perimeter

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.

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.

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.

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?