In a single semicircle, a red segment is drawn perpendicular to the diameter. The radius of a circle inscribed in one of the parts of the semicircle is twice the radius of the second circle. What is the length of the red segment?

The length of the red segment is `8/9R.`

Expert Answers

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Refer to the image below. We see `a^2 + b^2 = R^2 , ` `( -x - r + a )^2 + ( b - r )^2 = r^2 , ` `b / a = r / ( r + x ) ` or `b / R = r / sqrt ( r^2 + ( x + r )^2 ) `

From these equations we obtain `b = ( rR ) / sqrt ( r^2 + ( x + r )^2 ) , ` `a = ( (r+x) R ) / sqrt ( r^2 + ( x + r )^2 ) ` and

`(x+r)^2 ( R / sqrt ( r^2 + ( x + r )^2 ) - 1 )^2 + r^2 ( R / sqrt ( r^2 + ( x + r )^2 ) - 1 )^2 = r^2 `

This after simplification gives `( R - sqrt ( r^2 + ( x + r )^2 ) )^2 = r^2 , ` so `R - sqrt ( r^2 + ( x + r )^2 ) = r ` and `( x + r )^2 = R ( R - 2r ) .`

At the right, we similarly obtain

`(q-x)^2 ( R / sqrt ( q^2 + ( q - x )^2 ) - 1 )^2 + q^2 ( R / sqrt ( q^2 + ( q-x)^2 ) - 1 )^2 = q^2 `

and `R - sqrt ( q^2 + ( q - x )^2 ) = q and (q-x)^2 = R(R-2q).`

Now recall that `q=2r, ` so opening the parentheses and subtracting we obtain `r = 2x -2/3R. ` Substitute it back and obtain `81x^2=17R^2, ` or `x = sqrt(17)/9 R.`

Finally, the quantity we need is `sqrt (R^2-x^2)=8/9R.`

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