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perimeter

by crackgmat007 » Tue Oct 20, 2009 7:25 am
What is the perimeter of rectangle ABCD ?

The longer side of the rectangle is 2 meters shorter than its diagonal
The ratio of the shorter side of the rectangle to its diagonal is 1/3
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by mp2437 » Tue Oct 20, 2009 8:20 am
Choice B.

Statement (1): draw right triangle with hypotenuse x and side (x-2). Using pythagorean theory, the other side y is: x^2 = (x-2)^2 + y^2 --> y = sqrt(2x - 4).

Perimeter = 2y + 2x = 2 * sqrt(2x-4) + 2*(x - 2). Have not solved for x, so insufficient.

Statement 2: Ratio of short side / hypotenuse = 1/3, so again draw right triangle and apply pythagorean theory.

3^2 = 1^2 + y^2 --> y = sqrt(8), so perimeter is 2x + 2y = 2*sqrt(8) + 2(1) = 2*[sqrt(8) + 1], sufficient.
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by Stuart@KaplanGMAT » Tue Oct 20, 2009 10:48 am
mp2437 wrote:Choice B.

Statement (1): draw right triangle with hypotenuse x and side (x-2). Using pythagorean theory, the other side y is: x^2 = (x-2)^2 + y^2 --> y = sqrt(2x - 4).

Perimeter = 2y + 2x = 2 * sqrt(2x-4) + 2*(x - 2). Have not solved for x, so insufficient.

Statement 2: Ratio of short side / hypotenuse = 1/3, so again draw right triangle and apply pythagorean theory.

3^2 = 1^2 + y^2 --> y = sqrt(8), so perimeter is 2x + 2y = 2*sqrt(8) + 2(1) = 2*[sqrt(8) + 1], sufficient.
We can eliminate (2) by itself in about 3 seconds if we understand something fundamental about DS:

if you're asked to solve for an actual value, you need at least 1 actual number.

There's also a corollary, which is often helpful:

if you're asked to solve for a ratio/percent/proportion, you do not necessarily need any actual numbers, other proportions may suffice.

While (2) gives us the ratio of the short side to the long side, it doesn't tell us if we have the world's smallest rectangle or the universe's biggest.

When we combine the statements, however, we can use statement (1) to isolate a variable, sub in to statement (2) and then solve for both length and width.

Accordingly, (c) is the correct answer.
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by mp2437 » Tue Oct 20, 2009 12:05 pm
Stuart,

You are correct. I actually thought about it and came back to the computer to see your reply there. My solution shows only sides 1 and 3, of course they could be larger and smaller.

Thanks for clarifying.
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by crackgmat007 » Tue Oct 20, 2009 4:20 pm
Stuart Kovinsky wrote:
mp2437 wrote:Choice B.

Statement (1): draw right triangle with hypotenuse x and side (x-2). Using pythagorean theory, the other side y is: x^2 = (x-2)^2 + y^2 --> y = sqrt(2x - 4).

Perimeter = 2y + 2x = 2 * sqrt(2x-4) + 2*(x - 2). Have not solved for x, so insufficient.

Statement 2: Ratio of short side / hypotenuse = 1/3, so again draw right triangle and apply pythagorean theory.

3^2 = 1^2 + y^2 --> y = sqrt(8), so perimeter is 2x + 2y = 2*sqrt(8) + 2(1) = 2*[sqrt(8) + 1], sufficient.
We can eliminate (2) by itself in about 3 seconds if we understand something fundamental about DS:

if you're asked to solve for an actual value, you need at least 1 actual number.

There's also a corollary, which is often helpful:

if you're asked to solve for a ratio/percent/proportion, you do not necessarily need any actual numbers, other proportions may suffice.

While (2) gives us the ratio of the short side to the long side, it doesn't tell us if we have the world's smallest rectangle or the universe's biggest.

When we combine the statements, however, we can use statement (1) to isolate a variable, sub in to statement (2) and then solve for both length and width.

Accordingly, (c) is the correct answer.
Thanks for clarifying Stuart. Can you also solve the question pls?
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