LM=KJ[2 positions of same board]

stmt1,

LN=sqrt(2)

In triangle LMN,the sides are in ratio 1:1:sqrt(2)[45:45:90 traingle]

-->LN:MN:LM=1:1:sqrt(2)=sqrt(2):sqrt(2):2

LM=2

-->KJ=2

In traingle KJN,

KJ=2

JN:KN:KJ=1:sqrt(3):2[30:60:90 traingle]

-->JN:KN:KJ=1:sqrt(3):2[as KJ=2]

Now we got MN=sqrt(2) and KN=sqrt(3)

From this we can find the difference.

From option B we can find the lengths in the same way.

If this is DS question Pick D

Hope this clarify.

I believe answer should be D.

Given an angle(not right angle) and a side length in right triangle, we can calculate other sides and angles.

stmt 1: LN is known
From Triangle LMN, MN can be calculated.
LM can also be calculated.

Now, LM = JK.
Hence from triangle JKN, KN can be calculated.
SUFF

stmt2: JN is known
From triangle JKN. JK can be found. and also KN can be found.

JK = LM.
Hence from triangle LMN, MN can be found.
SUFF

pick D

Hope this helps!!

selango wrote:LM=KJ[2 positions of same board]

stmt1,

LN=sqrt(2)

In triangle LMN,the sides are in ratio 1:1:sqrt(2)[45:45:90 traingle]

-->LN:MN:LM=1:1:sqrt(2)=sqrt(2):sqrt(2):2

LM=2

-->KJ=2

In traingle KJN,

KJ=2

JN:KN:KJ=1:sqrt(3):2[30:60:90 traingle]

-->JN:KN:KJ=1:sqrt(3):2[as KJ=2]

Now we got MN=sqrt(2) and KN=sqrt(3)

From this we can find the difference.

If this is DS question Pick A

Hope this clarify.

Why didnt you consider stmt2?

Yes praveen Its D only..I typed it by mistake.

Now edited it..

priya_kapoor wrote: I guess what I would like to see if this wasn't a data sufficiency question, how would it be solved? Any and all help is welcomed!

Hi Priya,

GLad you saw the answer. Of course, on Test-day, we'd mark D and move on--if we have all the angles and at least one side of a triangle, that's always sufficient to answer for all the sides, and if we can find all the sides of one triangle then we have found the length of the board.

That being said, you're right--this type of problem could have shown up in the problem section. If it did, here's how we'd solve it:

The key deduction here is that we have two special right triangles. When the board is in the first position, it forms a 45 degree angle with the 90 degree wall--that means the last angle is 180 - (90 + 45) = 45 degrees. So, we have a right isosceles triangle, and the sides are always in the ratio of X:X:X root 2. Since one of the short sides is root 2, we know the other short side has the same measure and the long side (the board) is 2.

Similarly, when the board is in the second position, it forms a 30/60/90 triangle. This always has sides in the ratio of X : X root 3 : 2X. Given the shortest side has a measure of 1, the longest side, the board, has a measure of 2 (agreeing with our previous calculations!) and the middle side has a measure of root 3.

Thus, KN - MN = root 3 - root 2.