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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.

 

 Jun 13, 2021
 #1
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Since EF  ll  GH.......then  EH  =  GF

 

Then 

 

x^2 - 2x  =  48  - 4x

 

x^2 + 2x  - 48  =  0

 

(x + 8)  ( x - 6)  =  0

 

The second  factor  set to  0 ad solved  for  x gives  x =   6

 

So

EF  = (6)^2  - 2(6)  =   24°  =   GH

 

Then  arc  EPF  =  360  - 2*24  -  70   =  242° 

 

 

cool cool cool

 Jun 13, 2021

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