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Manager  Joined: 14 Dec 2011
Posts: 55
GMAT 1: 630 Q48 V29 GMAT 2: 690 Q48 V37 In the figures above, if the area of the triangle on the  [#permalink]

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14 00:00

Difficulty:   65% (hard)

Question Stats: 57% (01:45) correct 43% (02:01) wrong based on 462 sessions

### HideShow timer Statistics 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. $$2s$$

Attachment: Untitled.jpg [ 8.32 KiB | Viewed 7896 times ]

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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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
In the figures above, if the area of the triangle on the  [#permalink]

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Impenetrable wrote: 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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Director  Joined: 29 Nov 2012
Posts: 740
Re: In the figures above, if the area of the triangle on the  [#permalink]

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How would you solve this question if you didn't know that property?
Director  Joined: 29 Nov 2012
Posts: 740
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!
Verbal Forum Moderator B
Joined: 10 Oct 2012
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Re: In the figures above, if the area of the triangle on the  [#permalink]

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fozzzy wrote:
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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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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the figures above, if the area of the triangle on the  [#permalink]

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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:

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

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In two similar triangles, the ratio of their areas is the square of the ratio of their sides. Figure below illustrates on such example
Attachments st.JPG [ 19.49 KiB | Viewed 5185 times ]

Intern  B
Joined: 06 Oct 2017
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In the figures above, if the area of the triangle on the  [#permalink]

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Bunuel wrote:
Impenetrable wrote: 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!
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: In the figures above, if the area of the triangle on the  [#permalink]

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

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