Data Sufficiency

attached

Since all the corresponding angles are same in both the triangles, they are similar. For similar triangles, the length of the sides will be in same ratio.

Check the attachment. Suppose the ratio is a value 'k'.

Now we have the area of small triangle = (1/2) * s * h

and area of big triangle = (1/2) * ks * kh

Now, its given area of big triangle is twice the area of small triangle, we have

[(1/2) * s * h ] * 2 = (1/2) * ks * kh

k^2 = 2

k = sqrt (2)

S = ks = sqrt (2) * s which is option C.

hi could someone else elaborate e bit more?

thanks!!

mjjking-

I'll try to break down what Alara533 wrote, but it's pretty clear. So, you might need to be more specific on your confusion.

Similar triangles have the same interior angles and their lengths are all proportional by the same factor (as Alara wrote "k"). This tells us:
ks=S (side s of the small triangle * a number k is equal to side S of the large triangle)
kh=H (height h of the small triangle * a number k is equal to height H of the large triangle)

The initial statement tells us that the proportion of the AREAS is 2. Don't confuse this with the proportion of the sides.

The area of the smaller triangle is A=(1/2)bh. Since the b ("base") equals "s", we can state A=(1/2)sh.

The larger triangles area is A=(1/2)bh. The "base" is "S". And the height is "H" (I'm simply stating that the second triangles height is different from the smaller "h" of the small triangle).

Since the intial question tells us that the larger similar triangle is twice the AREA of the smaller similar triangle, we can set our two areas equal to each other as such:
2*((1/2)*s*h)=(1/2)*ks*kh

When you factor the right side, you get:
2*((1/2)*s*h)=((1/2)*s*h)*k^2

Divide (and remove) (1/2)*s*h) from each side, and you are left with:
2=k^2

Take the sqrt of each side, and you have:
k= sqrt 2

Since S= k * s, S= sqrt (2) * s.

Hope that helps clarify.

Here is a mnemonic I use when comparing areas of similar triangles (Compliments of Eric's flashcards):

The ratio of the areas of two similar triangles is the
square of the ratio of corresponding lengths.


See attached for the formula and solution.

This is a simple formula that is much easier to understand, imo. Just make note of the "S" and "s" in the formula.

Hope this helps.