I answered this quesiton incorrectly during my practice test. I tried to solve this question aftwerwards, but I am not able to come up with a solution that gives me the correct answer, C. Can you please help?

Thanks!

Then the sides of big traingle are as such S=sq.rt(3) * s ?

Lets say sides of the smaller triangles are a,b,s and of bigger one are A,B,S .Now since the triangles are similar (angles are same)
a b s a+b+c
- = - = - = -------- = k
A B S A+B+C

and now use Heron's formula to get

S=sqrt(2)s

This is a lengthy way, I m still trying to find a faster way of getting the result.
If you have any, please let me know

Area of a triangle = (bh)/2

If the two triangles are similar, then the legs are in proportion. In addition, the heights will be in proportion. For the smaller, call the base s and the height h. For the larger, call the base S and the height H.

If the two sets of variables (for the smaller and the larger) are in proportion to one another, then they increase at the same rate. In other words, s/h = S/H. Below is the math; after that is the conceptual (if you understand conceptually, then you can avoid the math).

We can also set up formulas for the two areas:
smaller = sh/2
larger = SH/2
and set them equal (b/c the larger is twice the smaller)
2(sh/2) = SH/2
sh = SH/2

The problem asks us to solve for S.
2sh/H = S
Now, remember that s/h = S/H. Can rearrange to s/S = h/H. Substitute h/H into above equation to get:
(2s^2)/S = S
2s^2 = S^2
sqrt2*s = S

Conceptually, because the bases and heights are in proportion to one another, the fact that the area of the larger = twice the area of the smaller is all we need to know. Area is a "squared" quantity - we multiple two variables together to get the area. If the two triangles are in proportion, the area increase is also in proportion - so I can just use that increase to figure out the proportional increase in the lengths of the sides. Because I'm trying to use area (two variables multiplied, which gives me a "square") to get sides (those two individual variables), I take the square root of the proportional increase in the area - or sqrt2. Sqrt2 is the multiplier I would use with any of the legs, or the height, to get the corresponding leg (or height) in the larger triangle.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

Thanks, Stacey. I definitely understand this problem now! :)

will it not be simple to solve it with the similar triangle property.

the square of the sides are proportional to the area of the triangles.

s^2 / S^2 = 1/2

s/S = 1 / √2

S = s√2

absolutely.

for those of you who don't know this property, here it is:

FOR SIMILAR FIGURES:

if the AREAS of the figures are in the ratio A:B, then the LENGTHS of anything in the figures are in the ratio √A:√B.
conversely,
if the LENGTHS are in the ratio A:B, then the AREAS are in the ratio A^2:B^2.

you can also include volume in this statement, if the figures are solids - you just cube / cube root, rather than square / square root - but that topic doesn't seem to come up nearly as often as do comparisons between similar 2-dimensional figures.