Q3 of 59 Page 83

In figure 3.83, M is the centre of the circle and seg KL is a tangent segment.

If MK = 12, KL = then find -


(1) Radius of the circle.


(2) Measures of K and M.



(1) Here LM is the radius of the circle


KML = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.


In triangle MLK right-angled at L,


Given MK = 12, KL = 6√3,


MK2 = LM2 + KL2 {Using Pythagoras theorem}


LM2 = 122 -6√32


LM2 = 144 - 108


LM = √36


LM = 6 cm


(2)



⇒∠K = 30°


M + K + L = 180° {Angle sum property of the triangle}


So, M = 180° - K - L


M = 180° - 30° - 90° = 60°


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