Question 1199394
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A rectangle is inscribed in a circle of radius r. If the rectangle has length 3x, 
write the area of the  rectangle as the product of two functions.  
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<pre>
For any rectangle, inscribed in a circle, the diagonals of the rectangle are diameters of the circle,
i.e. the diagonals pass through the center of the circle.


In other words, the diagonals of a inscribed rectangle intersect each other at the center of the circle.


Hence, in our case, the hypotenuse of the right angled triangle is 2r, while one of the leg is 3x 
(= the length of the rectangle).


It implies that the width of the rectangle is  {{{sqrt((2r)^2 - (3x)^2)}}} = {{{sqrt(4r^2-9x^2)}}}.


Then the area of the rectangle is the product of its length by its width


    Area of the rectangle = {{{3x*sqrt(4r^2-9x^2)}}},


giving the answer to your question.
</pre>

Solved.