December 10, 2018 December 10, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section  Integer properties, and rapidly improve your skills. December 09, 2018 December 09, 2018 07:00 AM PST 09:00 AM PST Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 30 May 2010
Posts: 29
Schools: YALE SOM

Equilateral triangle BDF is inscribed in equilateral triangl
[#permalink]
Show Tags
19 Jun 2010, 07:17
Question Stats:
42% (02:35) correct 58% (02:31) wrong based on 322 sessions
HideShow timer Statistics
Attachment:
Captura.jpg [ 108.93 KiB  Viewed 15521 times ]
Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE? (1) Angle DFE is 90º (2) The length of AF is 10*sqrt(3)
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 51035

Re: DS: Equilateral triangle inscribed in triangle
[#permalink]
Show Tags
19 Jun 2010, 08:09
Nusa84 wrote: Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE? 1. Angle DFE is 90º 2. The length of AF is 10*sqrt(3) Can you help me with this one? Thanks!!! The problem is easier to solve than to explain, but anyway: In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side. (1) Angle DFE is 90º > angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180\angle {BDF}\angle{EDF}=1806030\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 306090 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)). So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\) Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\); \(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\). Sufficient. (2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient. Answer: A.
_________________
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? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 30 May 2010
Posts: 29
Schools: YALE SOM

Re: DS: Equilateral triangle inscribed in triangle
[#permalink]
Show Tags
19 Jun 2010, 08:27
Bunuel wrote: Nusa84 wrote: Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE? 1. Angle DFE is 90º 2. The length of AF is 10*sqrt(3) Can you help me with this one? Thanks!!! Attachment: Captura.jpg The problem is easier to solve than to explain, but anyway: In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side. (1) Angle DFE is 90º > angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180\angle {BCD}\angle{EDF}=1806030\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 306090 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)). So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\) Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\); \(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\). Sufficient. (2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient. Answer: A. As you said, easier to solve, many thanks!!!!!



Manager
Joined: 06 Mar 2010
Posts: 88

Re: DS: Equilateral triangle inscribed in triangle
[#permalink]
Show Tags
27 Jun 2010, 14:19
Good solution in steps given.



Math Expert
Joined: 02 Sep 2009
Posts: 51035

Re: Equilateral triangle BDF is inscribed in equilateral triangl
[#permalink]
Show Tags
06 Mar 2014, 01:37



Intern
Affiliations: CA, SAP FICO
Joined: 22 Nov 2012
Posts: 35
Location: India
Concentration: Finance, Sustainability
GMAT 1: 620 Q42 V33 GMAT 2: 720 Q47 V41
GPA: 3.2
WE: Corporate Finance (Energy and Utilities)

Re: DS: Equilateral triangle inscribed in triangle
[#permalink]
Show Tags
06 Mar 2014, 18:12
Bunuel wrote: Nusa84 wrote: Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE? 1. Angle DFE is 90º 2. The length of AF is 10*sqrt(3) Can you help me with this one? Thanks!!! Attachment: Captura.jpg The problem is easier to solve than to explain, but anyway: In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side. (1) Angle DFE is 90º > angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180\angle {BCD}\angle{EDF}=1806030\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 306090 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)). So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\) Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\); \(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\). Sufficient. (2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient. Answer: A. \(\angle {BDC}=180\angle {BCD}\angle{EDF}=1806030\)  Could you explain the logic for taking EDF?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8649
Location: Pune, India

Re: Equilateral triangle BDF is inscribed in equilateral triangl
[#permalink]
Show Tags
06 Mar 2014, 22:55
Nusa84 wrote: Attachment: The attachment Captura.jpg is no longer available Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE? (1) Angle DFE is 90º (2) The length of AF is 10*sqrt(3) Note that in a given equilateral triangle, you can inscribe many other equilateral triangles. The areas of these inscribed triangles will be different hence the leftover areas will be different. Attachment:
Ques3.jpg [ 13.11 KiB  Viewed 10615 times ]
So depending on what type of triangle is inscribed, the shaded region will be a fraction of the area of triangle ACE. (1) Angle DFE is 90º All internal angles of Equilateral triangles are 60. Angle DFE is 90 so triangle DFE is a right triangle with angles 306090 and sides in the ratio 1:√3:2. Attachment:
Ques4.jpg [ 15.69 KiB  Viewed 10614 times ]
If we assume that side of inscribed triangle is s, we get the side of the large triangle in terms of s too. We also get the area of the right triangle (shaded region) in terms of s. Hence we can easily get the area of the shaded region as a fraction of area of triangle ACE (since they will both be in terms of s^2). This statement alone will be sufficient. (2) The length of AF is 10*sqrt(3) Look at the three triangles shown above again. AF could be 10√3 in any of the three cases. The area of shaded region as a fraction of the area of ACE will be different in each case. Hence this statement alone is not sufficient to answer the question. Answer (A)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 01 May 2013
Posts: 61

