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lukaswelker: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Mar 26, 2014 6:56 amFollowed by: 1 member

by lukaswelker » Thu Apr 17, 2014 8:42 am

Hello guys

I'm going to far in calculations and I get it wrong. There must be an easier way.

Here goes the 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?

19200 : 19600 : 20000 : 20400 : 20800.

Please let me know if you see an easier way.

Many thanks
Lukas

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Source: Beat The GMAT — Problem Solving | Thu Apr 17, 2014 8:42 am

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Thu Apr 17, 2014 9:14 am

560 = 2 * (L + B)

L + B = 280

To find: L*B

L^2 + B^2 = (200)^2
(L+B)^2 -2*L*B = 200^2
280^2 - 200^2 = 2 * L * B
480 * 80 = 2 * L * B
L * B = 19200

R A H U L

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Apr 17, 2014 9:26 am

lukaswelker wrote: 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) 19200
B) 19600
C) 20000
D) 20400
E) 20800.

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 LÂ² + WÂ² = 200Â²

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)Â² = 280Â²
If we expand this, we get: LÂ² + 2LW + WÂ² = 280Â²
Notice that we have LÂ² + WÂ² "hiding" in this expression.
That is, we get: LÂ² + 2LW + WÂ² = 280Â²

We already know that LÂ² + WÂ² = 200Â², so, we'll take LÂ² + 2LW + WÂ² = 280Â² and replace LÂ² + WÂ² with 200Â² to get:
2LW + 200Â² = 280Â²
So, 2LW = 280Â² - 200Â²
Evaluate: 2LW = 38,400
Solve: LW = [spoiler]19,200 = A[/spoiler]

For extra practice, here's a similar question: https://www.beatthegmat.com/area-of-a-re ... 00119.html

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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osama_salah: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Wed Jun 15, 2016 2:05 am

by osama_salah » Mon Jun 27, 2016 5:29 am

How did you calculate 280Â² - 200Â² = 38,400 ?

Brent@GMATPrepNow wrote:
lukaswelker wrote: 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) 19200
B) 19600
C) 20000
D) 20400
E) 20800.
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 LÂ² + WÂ² = 200Â²

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)Â² = 280Â²
If we expand this, we get: LÂ² + 2LW + WÂ² = 280Â²
Notice that we have LÂ² + WÂ² "hiding" in this expression.
That is, we get: LÂ² + 2LW + WÂ² = 280Â²

We already know that LÂ² + WÂ² = 200Â², so, we'll take LÂ² + 2LW + WÂ² = 280Â² and replace LÂ² + WÂ² with 200Â² to get:
2LW + 200Â² = 280Â²
So, 2LW = 280Â² - 200Â²
Evaluate: 2LW = 38,400
Solve: LW = [spoiler]19,200 = A[/spoiler]

For extra practice, here's a similar question: https://www.beatthegmat.com/area-of-a-re ... 00119.html

Cheers,
Brent

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by Brent@GMATPrepNow » Mon Jun 27, 2016 5:32 am

osama_salah wrote:How did you calculate 280Â² - 200Â² = 38,400 ?

Aside: 28Â² = 784, which means 280Â² = 78,400

So, 280Â² - 200Â² = 78,400 - 40,000
= 38,400

Brent Hanneson - Creator of GMATPrepNow.com

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setiavipul: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Tue Jun 06, 2017 9:29 am

by setiavipul » Wed Jun 21, 2017 7:15 am

Brent@GMATPrepNow wrote:
lukaswelker wrote: 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) 19200
B) 19600
C) 20000
D) 20400
E) 20800.
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 LÂ² + WÂ² = 200Â²

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)Â² = 280Â²
If we expand this, we get: LÂ² + 2LW + WÂ² = 280Â²
Notice that we have LÂ² + WÂ² "hiding" in this expression.
That is, we get: LÂ² + 2LW + WÂ² = 280Â²

We already know that LÂ² + WÂ² = 200Â², so, we'll take LÂ² + 2LW + WÂ² = 280Â² and replace LÂ² + WÂ² with 200Â² to get:
2LW + 200Â² = 280Â²
So, 2LW = 280Â² - 200Â²
Evaluate: 2LW = 38,400
Solve: LW = [spoiler]19,200 = A[/spoiler]

For extra practice, here's a similar question: https://www.beatthegmat.com/area-of-a-re ... 00119.html

Cheers,
Brent

Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Jun 21, 2017 7:22 am

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/ 19200
b/ 19600
c/ 20000
d/ 20400
e/ 20800

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

Since diagonal AD is a multiple of 5, check whether âˆ†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, âˆ†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.

The correct answer is A.

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by [email protected] » Wed Jun 21, 2017 1:57 pm

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

Contact Rich at [email protected]

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Jun 22, 2017 5:09 am

setiavipul wrote: Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?

That rule doesn't apply to rectangles. It does, however apply to squares

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Thu Jun 22, 2017 8:41 am

Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?

Note also, that if the figure were a square, and the perimeter were 560, each side would be 560/4 = 140. If each side of a 45/45/90 triangle were 140, the hypotenuse, or diagonal, would be 140 *rt2. So the fact that the diagonal is 200 tells us that we're not dealing with a square (or 45:45:90 triangles.)

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Thu Jun 22, 2017 8:44 am

DavidG@VeritasPrep wrote:
Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?
Note also, that if the figure were a square, and the perimeter were 560, each side would be 560/4 = 140. If each side of a 45/45/90 triangle were 140, the hypotenuse, or diagonal, would be 140 *rt2. So the fact that the diagonal is 200 tells us that we're not dealing with a square (or 45:45:90 triangles.)

One more fun note. For a set perimeter, the rectangle with the greatest area is a square. If the shape in question had been a square, and the sides had each been 140, the area would have been 140*140 = 19,600. Because we know the shape is not a square, we know the area would be less than 19,600. The answer must be A

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Jun 22, 2017 4:50 pm

DavidG@VeritasPrep wrote:
DavidG@VeritasPrep wrote:
Why we can not use 45,45,90 degree property here? Diagonal of a rectangle divides two sides into 45 and 45 degree angle. Am i wrong here?
Note also, that if the figure were a square, and the perimeter were 560, each side would be 560/4 = 140. If each side of a 45/45/90 triangle were 140, the hypotenuse, or diagonal, would be 140 *rt2. So the fact that the diagonal is 200 tells us that we're not dealing with a square (or 45:45:90 triangles.)
One more fun note. For a set perimeter, the rectangle with the greatest area is a square. If the shape in question had been a square, and the sides had each been 140, the area would have been 140*140 = 19,600. Because we know the shape is not a square, we know the area would be less than 19,600. The answer must be A

I love that, nice hack!

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