Semicircular paper and creasing of a chord A semicircular piece of paper with radius $2$ $cm$ is folded along a chord so that the arc is tangent to the diameter.If the contact point of the arc divides the diameter in the ratio $3:1$,determine the length of the crease.

Sorry I can't show any work as I can't develop any approach to solve this question. Any suggestions?

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Hint: Let the  semicircle be centered at the origin with radius 2, and lie above the x axis, so that the point $P=(1,0)$ cuts the diameter in a 3:1 ratio. Then consider another circle also of radius 2 but centered at $(1,2)$ so it would touch the x axis (diameter) at P so be tangent to the diameter there. 

Figure out by algebra the two points where the new circle goes through the original semicircle, and get the length of the chord via distance formula.
Calculation has two parts. Finding the segment on outside circle and the angle  $ \alpha $ subtended between fold/diameter.

By creasing or folding T falls on t

$$ r = nK = 2 ( 2 cm) ;\, OK = k ;$$

$$(k+r) ( k-r ) = OT^2 = Ot^2 = ( k-r/2)^2 \rightarrow  k = 5 r /4 $$

$$ on = om = OK- nK= r/4 ; \,  OT = Ot= on + nt= r/2 + r/4 = 3r/4 $$

dot marked angle $= \alpha , $

$$  \tan 2 \alpha = \ tan TOK  = TK/OT = 1/(3/4) = 4/3 $$

$$ \cos 2 \alpha = 3/5 ;\, \cos \alpha = 2/\sqrt 5 ;\, $$

$$ OM = 5 r/4 \cdot 2/\sqrt 5 = \sqrt 5 r/2 ; $$

By Circle segment product property,

$$ OT^2 = OP \cdot OQ = (OM + MQ) (OM-MP) = OM^2 - x^2 $$ 

$$ (if \, MP = MQ = x) $$

$$ x^2 = OM^2 - OT^2  = 5 r^2/4 - (3 r/4)^2 ;\, \rightarrow x = \sqrt {11}/4 $$

$$ PQ = 2 x  = r \sqrt{11}/2 = \sqrt{11} cm \;  (Answer).\,$$