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Trapezoid EFGH is inscribed in a circle, with EF parallel to 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.

 

 Apr 11, 2020
 #1
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Parallel chords in a circle cut off equal arcs.

 

Therefore, x2 - 2x  =  56 - 3x

--->      x2 + x - 56  =  0 

         (x + 8)(x - 7)  =  0

Either  x = -8  or  x = 7

 

Since the problem states that x is positive, x = 7

--->   x2 - 2x  =  72 - 2·7  =  35

        56 - 3x  =  56 - 3·7  =  35  

 

--->  arc(EHGF)  =  35o + 70o + 35o  =  140o

--->  arc(EPF)  =  360o - 140o  =  220o

 Apr 11, 2020
 #2
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TYSM!!!!!!!!!

Guest Apr 11, 2020

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