Question 1151182
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Circles M & N (centered at points M and N respectively) are tangent to each other and to {{{cross(AC)}}} AB & BC. If the radius of the circle with center N is 5/(1+sqrt2) m, what is the radius of M, in meters?<br>
Let r be the radius of circle N and R be the radius of circle M.<br>
Let D and E be the points of tangency of circles N and M, respectively, with BC; let F be the point of tangency of the two circles.<br>
Triangles BDN and BEM are isosceles right triangles.<br>
{{{MB = MF+FN+NB = (R)+(r)+(r*sqrt(2)) = R*sqrt(2)}}}<br>
Then<br>
{{{(R)+(r)+(r*sqrt(2)) = R*sqrt(2)}}}
{{{R+r(1+sqrt(2)) = R*sqrt(2)}}}<br>
Given that {{{r=5/(1+sqrt(2))}}},<br>
{{{R+(5/(1+sqrt(2)))*(1+sqrt(2)) = R*sqrt(2)}}}<br>
{{{R+5 = R*sqrt(2)}}}<br>
{{{5 = R*sqrt(2)-R = R(sqrt(2)-1)}}}<br>
{{{R = 5/(sqrt(2)-1) = (5*(sqrt(2)+1))/((sqrt(2)-1)(sqrt(2)+1)) = 5*(sqrt(2)+1)}}}<br>
ANSWER: The radius of circle M in meters is {{{5*(sqrt(2)+1)}}}<br>