Jul 16 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Tuesday, July 16th at 8 pm EDT Jul 16 03:00 PM PDT  04:00 PM PDT Join a free live webinar and find out which skills will get you to the top, and what you can do to develop them. Save your spot today! Tuesday, July 16th at 3 pm PST Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 27 Oct 2017
Posts: 1229
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
Updated on: 16 Oct 2018, 01:52
Question Stats:
50% (02:03) correct 50% (02:50) wrong based on 46 sessions
HideShow timer Statistics
2 equilateral triangles XYZ and ABC are shown as above figure. If YX = XA=m and AB=n, what is the area of a square CKHZ, in terms of m and n? A. \(\sqrt{3m^2+n^2}\) B. \(3m^2+n^2\) C. \(\sqrt{3}m^2+n^2\) D. \(\sqrt{m^2+3n^2}\) E. \(m^2+3n^2\) Weekly Quant Quiz #3 Question No 9 Attachment:
PS Question 9.jpg [ 16.87 KiB  Viewed 1229 times ]
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Originally posted by gmatbusters on 06 Oct 2018, 10:37.
Last edited by Bunuel on 16 Oct 2018, 01:52, edited 3 times in total.
Renamed the topic and edited the question.



Retired Moderator
Joined: 27 Oct 2017
Posts: 1229
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
06 Oct 2018, 10:38
Official Explanation If you enlarge the diagram above, we get below figure Attachment:
Q1.jpg [ 20.65 KiB  Viewed 953 times ]
Here, the angles of the triangle is 90deg:60deg:30deg, so the corresponding ratio of the side lengths is 1:√3:2, as shown below. Attachment:
Q2.jpg [ 9.76 KiB  Viewed 953 times ]
Then, you get ZA= 2TA = 2((√3 m)/2)=√3m. Since ∠ZAC=90deg, according to the Pythagorean Theorem, We get \(ZC^2 =ZA^2 + CA^2\) = \((√3m)^2+n^2=3m^2+n^2\), and the area of the square CKHZ = \(ZC^2 = 3m^2+n^2\). Therefore, the answer is B.
_________________



Manager
Joined: 16 Sep 2011
Posts: 91

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
06 Oct 2018, 10:52
triangle AZC is a right angled triangle
and AC= n
XZ= n ,AX=n so in AZ = ?
sin 120/ AZ = sin 30/n
so sqrt 3/2AZ = 1/2n
so AZ= n sqrt 3
so CZ ^2= sqrt (AZ^2 + AC^2) = 3n^2 + m^2= area of square
E is the answer



Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 751
GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
06 Oct 2018, 11:03
In my opinion (B) is the answer. I have made some constructions and the solution is in image. Best Regards, G
Attachments
Q9.jpeg [ 98.41 KiB  Viewed 1103 times ]
_________________
Regards, Gladi
“Do. Or do not. There is no try.”  Yoda (The Empire Strikes Back)



Intern
Joined: 17 Dec 2016
Posts: 19
Location: Viet Nam

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
06 Oct 2018, 11:09
Draw a line to connect Z and A
Angle ZXA = 180∘  angle ZXY = 90∘
Triangle ZXA is isoceles > Angle XAZ = 30∘, ZA = 2 x (m\sqrt{3}/2) = m\sqrt{3}
Angle ZAC = 180∘  Angle XAZ  Angle CAB = 180∘  30∘  60∘ = 90∘
> ZC = \sqrt{\(ZA^{2}\) + \(AC^{2}\)} = \sqrt{\((m[square_root]3})^{2}\) + \(n^{2}\)[/square_root]
> Area of square CKHZ = \((m\sqrt{3})^{2}\) + \(n^{2}\)
> Option B



Intern
Joined: 06 Feb 2018
Posts: 16

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
Updated on: 11 Oct 2018, 04:42
Use Pythagoras theorem
How?
Find the base and height of the triangle formed whose side is the side of square
Height(the horizontal part) comes out (after a few calculations) as (3n+m)/2 and base(upright( as root(3)/2(mn)
Use Pythagoras theorem (result turns out nicely as 3m^2+n^2
B
Originally posted by hargun3045 on 06 Oct 2018, 11:20.
Last edited by hargun3045 on 11 Oct 2018, 04:42, edited 1 time in total.



RC Moderator
Joined: 24 Aug 2016
Posts: 802
Location: Canada
Concentration: Entrepreneurship, Operations
GMAT 1: 630 Q48 V28 GMAT 2: 540 Q49 V16

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
06 Oct 2018, 13:07
I am glad that finally I could solve it.......... nice one. Worth the time. But if this shows up in the exam ........ ( I hope I can figure out the effort required in 1 min and move on (unless I can borrow some gray matter from the amazing Quant minds of the GC Community)) ........... Thanks again .
Attachments
Quiz_3 Question No 9_Oct 06 2018.jpg [ 1.63 MiB  Viewed 1048 times ]
_________________
Please let me know if I am going in wrong direction. Thanks in appreciation.



