Given a semicircle of radius r, the task is to find the largest trapezoid that can be inscribed in the semicircle, with base lying on the diameter.
Input: r = 5 Output: 32.476 Input: r = 8 Output: 83.1384
Approach: Let r be the radius of the semicircle, x be the lower edge of the trapezoid, and y the upper edge, & h be the height of the trapezoid.
Now from the figure,
r^2 = h^2 + (y/2)^2
or, 4r^2 = 4h^2 + y^2
y^2 = 4r^2 – 4h^2
y = 2√(r^2 – h^2)
We know, Area of Trapezoid, A = (x + y)*h/2
So, A = hr + h√(r^2 – h^2)
taking the derivative of this area function with respect to h, (noting that r is a constant since we are given the semicircle of radius r to start with)
dA/dh = r + √(r^2 – h^2) – h^2/√(r^2 – h^2)
To find the critical points we set the derivative equal to zero and solve for h, we get
h = √3/2 * r
So, x = 2 * r & y = r
So, A = (3 * √3 * r^2)/4
Below is the implementation of above approach:
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