The measure of the angle that would maximize the area of this isosceles trapezoid is equal to 0.4395 rad.
Given the following data:
Mathematically, the area of a trapezium is given by this formula:
A = ½ × (a + b) × h
A = ½ × (12 + 2l) × h
A = h(6 + l)
Next, we would derive a mathematical expression for A in terms of h as follows;
A = (6 + 4sin(θ)) × 4cosθ
In order to determine the value of θ for which the area of this isosceles trapezoid is maximized, we would differentiate the area (A) with respect to angle (θ):
Note: sin²θ + cos²θ = 1 ⇒ cos²θ = 1 - sin²θ.
[tex]\frac{dA}{d\theta} =16 cos^{2} \theta - 4sin \theta(6+4sin \theta)\\\\\frac{dA}{d\theta} = 16 cos^{2} \theta - 16 sin^{2} \theta - 24sin\theta\\\\\frac{dA}{d\theta} =16(1-sin^{2} \theta)- 16 sin^{2} \theta - 24sin\theta\\\\\frac{dA}{d\theta} = - 32 sin^{2} \theta - 24sin\theta+16\\\\32 sin^{2} \theta + 24sin\theta-16=0[/tex]
Next, we would use the quadratic formula to solve for the value of sinθ.
Mathematically, the quadratic formula is given by this equation:
[tex]sin\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
Where:
Substituting the parameters into the formula, we have;
[tex]sin\theta = \frac{-24\; \pm\; \sqrt{24^2 - 4(32)(-16)}}{2(32)}\\\\sin\theta = \frac{-24\; \pm\; \sqrt{2624}}{64}\\\\sin\theta = \frac{-24\; \pm\; 51.23}{64}\\\\sin\theta = \frac{-24\;+\; 51.23}{64}\\\\sin\theta = \frac{27.23}{64}\\\\sin\theta = 0.4255\\\\\theta = sin^{-1}(0.4255)[/tex]
θ = 0.4395 rad.
Note: We would only consider the positive value of the quadratic root.
For the obtuse interior angles of the trapezoid, we have [tex](\frac{\pi}{2} +0.4395)[/tex]
Similarly, the measure of the acute interior angles of the trapezoid is [tex](\frac{\pi}{2} -0.4395)[/tex]
Read more on isosceles trapezoid here: https://brainly.com/question/4758162
