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A samll, rectangular park has a perimeter of 560 feet and a diagonal measurement...

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by AAPL » Wed Jul 29, 2020 7:20 am

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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

OA A
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Re: A samll, rectangular park has a perimeter of 560 feet and a diagonal measurement...

by Brent@GMATPrepNow » Wed Jul 29, 2020 7:25 am
AAPL wrote: ↑
Wed Jul 29, 2020 7:20 am
 GMAT Prep

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

OA A
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
Brent Hanneson - Creator of GMATPrepNow.com
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Re: A samll, rectangular park has a perimeter of 560 feet and a diagonal measurement...

by Scott@TargetTestPrep » Fri Jul 31, 2020 6:40 am
AAPL wrote: ↑
Wed Jul 29, 2020 7:20 am
 GMAT Prep

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

OA A
Solution:

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

Scott Woodbury-Stewart
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