GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 May 2019, 02:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

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]

### Show Tags

Updated on: 05 May 2016, 07:47
1
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.
Math Expert
Joined: 02 Sep 2009
Posts: 55271
In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

12 Mar 2012, 03:05
5
4
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 $$\frac{AREA}{area}=\frac{S^2}{s^2}=2$$ --> $$S=s\sqrt{2}$$.

For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
_________________
##### General Discussion
Director
Joined: 29 Nov 2012
Posts: 740
Re: In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

09 Mar 2013, 22:13
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]

### Show Tags

12 Mar 2013, 01:31
Please, can anyone help out in this particular problem? Thanks!
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 611
Re: In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

12 Mar 2013, 01:44
2
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 = $$\sqrt{2}s.$$
_________________
Manager
Joined: 24 Jan 2013
Posts: 72
Re: In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

15 Mar 2013, 15:26
1
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: $$S=\sqrt{2}*s$$

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14198
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

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)

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Intern
Joined: 03 Feb 2017
Posts: 25
Re: In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

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
Attachments

st.JPG [ 19.49 KiB | Viewed 5185 times ]

Intern
Joined: 06 Oct 2017
Posts: 9
In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

20 Dec 2017, 15:16
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 $$\frac{AREA}{area}=\frac{S^2}{s^2}=2$$ --> $$S=s\sqrt{2}$$.

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

### Show Tags

20 Dec 2017, 20:55
1
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 $$\frac{AREA}{area}=\frac{S^2}{s^2}=2$$ --> $$S=s\sqrt{2}$$.

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 $$S=s*\sqrt{2}=\sqrt{2}*\sqrt{2}=2$$.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 11012
Re: In the figures above, if the area of the triangle on the  [#permalink]

### Show Tags

29 Dec 2018, 04:41
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.
_________________
Re: In the figures above, if the area of the triangle on the   [#permalink] 29 Dec 2018, 04:41
Display posts from previous: Sort by