SOLUTION: A rectangle is inscribed in a circle of radius r. If the rectangle has a length of 3x, write the area of the rectangle as a function of both x and r

Algebra ->  Functions -> SOLUTION: A rectangle is inscribed in a circle of radius r. If the rectangle has a length of 3x, write the area of the rectangle as a function of both x and r      Log On


   

Question 1193178: A rectangle is inscribed in a circle of radius r. If the rectangle has a length of 3x, write the area of the rectangle as a function of both x and r
Found 2 solutions by greenestamps, ankor@dixie-net.com:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!




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

sqrt%28%282r%29%5E2-%283x%29%5E2%29=sqrt%284r%5E2-9x%5E2%29

Then the area of the rectangle is length times width:

ANSWER: The area of the rectangle is %283x%29%28sqrt%284r%5E2-9x%5E2%29%29

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle is inscribed in a circle of radius r.
If the rectangle has a length of 3x, write the area of the rectangle as a function of both x and r
:
let s = the short side of the rectangle
then
A = 3x * s
find s as a function of r and x
The diagonal of the rectangle = 2r
Using pythag
s^2 + (3x)^2 = (2r)^2
s^2 = (2r)^2 - (3x)^2
s = sqrt%28%282r%29%5E2+-+%283x%29%5E2%29
:
A = 3x%2Asqrt%28%282r%29%5E2+-+%283x%29%5E2%29
or
3x%2Asqrt%28%284r%5E2%29+-+%289x%5E2%29%29