Question

Trapezoid EFGH is inscribed in a circle, with
`EF \parallel GH`. If arc GH is 70 degrees, arc EH is x^2 - 2x degrees, and arc FG is 56 - 3x degrees, where x > 0, find arc EPF, in degrees.

Answer 1

Since trapezoid EFGH is inscribed in a circle, with EF parallel to GH, the measure of arc EPF is equal to 220°.

In Mathematics and Euclidean Geometry, parallel chords that situated within the same circle always cut congruent arcs. This ultimately implies that, the measures of the arcs between parallel chords of a circle must be equal.

In this context, we have the following equal arcs;

Arc EH = Arc FG

`x^2 - 2x = 56 - 3x\\\\x^(2) -2x+3x-56=0\\\\x^(2) +x-56=0\\\\x^(2) +8x-7x-56=0\\\\(x + 8)(x - 7) = 0`

x = -8 or x = 7

Since x is a positive integer (x > 0), the value of x must be 7.

For the measure of arc EPF, we have:

arc EPF + arc EHGF = 360°

arc EPF = 360° - arc EHGF

arc EPF = 360° - (35° + 70° + 35°)

arc EPF = 360° - 140°

arc EPF = 220°.

Answer 2

Arc EPF is 240°

Explanation:

Since the quadrilateral, EFGH is a trapezoid and EF is parallel to GH, we have;

∠HGF + ∠GFE = 180°

∠GHE + ∠GFE = 180°

∠HGF + ∠HEF = 180°

∴∠HEF = ∠GFE

In ΔHEF and ΔGFE

∠EHF = ∠EGF (Angles subtending the same segment)

With side EF common to both triangles and ∠HEF = ∠GFE , we have;

ΔHEF ≅ ΔGFE (Angle Angle Side rule)

Hence, side FG = EH

For cyclic trapezoid side FG = EH

The base angles subtended by GH = 70

Arc EH = x² - 2·x

Arc FG = 56 - 3·x

Therefore;

70 + x² - 2·x + 56 - 3·x + arc EPF = 360 .............(1)

Also since the equation of a circle is (x-h)² + (y-k)² = r², where the center of the circle is (h, k), then as EF is a displacement of say z from GH, then arc EH = FG which gives;

x² - 2·x = 56 - 3·x

x² - 2·x - 56 + 3·x = 0

x² + x - 56 = 0

(x - 7)(x + 8) = 0

Therefore, since x > 0 we have x = 7

Plugging in the value of x into the equation (1), we have

70 + 7² - 2·7 + 56 - 3·7 + arc EPF = 360 .............(1)

70 + 70 + arc EPF = 360

arc EPF = 360 - 140 = 240°.