Q5 of 12 Page 164

A circle with centre O touches all the four sides of a quadrilateral ABCD. If the point of contact divides AB in the ratio 3 : 1 and AB = 8 cm, then find the radius of the circle if it is given OA = 10 cm.

Let the diagram be as follows:


P divides the line AB in the ratio 3:1



Adding 1 both side, we get






⇒ BP = 2 cm


And hence,


AP = 3BP = 3(2) = 6 cm


Let the centre be O.


From this figure it can be interpreted that OP will be the radius of the circle.


Radius OP is perpendicular to the tangent AB as radius is perpendicular to the tangent.


If OP is ⊥ to AB which implies that OP is ⊥ AP and OP is ⊥ PB.


Hence Δ OPA is right angled triangle


∠ OPA = 90°


OA2 = AP2 + OP2


OP2 = OA2 – AP2


OP2 = 102 – 62


OP2 = 100 – 36


OP2 = 64


∴ OP = 8 cm


Therefore the radius of circle = 8 cm.


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