can someone share the solution pls
 

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Dear friend,
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given: area of quad OBAC = 60 sm^2 
AB = 12 cm 
to find: circle's radius and
distance between a and circles center
construction: join AO
sol: in triangle ABO and triangle ACO
(R)<ABO=<ACO    ( radius is perpendicular on tangent )
(H)AO=AO             ( common)
(S)OB=OC             ( radii of same circle )
therefore by RHS congruency triangle ABO is congruent to triangle ACO 
this implies,
area of triangle ABO is equal to the area of triangle ACO 
it is given that area of the quad obac is 60 cm^2
since both triangles have equal areas we can say that the area of each triangle is 30 cm^2
in triangle ABO 
area=1/2 x base x height
30 cm^2= 1/2 x OB x AC
1/2 xOB x 12 = 30
OB = 30 x 2/12
OB = 5 cm
hence radii of the circle is 5 cm
 in triangle ABO 
AB^2 + OB^2 = ao^2                 ( Pythagoras theorem is applied since it is a right angled triangle( radius OB perpendicular on tangent AB)
12^2 + 5^2 =AO^2
144+25=AO^2
AO^2 = 169
AO = 13
hence the distance between point A and the center of the circle O is 13 cm
 
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?Radii of the circle is 5 cm.
?The distance between point A and the center of the circle O is 13 cm.
?
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