A small, rectangular park has a perimeter of 560 feet and a

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

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.

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

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

We can let L = the length of the park and W = the width of the park. Our goal is to find the area of the park, i.e., the value of LW.

We can create the equations:

Perimeter = 560

2L + 2W = 560

L + W = 280

and

L^2 + W^2 = 200^2

Squaring the first equation, we have:

(L + W)^2 = 280^2

L^2 + W^2 + 2LW = 280^2

Substituting, we have:

200^2 + 2LW = 280^2

2LW = 280^2 - 200^2

2LW = (280 - 200)(280 + 200)

2LW = 80 x 480

LW = 40 x 480 = 19,200

Answer: A

Shouldn't L square plus W square be equal to 200 rather than being equal to 200 square? Please assist.

200^2 is correct since it’s based on the Pythagorean theorem: a^2 + b^2 = c^2. Here, L^2 + W^2 = D^2 where D is the diagonal. Notice that on a rectangle, the short and long side together with the diagonal form a right triangle; that’s why the Pythagorean theorem is applicable.

ALWAYS LOOK FOR SPECIAL TRIANGLES.
Draw the rectangle and its diagonal:

Since diagonal AD is a multiple of 5 -- and every value in the problem is an INTEGER -- check whether triangle ABD is a multiple of a 3:4:5 triangle.
If each side of a 3:4:5 triangle is multiplied by 40, we get:
(40*3):(40*4):(40*5) = 120:160:200
The following figure is implied:

Check whether the resulting perimeter for rectangle ABCD is 560:
120+160+120+160 = 560
Success!
Implication:
For the perimeter of rectangle ABCD to be 560, triangle ABD must be a multiple of a 3:4:5 triangle with sides 120, 160 and 200.

Thus:
Area of rectangle ABCD = L * W = 160 * 120 = 19200

Video solution from Quant Reasoning (starts at 0:15:06):
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Bunuel why have you squared the diagonal?

The diagonal of a rectangle is also the hypotenuse of a right triangle that has the length and the width of the rectangle as its legs.



So, in order to express D in terms of l and we w can write \(D^2 = l^2 + w^2\).

Hope it's clear.

Isnt area of rectangle = 0.5*D1*D2 where D1 and D2 are diagonal of rectangle?? Please Confirm??

The formula above is valid for a KITE (a quadrilateral in which two pairs of ADJACENT sides are equal):
Note that a rhombus (with four equal sides) is a type of kite, as is a square (with four equal sides and four equal angles).
Thus, the formula can be used to find the area of a rhombus or square.
In the posted problem, no adjacent sides are equal, so the formula is not applicable.