Re: Equilateral triangle BDF is inscribed in equilateral triangl
[#permalink]
Show Tags
06 Mar 2014, 23:05
Since you asked for further discussion, here goes.
I more intuited the answer than solved. I know that each interior angle of both equilateral triangles is 60. If I am told that DFE = 90, I can figure out that the other 3 triangles are 306090 and congruent (Each 60degree angle is across from one side of the smaller equilateral triangle; thus all corresponding sides for the right triangles are the same "x(sqrt3)"). Because I know the relative dimensions of triangles BCD, ABF, DFE, and BDF, I can figure out the relative areas of the triangles. For example, I know I can make BD sqrt3, BC 2, and CD 1. I can figure out the areas of all 4 smaller triangles, add these areas to get ACE's area, and thus determine the fraction of ACE that BCD represents. I didn't actually bother to assign values or do the calculations because it was clear Statement 1 is sufficient.
Eliminate B, C, and E.
Statement 2 only tells me AF. It doesn't provide me with any information about any other sides or angles in the figure. Thus, I would not be able to solve for area. The information is useless. Insufficient.
Eliminate D. The answer is A.



Math Expert
Joined: 02 Sep 2009
Posts: 51035

Re: DS: Equilateral triangle inscribed in triangle
[#permalink]
Show Tags
07 Mar 2014, 00:34
X017in wrote: Bunuel wrote: Nusa84 wrote: Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE? 1. Angle DFE is 90º 2. The length of AF is 10*sqrt(3) Can you help me with this one? Thanks!!! Attachment: Captura.jpg The problem is easier to solve than to explain, but anyway: In equilateral triangle all angles equal to 60 degrees and \(Area_{equilateral}=\frac{a^2\sqrt{3}}{4}\), where \(a\) is the length of a side. (1) Angle DFE is 90º > angles ABF and BDC must also be 90º (for example \(\angle {BDC}=180\angle {BCD}\angle{EDF}=1806030\) and the same for ABF). Also as \(\angle{DEF}=\angle{BCD}=\angle{BAF}=60\), then triangles DFE, BCD and BAF are 306090 trianlges. In such triangle sides are in the ratio: \(1:\sqrt{3}:2\) (smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°)). So if \(DE=2x\) (hypotenuse in right triangle DFE), then \(DC=x\) (smaller leg in right triangle BCD) and \(BD=\sqrt{3}x\) (larger leg in right triangle BCD, also the side of inscribed triangle). So the side of triangle ACE would be \(CE=DC+DE=x+2x=3x\) Area of the shaded region (right triangle BDC) would be \(Area_{BDC}=\frac{BD*DC}{2}=\frac{\sqrt{3}x*x}{2}=\frac{\sqrt{3}x^2}{2}\) and the area of equilateral triangle \(Area_{ACE}=\frac{a^2\sqrt{3}}{4}=\frac{(3x)^2*\sqrt{3}}{4}=\frac{9x^2\sqrt{3}}{4}\); \(\frac{Area_{BDC}}{Area_{ACE}}=\frac{\sqrt{3}x^2}{2}*\frac{4}{9x^2\sqrt{3}}=\frac{2}{9}\). Sufficient. (2) The length of AF is \(10*\sqrt{3}\). Multiple breakdowns are possible, hence multiple ratios of areas. Not sufficient. Answer: A. \(\angle {BDC}=180\angle {BCD}\angle{EDF}=1806030\)  Could you explain the logic for taking EDF? It should be \(\angle {BDC}=180\angle {BDF}\angle{EDF}=1806030\) (those 3 angles make up a straight line, which is 180 degrees). Hope it's clear.
_________________
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? Extrahard Quant Tests with Brilliant Analytics



NonHuman User
Joined: 09 Sep 2013
Posts: 9085

Re: Equilateral triangle BDF is inscribed in equilateral triangl
[#permalink]
Show Tags
23 Aug 2018, 07:17
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Equilateral triangle BDF is inscribed in equilateral triangl &nbs
[#permalink]
23 Aug 2018, 07:17






