Hello, midnightnurse!
We need more information.
A full description of the figure is required.
Presently, I am guessing the following "facts" . . .
We have a quadrant of a circle with radius 4 (the "4-circle").
There is a semicircle of radius 2 (the "2-circle").
The "4 cm" at the bottom is not the length of the chord.
. . It is obviously a radius of the 4-circle.
We have a mystery circle tangent to the 2-circle.
It appears to extend beyond the 4-circle.
With no further constraints, it could have any radius.
I assume the problem intends this:
P1) Rightmost semicircle is centered on that vertical line, has diameter 4 cm.
P2) Bottom shape is a semicircle whose center is on that horizontal line.
P3) The circles touch at a point (hence tangentially).
P4) The leftmost point of the bottom circle is 4 cm from the bottom-most point of the right semicircle.
In that case, you might set the problem up this way:
0) All distances in centimeters.
1) Bottom left point is the origin.
2) Bottom semicricle has radius r.
3) Right semicircle has radius 2 and center (4, 2).
Then by #2, bottom semicircle has center at (r, 0).
Here's the key observation: since they touch tangentially, the distance between two centers is the sum of their radii.
That becomes Distance( (r,0), (4,2) ) = 2 + r.
Thus:
$\displaystyle \sqrt{ (4-r)^2 + (2-0)^2} = 2+r$
$\displaystyle \sqrt{ (4-r)^2 + 4} = 2+r$
$\displaystyle (4-r)^2 + 4 = (2+r)^2$
$\displaystyle (16 - 2(4)(r) + r^2) + 4 = 4 + 2(2)(r) + r^2$
$\displaystyle r^2 - 8r + 20 = r^2 +4r + 4$
$\displaystyle - 8r + 20 = 4r + 4$
$\displaystyle - 8r -4r = 4 - 20$
$\displaystyle -12r = -16$
$\displaystyle r = \frac{-16}{-12}$
$\displaystyle r = \frac{4}{3}$.
The answer is that the bottom semicircle has a radius of $\displaystyle \frac{4}{3}$ cm.