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In the figure, M is the centre of the circle and seg KL is a tangent segment. If `MK = 12 , KL = 6 sqrt(3)` then find, (1) Radius of the circle, (2) Measures of `/_ K ` and `/_ M `. |
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Answer» In `Delta MLK`, `/_ MLK = 90^(@)` ….(Tangent theorem) `:.` by Pythagoras theorem, `MK^(2) = ML^(2) + LK^(2)` `:. 12^(2) = ML^(2) + ( 6 sqrt(3))^(2) ` `:. 144 = ML^(2) + 36 xx 3 ` `:. 144 = ML^(2) + 108` `:. ML^(2) = 144 - 108` `:. ML^(2) = 36` `:. ML= 6` ...(Takin quare roots of both the sides ) `:. ` radius of the circle = ML = 6 . In `Delta MLK`, `ML = (1)/(2) MK ` `:. /_ K = 30^(@)` ...(By converse of `30^(@) - 60^(@) - 90^(@)` triangle theorem ) In `Delta MLK`, `/_ M + /_ K + /_ L = 180^(@)` ...(Sum of all angles of a triangle is `180^(@)0` `:. /_ M + 30^(@) + 90^(@) = 180^(@0` `:. /_ M + 120^(@) = 180^(@)` `:. /_ M = 180^(@) - 120^(@)` `:. /_ M = 60^(@)` |
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