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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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???

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

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

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
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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
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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
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