Hi, and welcome to Gmat Club.

The question you posted can be solved as follows:
Given:
(1) \(2x+2y=560\) (perimeter) --> \(x+y=280\)
(2) \(x^2+y^2=200^2\) (diagonal, as per Pythagoras).

Question: \(xy=?\)

Square (1) --> \(x^2+2xy+y^2=280^2\). Now subtract (2) fro this: \((x^2+2xy+y^2)-(x^2+y^2)=280^2-200^2\) --> \(2xy=(280-200)(280+200)\) --> \(2xy=80*480\) --> \(xy=40*480=19200\).

Answer: A.

Hope it helps.
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The fastest way:

we know 560 is an integer (no fraction) and hence probability of sides of rectangle being integer is quite high
200 is diagonal - recognizing it from pythagorean patterns it seems to be a multiple of 10 (10 - 8 - 6)
= 10 * 2 * 10
hence other sides of the pythagorean triplet will be:
8 * 2 * 10
and
6 * 2 * 10
= 160
= 120
bingo - (160 + 120 ) * 2 = 560
hence area = 160 * 120 = 19200

I have a small confusion. The diagonal is given and it divides the rectangle into two 30-60-90 triangles. Can't we find the measure of two other sides? What am i missing here???

great explanation thanks

Does the diagonal of a rectangle always divide it into two 30-90-60 triangles ? Think again...!!!

Yeah there has to be a shorter way to solve this one but where did you get the 10*2*10 stuff for each side on the (10 - 8 - 6)?

No I don't think so buddy, there's nothing clear when it is a rectangle. If its a square I'm pretty sure that the two right triangles that are divided by the diagonal are in fact two 45-45-90 triangles, but with the rectangle I don't think one can be sure about the angle

Hope this helps
Cheers
J

It is a really good exercise.

Solving this one in a standard way brings a lot of confusion.
In my opinion it is a helpful here to know, that a square has a biggest area if the sum of length of bases is the same.
For example: If sum of bases is 8, the biggest possible area is 16, which is a square. (Other options 5x3=15; 6x2=12; 8x1=8)

Knowing this, we could easily eliminate answer choices that are other 200 because we know that diagonal is 200 and therefore the maximum area is 200^2/2
Moreover, knowing the rule of square diagonal we could remove 200, because square diagonal would be base * sqrt(2) (90-45-45 formula)

And also 196 is a (14*14) so this means that diagonal is not integer and we could remove it also.

Hence, one answer choice is left:)

A :D

Say length of the park = x, width = (280-x)

\(200^2 = x^2 + (280-x)^2\)

\(2x^2 - 560x + (280^2 - 200^2) = 0\)

\(2x^2 - 560x + (280 + 200)(280 - 200) = 0\)

\(2x^2 - 560x + 480 * 80 = 0\)

\(x^2 - 280x + 480*40 = 0\)

Dimensions = 160 & 120

Area = 19200

Answer = A

great solution, thanks. just wondering, why did you decide to square \(x+y=280\)? did you do it so that you could subtract \(x^2+y^2=200^2\) from it?

Yes. I wanted to get xy in the end.
_________________

you can avoid a lot of work in this problem by recognizing that, with the info provided, the diagonal forms a triangle inside the rectangle with sides that have a 3:4:5 ratio.


diagonal = 200
2x + 2y = 560, or x + y = 280
a^2 + b^2 = c^2 for each the sides of the triangle

using the ratio 3:4:5 for sides, and knowing c = 200, you can deduce the following

a=120
b=160

160x120=19,200

A is the answer.

as the user above, how can you deduce that the ratio of the the sides is 3:4:5 ? (rather than just assuming and hoping to be right)
is it always the case? will a diagonal cut the rectangle in these proportions?

I just need to say that I love the elegance of this answer. It's beautiful... Thanks, Bunuel.

Hi All,

It's important to remember that nothing about a GMAT question is ever 'random' - the wording and numbers/data are always carefully chosen. Thus, you can sometimes use the 'design' of a prompt to your advantage and spot the built-in patterns that are often there.

Here, notice how ALL of the numbers are relatively 'nice', round numbers - even the diagonal is a nice number (and that doesn't happen very often when dealing with rectangles).... Since the answer choices are also round numbers, it's likely that the triangles that are 'hidden' in this rectangle are based on one of the common right-triangle patterns (in this case, the 3/4/5 - since 200 is a multiple of 5).

Using that knowledge to our advantage, IF we had a 3/4/5 and the diagonal was 200, then that would be '40 times' 5... so the other two sides would be 40 times 4 and 40 times 3: 160 and 120. With those two side lengths, we'd have a perimeter of 2(160) + 2(120) = 560... and that is an exact MATCH for what we were told, so this MUST be the situation that we're dealing with.

At this point, the area can be calculated easily enough: (L)(W) = (160)(120) = 19,200

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

while you got the answer correct here, i wonder why you assume that just because 200 is a multiple of 5, that there is a hidden 3-4-5 triangle? does this need be true? is this based on a property/rule that I'm overlooking?

Hi LakerFan24,

GMAT questions are almost always built around a pattern (and sometimes more than one pattern), so part of approaching questions efficiently is to think about the inherent patterns that could be involved. In addition, the 'world of math' is full of patterns that you can take advantage of (Number Properties, formulas, etc.).

Here, we're dealing with a rectangular park, so if you 'cut' the park from corner to opposite-corner, you'll end up with two right triangles. As a thought experiment, I want you to choose two random numbers for the two legs of the right triangle. Using the Pythagorean Theorem (A^2 + B^2 = C^2), you can calculate the diagonal. Now, how often do you actually end up with an INTEGER for that diagonal. Choose a few pairs of random numbers for the legs and you'll see that usually the diagonal is NOT an INTEGER... but here, the diagonal IS an integer (its length is 200). The answer choices to this question are all integers, meaning that the two sides of the rectangle are almost certainly integers. So, we're dealing with triangles that have integers for ALL 3 SIDES. By extension, that means the two triangles are likely one of the common 'magic' Pythagorean Triplets (the 3/4/5, 5/12/13 or a multiple of one them - since those triplets have a diagonal side that's an integer).

GMAT assassins aren't born, they're made,
Rich
_________________

Let L and W equal the length and width of the rectangle respectively.

perimeter = 560
So, L + L + W + W = 560
Simplify: 2L + 2W = 560
Divide both sides by 2 to get: L + W = 280

diagonal = 200
The diagonal divides the rectangle into two RIGHT TRIANGLES.
Since we have right triangles, we can apply the Pythagorean Theorem.
We get L² + W² = 200²

NOTE: Our goal is to find the value of LW [since this equals the AREA of the rectangle]

If we take L + W = 280 and square both sides we get (L + W)² = 280²
If we expand this, we get: L² + 2LW + W² = 280²
Notice that we have L² + W² "hiding" in this expression.
That is, we get: L² + 2LW + W² = 280²

We already know that L² + W² = 200², so, we'll take L² + 2LW + W² = 280² and replace L² + W² with 200² to get:
2LW + 200² = 280²
So, 2LW = 280² - 200²
Evaluate: 2LW = 38,400
Solve: LW = 19,200

Answer: A

Cheers,
Brent
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Assuming that the 560 feet perimeter of the park was a square, we get that one side would be 140 feet. For any parallelogram (of equal perimeter), square has the greatest area.

With 140feet for one side, we get the area 19600 sq ft.

The park, however, is a rectangle and will thus have an area less than a square.

We have only one option for this.

A. 19200