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

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2 equilateral triangles XYZ and ABC are shown as above figure. If YX =  [#permalink]

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Updated on: 16 Oct 2018, 01:52
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Question Stats:

50% (02:03) correct 50% (02:50) wrong based on 46 sessions

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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 1229 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 =  [#permalink]

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

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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

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

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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
Attachments

Q9.jpeg [ 98.41 KiB | Viewed 1103 times ]

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Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =  [#permalink]

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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 =  [#permalink]

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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(m-n)

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 =  [#permalink]

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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 .
Attachments

Quiz_3 Question No 9_Oct 06 2018.jpg [ 1.63 MiB | Viewed 1048 times ]

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Please let me know if I am going in wrong direction.
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Re: 2 equilateral triangles XYZ and ABC are shown as above figure. If YX =  [#permalink]

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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,

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 =  [#permalink]

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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 =  [#permalink]

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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 =  [#permalink]

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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?

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

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25 Oct 2018, 07:29
1
Hi T1101
Please 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/medians-altitudes-and-angle-bisectors-in-special-triangles-on-the-gmat/

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?

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