The perimeter of rectangle ABCD is 520 and the length of the diagonal

Question

The perimeter of rectangle ABCD is 520 and the length of the diagonal AC is 258. What is the area of the rectangle?

(A) 518
(B) 1,918
(C) 6,760
(D) 16,641
(E) 16,900

av1901 · Answer

Perimeter = 520 =>  2(L+W) = 520 => L+W = 260 
So, we can say that  W = 260-L  - Eq(1)
Area of rectangle =  L*W = L*(260-L)  - Eq2

We are given the diagonal = 258
Now, we know that angles of a rectangle are 90 so a diagonal divides the rectangle into 2 right angled triangles, with the Length and Width as the sides of the right triangle and the diagonal as the hypotenuse. Then, as per Pythagoras Theorem

Diagonal^2 = L^2 + W^2
Substituting value of W from Equation 1 we get

258^2 = L^2 + (L-260)^2 
 258^2 = L^2 + 260^2 + L^2 - 520L 
 0 = 2L^2 - 520L + {260^2 - 258^2} 
 2L^2 - 520L + (2)(518) = 0 
 L^2 - 260L + 518 = 0 
 260L - L^2 = 518 
 L(260-L) = 518

Now, as you can see from Eq2, that area of the rectangle = L(260-L)

So,  A = 518

Answer - A

ThatDudeKnows · Answer

The first thing I notice is that, with exception of D and E, the answer choices are very spread out. Let's just see if we can get the right answer without doing the math.

Perimeter is 520. We have 260 to work with for the sum of the L and W.
Diagonal is 258. Woah, that's almost the full 260. We need something that's really close to 258x2. Even 257x3 won't be wide enough. 
Only one answer choice is close.

Answer choice A.

GKLemon · Answer

ABCD = 520, lets called the sides of the rectangle

side L and side W. This is a rectangle so there are two equal sides,

2L + 2W = 520
2 (L + W) = 520
L+W = 260

Now, from Pythagorean theorem of right triangles, we get  x^2 + y^2 = z^2  (sum of two sides squared = hypotonus squared)

In this case, the hypotonus of the triangle is AC at 258. The two sides are L (AB) and W (DA) and we know L+W = 260

so  L^2 + W^2 = 258^2

now how do we get that? Let's square L+W

(L + W)^2 = 260^2
L^2 + W^2 + 2LW = 260^2

Plugging in  L^2 +W^2 = 258^2

258^2 + 2LW = 260^2

solving for LW

2LW = 260^2 - 258^2  here we can use an algebraic principal that's commonly tested on the GMAT  a^2 - b^2 = (a + b) (a - b)  since we don't actually want to solve whatever 260^2 and 258^2 is

2LW = (260 + 258) (260 - 258) 
 2LW = (518)(2)  eliminating the 2 from both sides

LW = 518

And area of a rectangle = Length * Width = LW = 518

Hence answer A

The perimeter of rectangle ABCD is 520 and the length of the diagonal

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GMAT Problem Solving (PS) Questions

The perimeter of rectangle ABCD is 520 and the length of the diagonal

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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