Trapezoid HGFE is inscribed in a circle, with EF || GH. If arc GH is 70 degrees, arc EH is x^2 - 2x degrees, and arc FG is 48 - 4x degrees, where x > 0 find arc EPF, in degrees.

Guest Jan 6, 2021

#1**0 **

A trapezoid inscribed in a circle is isosceles.

Therefore, EH = GF, and arc (EH) = arc (GF).

EH is x^2-2x and FG is 48-4x.

So, x^2-2x = 48-4x => x^2+2x-48 = 0.

Factoring, we get (x+8)(x-6) = 0.

Since x = -8 wouldn't make sense, we are left with x = 6.

EH = 24, GH = 70, GF = 24.

EPF = 360-24-70-24= 212.

So I think the answer is 212.

I hope this helps!

Guest Jan 6, 2021

#2**+1 **

Trapezoid HGFE is inscribed in a circle, with EF || GH. If arc GH is 70 degrees, arc EH is x^2 - 2x degrees and arc FG is 48 - 4x degrees, where x > 0 find arc EPF, in degrees.

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x^{2} - 2x = 48 - 4x x = 6

48 - 4 * 6 = 24

**Arc EPF = 360 - (70 + 2 * 24) = 242 degrees**

Guest Jan 6, 2021