Question 1209520
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Square SQUR has sides of length x. If triangle SQE is equilateral, find the area of triangle QAU.
Link to diagram: https://ibb.co/C58rZ09R
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        The solution in the post by @CPhill is  INCORRECT.


        I came to bring a correct solution.



<pre>
I will use coordinate plane (x,y).  To avoid missing coordinate x with the side length of the square, 
I will use 'a' for the square side length.


Let's place the origin of the coordinate system (x,y) at point S.


The line EQ has an equation

    y = {{{sqrt(3)*x + b}}}    (1)

since its slope is, obviously,  {{{sqrt(3)}}}.  To find 'b' in this equation, substitute coordinates of 
the vertex Q = {a,0) of the square into this equation


    0 = {{{sqrt(3)*a + b}}}.


It gives  b = {{{-sqrt(3)*a}}}.  So, equation of the line EQ is 


    y = {{{sqrt(3)*x - sqrt(3)*a)}}} = {{{sqrt(3)*(x-a))}}}.    (2)


We want to find the point A as the intersection of the line SU and the line EQ.

The line SU has the equation  y = -x;  so, we substitute y = -x into equation (2).  It gives


    -x = {{{sqrt(3)*(x-a)}}} ,  

or  

    -x = {{{sqrt(3)*x - sqrt(3)*a}}},

    {{{sqrt(3)*a}}} = {{{sqrt(3)x + x}}},

    x = {{{(sqrt(3)*a)/(sqrt(3)+1)}}}.


Thus we know now the x-coordinate of the point A.



So, now we can find the height h of the triangle QAU as the difference a-x:


    h = a-x = a - {{{(sqrt(3)*a)/(sqrt(3)+1)}}} = {{{(sqrt(3)*a + a - sqrt(3)*a)/(sqrt(3)+1)}}} = {{{a/(sqrt(3)+1)}}}.



You can rationalize the denominator

    h = {{{(a/(sqrt(3)+1))*((sqrt(3)-1)/(sqrt(3)-1))}}} = {{{a*(sqrt(3)-1)/(3-1)}}} = {{{a*((sqrt(3)-1)/2)}}}.



Now the area of the triangle QAU is half the product of its base 'a' by its height h = {{{a*((sqrt(3)-1)/2)}}}


The  <U>ANSWER</U>  is :  the area of the triangle QAU is


    {{{(1/2)*a*h}}} = {{{(1/2)*a*(a*(sqrt(3)-1)/2))}}} = {{{a^2*((sqrt(3)-1)/4)}}}.
</pre>

Solved.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Regarding the post by @CPhill . . . 



Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.


The artificial intelligence is like a baby now. It is in the experimental stage 
of development and can make mistakes and produce nonsense without any embarrassment.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It has no feeling of shame - it is shameless.



This time, again, &nbsp;it made an error.



Although the @CPhill' solution are copy-paste &nbsp;Google &nbsp;AI solutions, &nbsp;there is one essential difference.


Every time, &nbsp;Google &nbsp;AI &nbsp;makes a note at the end of its solutions that &nbsp;Google &nbsp;AI &nbsp;is experimental
and can make errors/mistakes.


All @CPhill' solutions are copy-paste of &nbsp;Google &nbsp;AI &nbsp;solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, &nbsp;he &nbsp;NEVER &nbsp;SAYS &nbsp;TRUTH.


Every time, &nbsp;@CPhill embarrassed to tell the truth.

But I am not embarrassing to tell the truth, &nbsp;as it is my duty at this forum.



And the last my comment.


When you obtain such posts from @CPhill, &nbsp;remember, &nbsp;that &nbsp;NOBODY &nbsp;is responsible for their correctness, 
until the specialists and experts will check and confirm their correctness.


Without it, &nbsp;their reliability is &nbsp;ZERO and their creadability is &nbsp;ZERO, &nbsp;too.