Question 1209520
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Tutor @ikleyn has provided a response showing a valid solution using equations of the lines in the figure.<br>
Here is a very different solution.<br>
Draw segment AP parallel to SQ with P on QU.  Let a be the length of AP.<br>
Angle PUA is 45 degrees, so triangle APU is a 45-45-90 right triangle.  The length of PU is then a.<br>
Angle SQA is 60 degrees, so angle AQP is 30 degrees; that make triangle APQ a 30-60-90 right triangle.  So then the length of QP is a*sqrt(3).<br>
Use the lengths of PU and QP to find an expression for a in terms of the side length of the square.<br>
{{{a+a*sqrt(3)=x}}}
{{{a(1+sqrt(3))=x}}}
{{{a=x/(1+sqrt(3))}}}
{{{a=(x(sqrt(3)-1))/((sqrt(3)+1)(sqrt(3)-1))}}}
{{{a=(x/2)(sqrt(3)-1)}}}<br>
Use the standard formula one-half base times height to find the area of triangle QAU.  The base is x and the height is a.<br>
{{{A=(1/2)(x)((x/2)(sqrt(3)-1))=(x^2/4)(sqrt(3)-1)}}}<br>
ANSWER: {{{(x^2/4)(sqrt(3)-1)}}}<br>