Help, I've been staring at this problem for days...?

Question

I just can't seem to make sense of it.
A trapezoid is inscribed in the upper half of a unit circle.
a)write the area of the trapezoid as a function of the angle theta.
b)Find the value of the angle theta that maximizes the area of the trapezoid and the maximum area.

If somebody could just help explain how to get the first part (a) I would appreciate it.

It doesn't give an angle for theta. The x,y coor. are from (-1,0) to (1,0). There's a point x,y that looks to make a 45deg angle and there's an equaltion that says x^2+y^2=1, which means the radius is one, right?

That probably doesn't make any sense...sorry

It's from the origin up.

Answer 1

Where is the angle Θ? Is it at the origin, or is it one of the sides of the trapezoid?

Edit: from your description, it sounds like Θ is at the origin. And the point at the upper right of the trapezoid would be (x,y). This point is also (sin(Θ), cos(Θ))

You have an isosceles trapezoid formed by the points (1,0), (x, y), (-x, y), (-1, 0). And x = cos(Θ), y = sin(Θ).

The area of a trapezoid is given by the formula:
(b1 + b2)/2 * h

In your case:
Base 1 goes from from (-1,0) to (1,0) for a length of 2.
Base 2 goes from (-x,y) to (x,y) for a length of 2x or 2 cos(Θ).

So the average of these two:
(2 + 2 cos(Θ))/2
= 1 + cos(Θ)

And the height is y or just sin(Θ).

So putting it all together, a formula for the area of your trapezoid is:

f(Θ) = sin(Θ) * (1 + cos(Θ) )

As requested, there is part (a)...

For part (b) ... take the derivative and set it to zero. You will get a Θ between 0° and 90° (π/2 radians)... I suspect that the maximum area will be at Θ = 60° (π/3 radians). Just thinking about it visually, that would be the top half of a regular hexagon...

Answer 2

Let's assume that the base of the trapezoid is the diameter of the semicircle and coincident with x-axis. The trapezoid with the largest area has to be an isosceles trapezoid. I guess the angle Θ is formed by the x-axis and the line from the origin to the corner of the trapezoid in the first quadrant.

The radius of the semicircle is appapently 1. The height of the trapezoid is sin (Θ).

I accept Puzzling's solution to part a).

Allow me to take the first derivative of the formula for the area:

f(Θ) = sin(Θ) * (1 + cos(Θ) )
f'(Θ) = (sin(Θ))(-sin(Θ)) + (1 + cos(Θ))(cos(Θ))
= -(sin(Θ))^2 + cos(Θ) + (cos(Θ))^2

Set that equal to zero and solve for Θ.

Let's see if Puzzling is right about that 60 degrees.
cos 60 = 1/2
sin 60 = sqrt(3)/2

f'(60) = -3/4 + 1/2 + 1/4 = 0
Yes!!

And the area = sqrt(3)/2 * 3/2 = 3sqrt(3)/4

Answer 3

u need to give a picture then i can try