A parallelogram inscribed in a circle is a rectangle or a square. The perpendicular bisector
of two opposite sides of the rectangle corresponds to a diameter of the circle.
So, a rectangle inscribed in a semicircle rests on the diameter.
( Proving things mentioned above is skipped in here.)
Label each vertex of the rectangle A, B, C and D counterclockwise as placing base AB
on the diameter and vertices C and D on the circumference.
The midpoint of AB corresponds to the circle's center O.
Let angle∠AOD=x (0<x<π/2), the radius of circle=r (=4) and the area of rectangle=f(x).
OD=OC=r, so AD=BC=r·sinx and AB=CD=2r·cosx, so
f(x)=AD·BC=r·sinx·2r·cosx=2r²·sinx·cosx
Use the double-angle formula for sine: sin2x=2sinx·cosx, so we have:
f(x)=r²·sin2x Differentiate f(x) using the chain rule:
f'(x)=2r²·cos2x and f''(x)=-4r²·sin2x Here, 0<x<π/2 so 0<2x<π
Thus, if f'(x)=0, then cos2x=0, so x=π/4 and f''(π/4)=-4r²·sin(π/2)=-4r²<0
If f'(π/4)=0 and f''(π/4)<0, then f(x) takes its maximum at x=π/4.
We have: f(π/4)=r²·sin(π/2)=4²=16
AD=BC=4sin(π/4)=2√2 and AB=CD=8cos(π/4)=4√2
Therefore, the answers are:
1. The lagest area of rectangle is 16 unit squared.
2. The shorter sides are 2√2 (=approx. 2.83) units long.
3. The longer sides are 4√2 (=approx. 5.66) units long.