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

[BTGModeratorVI]

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

Answer: A
Source: official guide

[Brent@GMATPrepNow]

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

Answer: A
Source: official guide

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

[Scott@TargetTestPrep]

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

Answer: A
Source: official guide

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 (Eq. 1)

and

L^2 + W^2 = 200^2 (Eq. 2)

Squaring the first equation, we have:

(L + W)^2 = 280^2

L^2 + W^2 + 2LW = 280^2 (Eq. 3)

From Eq. 2, we substitute the value of 200^2 for L^2 + W^2 into Eq. 3, obtaining:

200^2 + 2LW = 280^2

2LW = 280^2 - 200^2 (Eq. 4)

We recognize the right side of Eq. 4 as a difference of squares, and so we have:

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

2LW = 80 x 480

LW = 40 x 480 = 19,200

Answer: A