Question 1193178
<br>
{{{drawing(400,400,-6,6,-6,6,circle(0,0,5)
,line(-4,-3,4,-3),line(4,-3,4,3),line(4,3,-4,3),line(-4,3,-4,-3),line(-4,-3,4,3)
,locate(0,-2.5,3x),locate(1,.5,2r)
)}}}<br>
The diagonal of the rectangle is a diameter of the circle; and it is the hypotenuse of a right triangle whose legs are the length and width of the rectangle.  With hypotenuse of length 2r and the length of the rectangle 3x, the width of the rectangle is<br>
{{{sqrt((2r)^2-(3x)^2)=sqrt(4r^2-9x^2)}}}<br>
Then the area of the rectangle is length times width:<br>
ANSWER: The area of the rectangle is {{{(3x)(sqrt(4r^2-9x^2))}}}<br>