90 degree triangle? If yes, then the length and width can be found out by using the 1 : root(3) : 2 ratios for the 3 sides of the triangle. And hence, answer choice A would be sufficient.

Hi saar,

Unfortunately, we do NOT know what the angles will be when we draw a diagonal line through that rectangle. While it's possible that the resulting triangles could 30/60/90s, it's also possible (and highly likely) that they will be some OTHER type of triangle. As such, neither of the two Facts provides enough information to answer the question on its own.

GMAT assassins aren't born, they're made,
Rich
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Bunnuel - Why isn't it B then ? I marked it as D. However, after you explained, I understood why A isn't valid. But why isn't it B?

2L + 2B = 14
L + B = 7
The only combination that suits this is 4 and 3, and hence we can find the area?

Also, given the logic you explained for Statement A, the reverse logic has been used in some of the problems in Manhattan GMAT and GMAT Prep. How can we solve this ? Thanks.

Hi torreadortorment,

Using the information in Fact 2, you've correctly deduced that...

L+B = 7

The question asks for the AREA of the rectangle though...

What's the area when....
L = 1, B = 6
L = 2, B = 7
L = 2.5, B = 4.5
L = 3, B = 4
Etc.
If the area changes, then the Fact is INSUFFICIENT.

Once you COMBINE the two Facts though, then you know that you're dealing with a 3/4/5.

GMAT assassins aren't born, they're made,
Rich
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

What is the area of rectangular region R?

(1) Each diagonal of R has length 5.
(2) The perimeter of R is 14.

There are 2 variables (width, length) and 2 equations are given by the conditions, so there is high chance (C) will be the answer,
Looking at the conditions together,
l^2+w^2=5^2, 2(l+w)=14
(l,w)=(3,4)=(4,3)
area=3*4=12.
This is sufficient and the answer becomes (C).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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I have one question here. let's say if the hypotenuse of any right angle triangle is 5 then can I safely assume the lengths of other two sides are 3 and 4? If not, why ? could some explain in detail examples.

Hi santro789,

The simple answer to your question is NO. The reason why is because there are two variables that you have to account for:

(A^2) + (B^2) = 25

There are LOTS of different combinations for A and B that would complete this equation.

eg.
A = 3, B = 4
A = 1, B = (root24)
A = 2, B = (root21)
A = (root3), B = (root22)
Etc.

Now, IF you have specific types of additional information, then that could limit the possibilities. For example, IF we're told that A and B are both INTEGERS, then the triangle would have to be a 3/4/5 right triangle (although we wouldn't know which variables was the 3 and which was the 4).

GMAT assassins aren't born, they're made,
Rich
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This one doesn't make sense to me. It says that both diagonals are of equal length. That is only possible if the rectangular is a square (a special form of a rectangle). If each diagonal is 5, then each side of the square can be calculated as root of (2.5^2+2.5^2). If each side can be determined, it is sufficient (A).

What am I missing.

1. All rectangles, not necessarily square, have equal diagonals.
2. If the diagonal of a square is 5, then x^2 + x^2 = 5^2, not what you've written.
3. You should brush-up fundamentals. Check the the links here: https://gmatclub.com/forum/what-is-the- ... l#p1903104

Hope it helps.
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I had read a formula somewhere which said if any quadrilateral has diagonals intersecting at 90 degrees, then their area can be found out by 1/2 * d1 * d2.
If we use this formula, then A alone is sufficient to answer the question.
Where am I going wrong Bunuel ? Is that formula right?
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Do we know that the diagonals intersect at 90 degrees from (1)?
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You're right, "rectangular" doesn't necessarily mean a rectangle. Missed that.
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No, that's not the point. Rectangular region does mean that the figure is a rectangle. But the diagonals of a rectangle do not always intersect at 90 degrees. This happnes only when a rectangle is a square.
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