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

Question

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200 
B. 19,600 
C. 20,000 
D. 20,400 
E. 20,800

Bunuel · Accepted Answer

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200
B. 19,600
C. 20,000
D. 20,400
E. 20,800

Given:

(1) The perimeter is 560 feet: 
  2x+2y=560 ;
 x+y=280

(2) The diagonal is 200 feet:
  x^2+y^2=200^2  ( diagonal, as per Pythagoras )

Question:

xy=?

Solution:

Square (1):

x^2+2xy+y^2=280^2 .

Now subtract (2) from 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.

sleekmover · Answer

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

Princas1 · Answer

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

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

PareshGmat · Answer

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

mbaboop · Answer

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.

EMPOWERgmatRichC · 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:  A

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

LakerFan24 · Answer

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?

EMPOWERgmatRichC · Answer

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

BrentGMATPrepNow · Answer

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

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  + 2  LW   +  W²  = 280²

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

Answer: A

Cheers,
Brent

MsInvBanker · Answer

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

ScottTargetTestPrep · Answer

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

Shbm · Answer

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

ScottTargetTestPrep · Answer

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.

GMATGuruNY · Answer

ALWAYS LOOK FOR SPECIAL TRIANGLES. Draw the rectangle and its diagonal: diagonal_of_200.png 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: diagonal_of_200_1 (1).png 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 A

avigutman · Answer

Video solution from Quant Reasoning (starts at 0:15:06): Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=cn-ojY-2Gj0

AbhaGanu · Answer

Bunuel  why have you squared the diagonal?

Bunuel · Answer

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. https://gmatclub.com/forum/download/file.php?id=62782 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. 1.png

vedikadadhich · Answer

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

GMATGuruNY · Answer

The formula above is valid for a KITE (a quadrilateral in which two pairs of ADJACENT sides are equal): area of a kite.png 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.

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