In the figures above if the area of the triangle on the right is twice

Bunuel has already provided a solution. Nonetheless, assume the given triangles to be equilateral. The area of the first triangle is\(\sqrt{3}/4*s^2\). The area of the other triangle would be \(\sqrt{3}/4*S^2\). Given that\(\sqrt{3}/4*S^2\) = \(2*\sqrt{3}/4*s^2\). Thus, S = \(\sqrt{2}s.\)

Hi All,

The concept of similar triangles is relatively rare on the GMAT (you likely won't see it more than once on Test Day and you probably won't see it at all). That having been said, the concept is about the ratios of the sides (and how side length effects other things - area, perimeter, other sides, etc.).

Here, we have two triangles with the exact same set of 3 angles, so we know that the two triangles are similar. The one ratio that we're given to work with is that the area of the larger triangle is exactly TWICE that of the smaller triangle. We can use this ratio, along with TESTing VALUES, to get to the solution.

Rather than deal with an abstract triangle, I'm going to say that the small triangle is a 3/4/5 right triangle...

Small Triangle
Area = (1/2)(Base)(Height)
Area = (1/2)(3)(4) = 6

Since the larger triangle has TWICE this area, we know that it's area is 6(2) = 12...

Large Triangle
Area = (1/2)(Base)(Height)
12 = (1/2)(Base)(Height)

Since the triangles are similar, each side of the larger triangle is the same proportionate larger than the corresponding side of the smaller triangle. This means that the two sides (the 3 and the 4) have to each be multiplied by the same number and the result has to DOUBLE the area. The only way to get to DOUBLE is if each side is multiplied by \sqrt{2}

12 = (1/2)(3\sqrt{2})(4\sqrt{2})

In this way, when you multiply everything, you get....

12 = (1/2)(12)(2)

Final Answer:

GMAT assassins aren't born, they're made,
Rich

In two similar triangles, the ratio of their areas is the square of the ratio of their sides. Figure below illustrates on such example

Hi Bunuel,

Would my solution make sense?

I turned the triangles into right isosceles triangles, the smaller one with legs "1" and 1", which gives an area of 1/2. This would mean the larger triangle would have an area of 1 (double a half). This would make each leg of the larger triangle \(\sqrt{2}\)?

Much appreciated!

Since a PS question can have only one correct answer then considering one particular case which satisfies the conditions given must also give correct answer.

So, yes, if the smaller triangle is \(1:1:\sqrt{2}\), then the larger triangle is \(\sqrt{2}:\sqrt{2}:2\), thus \(s=\sqrt{2}\) and \(S=s*\sqrt{2}=\sqrt{2}*\sqrt{2}=2\).

Solution:

We see that the two triangles are similar. Recall that if the ratio of the areas the two similar figures is a/b, then the ratio of the lengths of the corresponding sides of the two figures is √a / √b.

Here, we are given that the area of the larger triangle is twice that of the smaller; therefore, we can think of the ratio of the two areas as 2/1 and hence the ratio of lengths of the corresponding sides of the two figures is √2/√1 = √2. So, S = √2s.

Answer: C

Since the two triangles have the same combination of angles, they are similar.
Plug in a type of triangle whose characteristics are familiar.
Let each triangle be a 45-45-90 triangle.

Triangle ABC:
Let s=1
A = (1/2)(1)(1) = 1/2

Triangle DEF:
Since triangle DEF is twice the size, A=1.
Thus:
(1/2)(S)(S) = 1
S² = 2
S = √2

The correct answer must yield √2 when s=1.
Only C works:
\(\sqrt{2}s= \sqrt{2}*1 = \sqrt{2}\)

By AA theorem of similarity, the 2 triangles are similar.

By similarity, \(\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} = \frac{s}{S}\)


Ratio of areas of \(\frac{\triangle{ABC}}{\triangle{}EDF} = \frac{1}{2} = square \space of \space the \space ratio \space of \space the \space sides = \frac{s^2}{S^2}\)

Therefore \(S^2 = 2s^2\)

S = \(\sqrt{2}\)s



Option C

Arun Kumar

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.