Question #11993

Given a square ABCD with side 8.A circle is drawn through vertices A and D and tangent to the side BC.find the radius of the circle.

Expert's answer

Suppose M is the point where the circle tangents BC. BM=MC. Let's put point N on

AD, AN=ND. MN is the height of the triangle AMD. Let O is the center of the

circle, then AO=MO=DO. From the triangle AON, where ON is perpendicular AN,

AO^2=AN^2 ON^2. If r is the radius of the circle, then AO = r, AN = AD/2 = 4, ON

= 8 - r (MN=AB=8, MO = r). So, (8 - r)^2 4^2 = r^2,

64 - 16r r^2 16 =

r^2,

16r = 80,

r = 5

AD, AN=ND. MN is the height of the triangle AMD. Let O is the center of the

circle, then AO=MO=DO. From the triangle AON, where ON is perpendicular AN,

AO^2=AN^2 ON^2. If r is the radius of the circle, then AO = r, AN = AD/2 = 4, ON

= 8 - r (MN=AB=8, MO = r). So, (8 - r)^2 4^2 = r^2,

64 - 16r r^2 16 =

r^2,

16r = 80,

r = 5

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