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2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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Updated on: 16 Oct 2018, 01:52
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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 1491 times ]
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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.



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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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06 Oct 2018, 10:38
Official Explanation If you enlarge the diagram above, we get below figure Attachment:
Q1.jpg [ 20.65 KiB  Viewed 1175 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 1176 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.
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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



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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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



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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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.



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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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 .
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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



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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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.
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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.
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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!
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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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 851 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!
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =
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