Question

Find the maximum area of a trapezoid inscribed in a semicircle

Answer 1

Sure, let's approach this step by step.

First, let's comprehend the relationship between the trapezoid and the semicircle.

The maximum area of a trapezoid inscribed in a semicircle occurs when the trapezoid is essentially an isosceles triangle where the base of the triangle is perfectly aligned with the diameter of the semicircle.

Why?

The area of a triangle is given by the formula 1/2 * base * height. The highest possible height of a triangle inscribed in a semicircle is the radius of that semicircle. The longest possible base is the diameter, which is twice the radius. The area therefore maximizes when the inscribed trapezoid conforms to these conditions, i.e., it becomes an isosceles triangle.

Next, we would need to calculate the maximum area.

Since we have clarified that the maximum area is represented by an isosceles triangle with the triangle base as the semicircle's diameter (or 2*r where r represents the radius of the semicircle) and the height equivalent to the radius, we plug these values into the area formula of a triangle.

This means the formula becomes: A = 1/2 * base * height = 1/2 * 2*r * r = r^2.

Therefore, the maximum area of a trapezoid inscribed in a semicircle of radius r is represented by the formula A = r^2.

In conclusion, to find the maximum area of a trapezoid inscribed in semicircle, place an isosceles triangle within the semicircle such that the base of the triangle aligns with the semicircle's diameter and the height is the radius. The formula for the maximum area then would be A = r^2.

Answer 2

Without loss of generality, we can assume the semicircle has a radius of 1 and is described by
y = √(1 - x²)
Then the shorter base has length 2x and the longer base has length 2. The area of the trapezoid is
A = (1/2)(2x+2)√(1-x²) = (1+x)√(1-x²)

Differentiating with respect to x, we have
A' = √(1-x²) + (1+x)(-2x)/(2√(1-x²)
Setting this to zero, we get
0 = (1-x²) +(1+x)(-x)
0 = 2x² +x -1
(2x-1)(x+1) = 0
x = {-1, 1/2} . . . . . -1 is an extraneous solution that gives minimum area

So, for x = 1/2, the area is
A = (1 + 1/2)√(1 - (1/2)² = (3/2)√(3/4)
A = (3/4)√3

Of course, if the radius of the semicircle is "r", the maximum area is
A = (r²·3·√3)/4