Intern
Joined: 17 Aug 2018
Posts: 13

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
07 Oct 2018, 18:49
Hi..Congrats for getting it right.Could u please elaborate ur explanation? I am not clear with the image that u hve uploaded. Thanks & Regards, Gladiator59 wrote: In my opinion (B) is the answer. I have made some constructions and the solution is in image.
Best Regards, G Posted from my mobile device



Director
Joined: 27 May 2012
Posts: 814

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
16 Oct 2018, 01:49
gmatbusters wrote: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX = XA=m and AB=n, what is the area of a square CKHZ, in terms of m and n? A. \(\sqrt{3m^2+n^2}\) B. \(3m^2+n^2\) C. \(\sqrt{3}m^2+n^2\) D. \(\sqrt{m^2+3n^2}\) E. \(m^2+3n^2\) Weekly Quant Quiz #3 Question No 9 Attachment: PS Question 9.jpg Hi gmatbusters, The figure is not visible in the question , kindly look into this. Thank you.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 56244

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
16 Oct 2018, 01:53
stne wrote: gmatbusters wrote: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX = XA=m and AB=n, what is the area of a square CKHZ, in terms of m and n? A. \(\sqrt{3m^2+n^2}\) B. \(3m^2+n^2\) C. \(\sqrt{3}m^2+n^2\) D. \(\sqrt{m^2+3n^2}\) E. \(m^2+3n^2\) Weekly Quant Quiz #3 Question No 9 Attachment: PS Question 9.jpg Hi gmatbusters, The figure is not visible in the question , kindly look into this. Thank you. __________________ Fixed that. Thank you.
_________________



Manager
Joined: 07 Aug 2018
Posts: 111
Location: United States (MA)
GMAT 1: 560 Q39 V28 GMAT 2: 670 Q48 V34

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
25 Oct 2018, 07:19
gmatbusters wrote: Official Explanation If you enlarge the diagram above, we get below figure Attachment: Q1.jpg Here, the angles of the triangle is 90deg:60deg:30deg, so the corresponding ratio of the side lengths is 1:√3:2, as shown below. Attachment: Q2.jpg Then, you get ZA= 2TA = 2((√3 m)/2)=√3m. Since ∠ZAC=90deg, according to the Pythagorean Theorem, We get \(ZC^2 =ZA^2 + CA^2\) = \((√3m)^2+n^2=3m^2+n^2\), and the area of the square CKHZ = \(ZC^2 = 3m^2+n^2\). Therefore, the answer is B. gmatbusters, how do I know that the bisector of line \(ZA\) will divide \(∠ZXA\)into two eual angles? I somehow cannot get into my head how we can be sure of that... Do you mind explaining it in more detail? Thanks you in advance!
_________________



Retired Moderator
Joined: 27 Oct 2017
Posts: 1229
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
Show Tags
25 Oct 2018, 07:29
Hi T1101Please note that since YX = XA hence triangle ZXA is an isosceles triangle, hence the altitude XT is also the angle bisector. ( Geometric property) I hope it is clear now. Attachment:
Q1.jpg [ 15.29 KiB  Viewed 629 times ]
In an isosceles triangle (where base is the side which is not equal to any other side):
– the altitude drawn to the base is the median and the angle bisector;
– the median drawn to the base is the altitude and the angle bisector;
– the bisector of the angle opposite to the base is the altitude and the median. https://www.veritasprep.com/blog/2014/05/mediansaltitudesandanglebisectorsinspecialtrianglesonthegmat/T1101 wrote: gmatbusters wrote: Official Explanation If you enlarge the diagram above, we get below figure Attachment: The attachment Q1.jpg is no longer available Here, the angles of the triangle is 90deg:60deg:30deg, so the corresponding ratio of the side lengths is 1:√3:2, as shown below. Attachment: The attachment Q2.jpg is no longer available Then, you get ZA= 2TA = 2((√3 m)/2)=√3m. Since ∠ZAC=90deg, according to the Pythagorean Theorem, We get \(ZC^2 =ZA^2 + CA^2\) = \((√3m)^2+n^2=3m^2+n^2\), and the area of the square CKHZ = \(ZC^2 = 3m^2+n^2\). Therefore, the answer is B. gmatbusters, how do I know that the bisector of line \(ZA\) will divide \(∠ZXA\)into two eual angles? I somehow cannot get into my head how we can be sure of that... Do you mind explaining it in more detail? Thanks you in advance!
_________________




Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
[#permalink]
25 Oct 2018, 07:29






