The internal angles of 64.801° (2 in total) and 115.199° (2 in total) lead to a maximum area of 27.88 square units.
How to determine the maximum possible area of an isosceles trapezoid
Quadrilaterals are figures formed by four line segments and whose sum of internal angles equals 360°. By geometry we know that the area of a trapezoid is the average of the length of the two bases (B, b), in inches, multiplied by the height of the quadrilateral.
A = 0.5 · (B + b) · h (1)
By trigonometry the measure of the longer base and the height of the isosceles trapezoid are, respectively:
B = b + 2 · l · cos θ (2)
h = l · sin θ (3)
Where θ is an internal angle, in degrees.
By (2) and (3) in (1):
A = 0.5 · (2 · b + 2 · l · cos θ) · (l · sin θ)
A = (b + l · cos θ) · (l · sin θ)
A = b · l · sin θ + l² · sin θ · cos θ
A = b · l · sin θ + 0.5 · l² · sin 2θ (4)
If we know that b = 6 in and l = 4 in, then the area of the isosceles trapezoid is represented by the following function:
A = 24 · sin θ + 8 · sin 2θ (5)
Now we determine the angle associated to the maximum area by graphic approach.
According to this approach, the internal angles of 64.801° (2 in total) and 115.199° (2 in total) lead to a maximum area of 27.88 square units. [tex]\blacksquare[/tex]
To learn more on trapezoids, we kindly invite to check this verified question: https://brainly.com/question/8643562
