Trapezoid $HGFE$ is inscribed in a circle, with $\overline{EF} \parallel \overline{GH}$. If arc $GH$ is $66$ degrees, arc $EH$ is $x^2 + 5x$ degrees, and arc $FG$ is $22 - 4x$ degrees, where $x > 0,$ find arc $EPF$, in degrees.

[asy]
unitsize(2 cm);

pair A, B, C, D, E;

A = dir(170);
B = dir(30);
D = dir(130);
C = dir(70);
E = dir(280);

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

label("$E$", A, W);
label("$F$", B, dir(0));
label("$G$", C, NE);
label("$H$", D, NW);
dot("$P$", E, S);
[/asy]


We are given trapezoid $ HGFE $ inscribed in a circle, with $ \overline{EF} \parallel \overline{GH} $. We are asked to find the measure of arc $ EPF $, and we know the following information about the arcs of the circle:

- Arc $ GH $ is $ 66^\circ $,
- Arc $ EH $ is $ x^2 + 5x $ degrees,
- Arc $ FG $ is $ 22 - 4x $ degrees.

To solve the problem, we will use the fact that the sum of all the arcs in a circle is $ 360^\circ $.

### Step 1: Express total arc measure in terms of $ x $

Since the trapezoid is inscribed in a circle, and $ \overline{EF} \parallel \overline{GH} $, the arcs $ EF $ and $ GH $ are supplementary. This implies that:

\[
\text{Arc } EF + \text{Arc } GH = 180^\circ.
\]

Let’s now express the total measure of arcs $ EF, FG, GH, $ and $ EH $ in terms of $ x $. The sum of all these arcs should equal $ 360^\circ $.

The total measure of the arcs can be written as:

\[
\text{Arc } EH + \text{Arc } FG + \text{Arc } GH = 360^\  \

Trapezoid $HGFE$ is inscribed in a circle, with $\overline{EF} \parallel \overline{GH}$. If arc $GH$ is $66$ degrees, arc $EH$ is $x^2 + 5x$ degrees, and arc $FG$ is $22 - 4x$ degrees, where $x > 0,$ find arc $EPF$, in degrees.

Learn how to solve a geometry problem involving an inscribed trapezoid in a circle. The problem includes arc calculations, quadratic expressions, and the properties of trapezoids with parallel sides. Suitable for high school students in Grades 10-12.

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