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

Question

https://gmatclub.com/forum/download/file.php?id=68054 In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S= A. \frac{\sqrt{2}}{2}s B. \frac{\sqrt{3}}{2}s C. \sqrt{2}s D. \sqrt{3}s E. 2s 00.JPG Untitled.jpg 2022-08-14_17-01-57.png Untitled.png

Bunuel · Accepted Answer

Since in the given triangles the three angles are identical then they are similar triangles. Useful property of similar triangles: in two similar triangles, the ratio of their areas is the square of the ratio of their sides. Hence \frac{AREA}{area}=\frac{S^2}{s^2}=2 --> S=s\sqrt{2} . Answer: C. For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html Hope it helps.

johnwesley · Answer

You must know that property for GMAT:

"Similar triangles and areas: the ratio of the areas of two similar triangles is the  square  of the ratio of corresponding lengths" and the inverse... 
"The ratio of the lengths of two similar triangles is the  square root  of the ratio of corresponding areas"

(Area left)x2=(Area right) meaning that the ratio of the areas is  \frac{Area_{right}}{Area_{left}}=2

Therefore, the ratio of the lengths:  \frac{S}{s}=\sqrt{2}

Finally:  S=\sqrt{2}*s

Answer C

mau5 · Answer

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.

EMPOWERgmatRichC · Answer

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:  C

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

onamarif · Answer

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

YYZ · Answer

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!

Bunuel · Answer

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 .

ScottTargetTestPrep · Answer

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

GMATGuruNY · Answer

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}

C

CrackverbalGMAT · Answer

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{	riangle{ABC}}{	riangle{}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

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

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

Prep Toolkit

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