Question 739994
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a rectangle has length 2squarerootxcm and width squarerootxcm. the length of a diagonal of the rectangle is the squareroot of 45cm.
1 find the area of the rectangle
2 the area of the square is twice the area of the rectangle.
find the length of a side of a square.
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<pre>
             <U>P a r t   1.</U>


The length is  {{{2sqrt(x)}}}.  The width is  {{{sqrt(x)}}}.

The diagonal is  {{{sqrt((2sqrt(x))^2 + (sqrt(x))^2)}}} = {{{sqrt(4x + x)}}} = {{{sqrt(5x)}}}.


From the problem, we have this equation

    {{{sqrt(5x)}}} = {{{sqrt(45)}}}.


Square both sides

    5x = 45

and find  x = 45/5 = 9.


Then the area of this rectangle is  Length*Width = {{{(2sqrt(9))*sqrt(9)}}} = 2*9 = 18 cm^2.   <U>ANSWER</U>


Part 1 is complete.



             <U>P a r t   2.</U>


The area of the square is twice the area of the rectangle, i.e.  2*18 = 36 cm^2.


Hence, the side of the square is  {{{sqrt(36)}}} = 6 cm.    <U>ANSWER</U>


Part 2 is complete.
</pre>

The problem is fully solved.