Question

In figure, M is the centre of the circle and seg KL is a tangent segment. If MK = 12, KL = `6sqrt(3)`, then find

(i) Radius of the circle.
(ii) Measures of ∠K and ∠M.

Answer 1

(i)

Line KL is the tangent to the circle at point L and seg ML is the radius.                   ......[Given]

∴ ∠MLK = 90°      ......(i) [Tangent theorem]

In ∆MLK,

∠MLK = 90°

∴ MK² = ML² + KL²    .....[Pythagoras theorem]

∴ 12² = ML² + `(6sqrt(3))^2`

∴ 144 = ML² + 108

∴ ML² = 144 − 108

∴ ML² = 36

∴ ML = `sqrt(36)`

∴ ML = 6 units     ......[Taking square root of both sides]

∴ Radius of the circle is 6 units.

(ii)

We know that,

ML = `1/2` MK,

∴ ∠K = 30°     .....(ii) [Converse of 30°−60°−90° theorem]

In ∆MLK,

∠L = 90°     .....[From (i)]

∠K = 30°    .....[From (ii)]

∴ ∠M = 60°    ......[Remaining angle of ∆MLK]