Another take.
Let $(0,-r)$ and $(x, y)$ be the endpoints of the arc. Then
$h = |r - y|$ and $w = |x|$.
We know $x^2 + y^2 = r^2$ so $h = |r \pm \sqrt{r^2 - w^2}| = r \pm \sqrt{r^2 - w^2}$. And $w = |x| = \sqrt{r^2 - y^2} = \sqrt{r^2 - (h-r)^2} = \sqrt{2hr - h^2}$
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However not your interpretation of width and height are a little odd. You assume you are lying the arc on its tangent (like pressing one end of a clipped toenail to the floor) and measuring the displacement.
I think most people would assume the width is the measure from endpoint to endpoint and the height to be the displace of that. (Like leaving the toenail balancing one its center and measuring how high and far apart the ends are).
(The difference between the new picture and your picture is in the new picture the radius goes through the arc cutting it in half. The height is from the midpoint of the chord to the tip of the circle, and the width is the length of a chord.)
(In Your picture the chord is the hypotenuse slanting upward. Width is the horizontal leg and the height is the vertical leg. I think most would view the chord itself as the width, and the height as the perpendicular bisector of the chord to the circle.)
In this case the endpoints are $(-x,y)$ and $(x,y)$ and
$w = 2x$ and $h = r - y$. Using $x^2 + y^2 = r^2$ we get:
$w = 2\sqrt{r^2 - (r-h)^2} = 2\sqrt{2hr - h^2}$ and
$h = r \pm \sqrt{r^2 - ((1/2)w)^2} = r \pm \sqrt {r^2 - w^2/4}$.
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Note. Given an arc's width and length you can calculate the radius of the circle.
I probably should have included those as well.
In your picture: $(r-h)^2 + w^2 = r^2$ so $w^2 = 2hr - h^2$ so $r = (h^2 + w^2)/2h$.
In mine: $(r-h)^2 + (w/2)^2 = r^2$ so $w^2/4 = 2hr - h^2$ so $r = (h^2 + w^2/4)/2h$.
Best Answer
Two jokes I tell my students (though I tell them better in person) when width/height confusion strikes (don't feel bad ... it happens often!):
and
... and that, my friends, is why I always endeavor to phrase the formula for the area of a parallelogram as "base-times-altitude".
Importantly, in the formula "base-time-altitude", the "base" is indeed any side you choose, but "altitude" is the altitude perpendicular to the chosen base.
So, in your problem, while $8$cm is a perfectly-good base measurement, the $6$cm measurement is not perpendicular to that base, so these numbers don't fit together into the area formula. (The altitude for the $8$cm base doesn't appear in the diagram at all.) The $6$cm measure is of the altitude perpendicular to the base of length $10$cm, so that the area is $10\times 6 = 60\text{cm}^2$.