Question #00f2d

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Question #00f2d

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Answer 1 · 2017-04-08T04:46:20

$r = \sqrt{\frac{7}{3}}$ $r=sqrt(7/3)$

Explanation:

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I assume that $A B = B C = C D = 1, and D E = E F = F A = 2$ $AB=BC=CD=1, and DE=EF=FA=2$

As shown in the diagram,
$3 (x + y) = 360^{\circ} \Rightarrow x + y = 120^{\circ}$ $3(x+y)=360^@ => x+y=120^@$
$w = \frac{180 - x}{2}, z = \frac{180 - y}{2}$ $w=(180-x)/2, z=(180-y)/2$
$\Rightarrow w + z = \frac{360 - (x + y)}{2} = \frac{360 - 120}{2} = 120^{\circ}$ $=> w+z=(360-(x+y))/2=(360-120)/2=120^@$

In $DeltaCDE$ ,
$CE^2=CD^2+DE^2-2(CD)(DE)cos(w+z)$
$CE^2=1^2+2^2-2*1*2*cos120=7$

In $DeltaOCE$ ,
$CE^2=OC^2+OE^2-2(OC)(OE)cos(x+y)$
$7=r^2+r^2-2r^2cos120$
$=> 7=2r^2(1-cos120)=2r^2(1-(-1/2))=3r^2$
$=> r^2=7/3$
$=> r=sqrt(7/3)$

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