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In the figures above, if the area of the triangle on the

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In the figures above, if the area of the triangle on the [#permalink]

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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. \(2s\)

[Reveal] Spoiler:
Attachment:
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[Reveal] Spoiler: OA

Originally posted by Impenetrable on 12 Mar 2012, 02:44.
Last edited by Bunuel on 05 May 2016, 07:47, edited 3 times in total.
Edited the question.
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In the figures above, if the area of the triangle on the [#permalink]

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Impenetrable wrote:
Image
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. \(2s\)


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.
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Re: In the figures above, if the area of the triangle on the [#permalink]

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New post 09 Mar 2013, 22:13
How would you solve this question if you didn't know that property?
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Re: In the figures above, if the area of the triangle on the [#permalink]

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New post 12 Mar 2013, 01:31
Please, can anyone help out in this particular problem? Thanks!
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Re: In the figures above, if the area of the triangle on the [#permalink]

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Please, can anyone help out in this particular problem? Thanks!


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 = \(\sqrt{2}s.\)
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Re: In the figures above, if the area of the triangle on the [#permalink]

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New post 15 Mar 2013, 15:26
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
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Re: In the figures above, if the area of the triangle on the [#permalink]

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New post 20 Jan 2015, 20:51
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:
[Reveal] Spoiler:
C


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Re: In the figures above, if the area of the triangle on the [#permalink]

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New post 17 Feb 2017, 22:58
In two similar triangles, the ratio of their areas is the square of the ratio of their sides. Figure below illustrates on such example
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In the figures above, if the area of the triangle on the [#permalink]

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New post 20 Dec 2017, 15:16
Bunuel wrote:
Impenetrable wrote:
Image
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. \(2s\)


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: http://gmatclub.com/forum/math-triangles-87197.html

Hope it helps.


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!
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Re: In the figures above, if the area of the triangle on the [#permalink]

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New post 20 Dec 2017, 20:55
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YYZ wrote:
Bunuel wrote:
Impenetrable wrote:
Image
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. \(2s\)


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: http://gmatclub.com/forum/math-triangles-87197.html

Hope it helps.


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 \(S=s*\sqrt{2}=\sqrt{2}*\sqrt{2}=2\).
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: In the figures above, if the area of the triangle on the   [#permalink] 20 Dec 2017, 20:55
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