Area of circle segment using parallel trapezoid This is the circle with the parallel trapezoid

$$A_t = \frac{h(b+h)}{2}$$
Question of the task: Examine how good approximations this formula gives for different values for $b$ in the interval $0<b<2r$
I Used this circle and divided it like this:

$$x=\frac{b}{2}$$
I supposed $r=1$ and by then using Pythagoras theorem got $a=(1-x^2)^{1/2}$
then I calculated the value of "h" by using Pythagoras theorem and then $p-q$ formula
$$h=\frac{2}{5} \left(\sqrt{4+x^2} - 2\sqrt{1-x^2}\right)$$
this is how far I have come and don't know how to continue. I got a "hint" of how I could continue.
Hint: the trapezoid will be $A_t=\frac{h(2x+h)}{2}$, which after substitution could be expressed as a function of $x$.
with this hint I added in the value of h in the formula for the area of the trapezoid and got (this part I am unsure of if right):
$$
h(x) = \frac{(\frac{2}{5})((4+x^2)^{1/2} - 2(1-x^2)^{1/2}) (2x+(\frac{2}{5})((4+x^2)^{1/2} - 2(1-x^2)^{1/2}))}{2}
$$
Then the hint followed by: the circle segment area is
$$
\begin{split}
2\int_{(1-x^2)^{1/2}}^1 (1-t^2)^{1/2}dt 
 &= \frac{\pi}{2} - x(1-x^2)^{1/2}-\arcsin(1-x^2)^{1/2} \\
 &= \arcsin(x) - x(1-x^2)^{1/2}
\end{split}
$$
I did not understand this hint or how he came to the answer for it. Also how I could go on from this step now that I have a function for $h(x)$ and a formula for what I assume is the area of the circle segment.
 A: Everything you have written after

I used this circle and divided it like this:

is irrelevant to the original question.
The area of the orange shaded region in the first diagram is compared to the area of the trapezoid in that same diagram.  The area of the trapezoid is $$A = \frac{h(b+h)}{2}.$$  The area of the orange shaded region is equal to the area of the corresponding circular sector, minus the area of the isosceles triangle with base $b$ and height $r-h$.  Since the angle subtended by the sector is $$\theta = 2 \arcsin \frac{b}{2r},$$ it follows that the area of the orange shaded region is $$S = \frac{1}{2}r^2 \theta - \frac{1}{2} b(r-h) = r^2 \arcsin \frac{b}{2r} - \frac{1}{2} b(r-h).$$  However, since $A$ is not a function of $r$ but $S$ is a function of $r$, we must eliminate $r$ from this equation by noting that $r$, $b$, and $h$ are related; specifically, $$(b/2)^2 + (r-h)^2 = r^2.$$  I leave it as an exercise to perform the computation and express $S$ in terms of $b$ and $h$, after which a comparison of $A$ and $S$ will be possible.
A: I will also assume $r = 1$ since this has no effect on the relative accuracy of any of the calculations.
For the area of the circle segment, the hint is suggesting that you find the
area by integration.
Consider a horizontal slice of the circle segment at a distance $t$ above the center of the circle. The horizontal slice has total length
$2(1 - t^2)^{1/2}$,
of which $(1 - t^2)^{1/2}$ is in the left half of the segment and
$(1 - t^2)^{1/2}$ is in the right half.
The bottom slice is at the bottom of the segment, at a distance $a$ above the center of the circle, and the top slice is at the top of the circle, at a distance $1$ from the center.
So the hint is saying you can get the area of one half of the segment
(the right half, for example) by integrating the half-slice lengths from
$a$ (at the bottom) to $1$ (at the top):
$$ \int_a^1 (1 - t^2)^{1/2} \, \mathrm dt. $$
But in the hint they have not given the name $a$ to the distance from the center to the bottom of the segment, so instead they use this cacluated value:
$a = (1 - x^2)^{1/2}.$
But you don't need calculus for this. Look again at the original figure:

The right half of the circular segment lies between the two radial lines drawn in this figure. The angle between those radial lines is $\arcsin(x)$ where $x = \frac b2$.
The region of the the circle between the radial lines, which is a circular sector, therefore has area $\frac12 \arcsin(x).$
But the circular sector has two parts, a colored part with is half of the circular segment and a white triangle below that.
The area of the triangle is $\frac 12 ax.$
Therefore the area of half the circular segment is what you get after subtracting the triangle's area from the sector's area:
$$ \frac12 \arcsin(x) - \frac 12 ax.$$
Now double that in order to count both halves of the circular segment,
and remember that $a = (1 - x^2)^{1/2}$ so you get a formula with $x$ the only variable:
$$ \arcsin(x) - x(1 - x^2)^{1/2}.$$
This is indeed the exact area of the circular segment.
The next step would be to compare the area of the circular segment to the area of the trapezoid to see "how good approximations this formula gives".
The question asks you to do this comparison for different values of $b.$
It doesn't say how many values but I would suppose it should be more than two. It also doesn't say how to compare the areas, but I would use a ratio (close to $1$ is better) or a percentage difference; if you have done any other approximation questions you could see how the difference was described in those examples.

Finally, look again at the figure from the original question.
The top of the trapezoid is tangent to the circle, not a chord of the circle.
So the distance from the center of the circle to the top of the trapezoid,
which is $a + h,$ is also the radius of the circle.
That is, $a + h = 1,$ as pointed out in a comment.
This provides a much simpler formula for $h$,
even after the substitution $a = (1 - x^2)^{1/2}$,
than the one you derived from an incorrect diagram.
