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

Question

https://gmatclub.com/forum/download/file.php?id=44219 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 PS Question 9.jpg

GMATBusters · Answer

Official Explanation

If you enlarge the diagram above, we get below figure

Attachment:

Q1.jpg [ 20.65 KiB | Viewed 4735 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 4719 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.

pk123 · Answer

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

Gladiator59 · Answer

In my opinion (B) is the answer. I have made some constructions and the solution is in image.

Best Regards,
G

yenbh · Answer

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 3})^{2}  +  n^{2}

--> Area of square CKHZ =  (m\sqrt{3})^{2}  +  n^{2}

--> Option B

hargun3045 · Answer

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

u1983 · Answer

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 .

pankajpatwari · Answer

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,

Posted from my mobile device

stne · Answer

Hi  gmatbusters ,

The figure is not visible in the question , kindly look into this. Thank you.

Bunuel · Answer

__________________
Fixed that. Thank you.

T1101 · Answer

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!

GMATBusters · Answer

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. Q1.jpg 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/

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

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