SOLUTION: The rectangle ABCD is inscribed in the circle with center(0,0) and radius 6. The x-coordinate of the vertex A is expressed as x. (Shows a diagram with an x and y axis. A circle,

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Question 56077: The rectangle ABCD is inscribed in the circle with center(0,0) and radius 6. The x-coordinate of the vertex A is expressed as x. (Shows a diagram with an x and y axis. A circle, with a square inside it where each of the 4 corners of the square touches the circle. And a point A at the top right corner labeled as point A (x,y). Where A is the vertex.)
Then express the area S of the rectangle ABCD as a function of x in the form: A(x) = b*x*sqr(c-x^2)
Where b and c are coefficients.
I have tried area=base*height which gives the formula area= x^2+Y^2=12^2. I just am not sure where to go from here or if I am even on the right track.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
The rectangle ABCD is inscribed in the circle with center(0,0) and radius 6. The x-coordinate of the vertex A is expressed as x. (Shows a diagram with an x and y axis. A circle, with a square inside it where each of the 4 corners of the square touches the circle. And a point A at the top right corner labeled as point A (x,y). Where A is the vertex.)
OK ..BUT THE DIAGRAM IS NEEDED AS THE DATA GIVEN BY YOU IS NOT FULLY CLEAR.IS THE RECTANGLE AND CIRCLE HAVE SAME CENTRE IS A VERY IMPORTANT POINT.ANY WAY ASSUMING THAT CENTRE IS SAME AND THE AXES ARE PARALLEL TO SIDES,I AM GIVING THE ANSWER.PLEASE CONFIRM THE ASSUMPTIONJS OR GIVE THE DRAWING.
SO AS PER OUR ASSUMPTION
AB || DC || Y AXIS
BC || AD || X AXIS
CENTRE OF RECTANGLE IS O(0,0) AS IS THE CENTRE OF CIRCLE.
HENCE DIAGONAL = AC = DIAMETER = 2*6=12 AND OA = RADIUS = 6
SINCE A IS (X,Y)
OA^2 = 6^2 = X^2+Y^2
Y=SQRT(36-X^2)
SIDES OF RECTANGLE ARE
AB = 2Y=2SQRT(36-X^2)
BC = 2X
AREA OF RECTANGLE = AB*BC = 2X*2SQRT(36-X^2)=4XSQRT(36-X^2)
B=4.....C=36 ...IN YOUR EQN.



Then express the area S of the rectangle ABCD as a function of x in the form: A(x) = b*x*sqr(c-x^2)
Where b and c are coefficients.
I have tried area=base*height which gives the formula area= x^2+Y^2=12^2. I just am not sure where to go from here or if I am even on the right track.