Question 1180067
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            The answer and the solution that you got from @MathLover1,  are  ABSURDIST.



            Indeed, for the area of the inscribed square,  she gives you   A = {{{2*(10-x)^2}}}.


            Substitute here  x= 0: you will get  A = 200 square units for the area of the inscribed square,
            i.e. is  TWO  TIMES  the area of the original square.



            Similarly,  for the perimeter of the inscribed square,  she gives you  P = {{{4*(10-x)*sqrt(2)}}}.


            Substitute here  x= 0: you will get  P = {{{40sqrt(2)}}} units for the perimeter of the inscribed square,
            i.e. is  1.41  times the perimeter of the original square.



            So,  her solutions,  OBVIOUSLY,  are only good to throw them all to a  TRASH  BIN.



Ok,  don't worry - - - you are in good hands.   I came to bring a correct solution.



<pre>
The square of the side of the inscribed triangle is 

    {{{a^2}}} = {{{x^2}}} + {{{(10-x)^2}}} = {{{2x^2 - 20x + 100}}} = {{{2*(x^2-10x+50)}}}.



THEREFORE, the area of the inscibed square

    area = {{{a^2}}} = {{{2x^2 - 20x + 100}}} = {{{2*(x^2-10x+50)}}}.      <U>ANSWER</U>



The perimeter is 4 (four) times the side lenght "a"

    perimeter = {{{4*sqrt(2x^2 - 20x + 100)}}} = {{{4*sqrt(2)*sqrt(x^2-10x+50)}}}.



Regarding the domain and the range of these functions, the question is posed in ambiguous way:


    It has one answer for formal functions and another answer 
    if to base it on geometric meaning of variables.


THEREFORE, I prefer do not answer this question, until it is posed in a formally correct way.
</pre>