Answer:
Arc EPF is 240°
Step-by-step 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°.
