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# CHALLENGE FOR GEOMETRY LoVeRs (or math lovers)

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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 56 - 3x degrees, where x > 0 find arc EPF, in degrees.

(btw i will post the next qustions after I see the answer. May the best player win!!)

Apr 14, 2020

#1
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By equal arcs, angle EPF is 212 degrees.

Apr 14, 2020
#2
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arc EH = arc FG

$$x^2 - 2x = 56 - 3x\\ x^2 + x - 56 = 0\\ (x + 8)(x - 7) = 0\\ x = -8\text{(rej.) or }x = 7\\$$

Both arcs are 35 degrees.

arc EPF = $$360^\circ - 70^\circ - 2\cdot 35^\circ = \boxed{220^\circ}$$

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Apr 14, 2020
#3
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$$\frac{\angle P}{2} = \frac{\angle Q}{3} = \frac{\angle R}{4}.$$

Find the largest angle of quadrilateral PQRS in degrees.

GO GO Go GO GO

godmathguy  Apr 14, 2020
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Angle S? 162.85 degrees

Guest Apr 14, 2020