Finding the area of a semicircle knowing a chord and other small semicircle tangent to it. In a semicircle of diameter $CD$ there's a chord $AB$ of length 7, and it's parallel to the diameter. There's also a small semicircle that is tangent to $AB$ and its diameter is a segment in $CD$ . Find the area of the semicircle without the small semicircle. I'm pretty curious about this problem, i've tried many things, like drawing triangles that are similar and also rectangles, so i tries with pithagorean theorem but i got nothing. I'd really like to know how to solve it, thanks.

Place the small semicircle center on the center of the big semicircle, for simplicity. This won't change the area you are looking for.
Let us call $r$, $R$ and $d$ the radius of the small semicircle, the radius of the big semicircle and half the chord, respectively.
Using your notation, join $A$ with the center $O$ of the semicircles: that is of course a radius of the big semicircle. If you erect a perpendicular from the center to the chord, intersecting at $H$, you draw a right triangle $AOH$, whose hypotenuse is the radius $R$ and whose cathetuses are $r$ and $d$. The area you are looking for is: $$\frac{\pi}{2}R^2 - \frac{\pi}{2} r^2$$ Though, by Pythagorean Theorem, you have that $R^2 = r^2 + d^2$. By plugging this into the preivous expression, you find: $$A= \frac{\pi}{2}(r^2 + d^2) - \frac{\pi}{2}r^2 = \frac{\pi}{2}d^2$$ Now you can calculate that area, since you have the length of the chord and, thus, $d=7/2$. As a result, $$A = \frac{49 \pi}{8}$$