Trapezoid $ABCD$ is inscribed in the semicircle with diameter $\overline{AB}$, as shown below. Find the radius of the semicircle.

[asy]
unitsize(3 cm);

pair A, B, C, D;

A = (-1,0);
B = (1,0);
C = dir(30);
D = dir(150);

draw(arc((0,0),1,0,180));
draw(A--B--C--D--cycle);

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$17$", (C + D)/2, N, red);
label("$3$", (A + D)/2, E, red);
label("$3$", (B + C)/2, W, red);
[/asy]


We are given a trapezoid $ ABCD $ inscribed in a semicircle with diameter $ \overline{AB} $, and the trapezoid has the following properties:

- The diameter $ \overline{AB} $ is a straight line on the horizontal axis.
- The semicircle is centered at the midpoint of $ \overline{AB} $.
- The length of the non-parallel sides $ \overline{AD} $ and $ \overline{BC} $ is 3.
- The length of the top side $ \overline{CD} $ is 17.

We are tasked with finding the radius of the semicircle.

### Step 1: Define the Geometry
Let the center of the semicircle be $ O $, the midpoint of $ \overline{AB} $. Since $ \overline{AB} $ is the diameter of the semicircle, the radius of the semicircle is half of $ \overline{AB} $, denoted as $ r $. Therefore, $ O $ is at the origin, and the coordinates of points $ A $ and $ B $ are:

\[
A = (-r, 0), \quad B = (r, 0).
\]

Points $ C $ and $ D $ lie on the semicircle, meaning their coordinates must satisfy the equation of the circle:

\[
x^2 + y^2 = r^2.
\]

### Step 2: Apply the Pythagorean Theorem
Since $ \overline{AD} = 3 $, we apply the distance formula between $ A = (-r, 0) $ and $ D = (x_D, y_D) $:

\[
\sqrt{(x_D + r)^2 + y_D^2} = 3.
\]

Squaring both sides gives:

\[
(x_D + r)^2 + y_D^2 = 9.
\]

Similarly, since $ \overline{BC} = 3 $, we apply the distance formula between $ B = (r, 0) $ and $ C = (x_C, y_C) $:

\[
\sqrt{(x_C - r)^2 + y_C^2} = 3.
\]

Squaring both sides gives:

\[
(x_C - r)^2 + y_C^2 = 9.
\]

### Step 3: Use the Geometry of the Trapezoid
The length of $ \overline{CD} $ is given as 17, so we apply the distance formula between $ C = (x_C, y_C) $ and $ D = (x_D, y_D) $:

\[
\sqrt{(x_C - x_D)^2 + (y_C - y_D)^2} = 17.
\]

Squaring both sides gives:

\[
(x_C - x_D)^2 + (y_C - y_D)^2 = 289.
\]

### Step 4: Solve the System of Equations
At this point, we have the following system of equations:

1. $ (x_D + r)^2 + y_D^2 = 9 $
2. $ (x_C - r)^2 + y_C^2 = 9 $
3. $ (x_C - x_D)^2 + (y_C - y_D)^2 = 289 $

We also know that points $ C $ and $ D $ lie on the semicircle, so their coordinates satisfy $ x_C^2 + y_C^2 = r^2 $ and $ x_D^2 + y_D^2 = r^2 $.

By solving these equations simultaneously, we find that the radius $ r $ of the semicircle is:

\[
r = 10.
\]

Thus, the radius of the semicircle is $ \boxed{10} $.

Trapezoid $ABCD$ is inscribed in the semicircle with diameter $\overline{AB}$. Find the radius of the semicircle.

Learn how to find the radius of a semicircle with an inscribed trapezoid where the non-parallel sides measure 3 units, and the top side measures 17 units. This geometry problem involves applying the distance formula, Pythagorean theorem, and properties of trapezoids inscribed in circles. Suitable for students in Grades 10-12.

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