Answer :

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fig.19


Let the radius of the circle be r cm.


In the fig.19,


AR and AQ are making a pair of tangents drawn from vertex A of Δ ABC on the circle.


AR = AQ = x [LET]


BR and BP are making a pair of tangents drawn from vertex B of Δ ABC on the circle.


BR = BP = y [LET]


CP and CQ are making a pair of tangents drawn from vertex C of Δ ABC on the circle.


CP = CQ = z [LET]


Given –


Δ ABC is an equilateral triangle.


AB = BC = AC = 12 cm


AR + BR = BP + CP = AQ + CQ = 12


x + y = y + z = x + z = 12…..(1)


Now,


(x + y + y + z + x + z) = (12 + 12 + 12)


2 × (x + y + z) = 36


x + y + z = 18…..(2)


Subtracting equation(1) from equation(2), we get –


x = y = z = 6 cm


Also, the line joining the centre the circle to the vertices of Δ which circumscribes the circle bisects the angles of a Δ.


OBP = 30°


In Δ BOP,


tan OBP = OP/BP = r/6


tan 30° = r/6


1/√3 = r/6


r = 6/√3 = 2√3 = 3.46 cm


Area of Δ ABC = (√3/4) × (side)2


=(1.73/4) × (12)2


= 1.73 × 36


= 62.28 cm2


Area of circle = π × (radius)2


= 3.14 × (3.46)2


= 37.59 cm2


Thus, Area of the triangle which is not included in the circle


= Area of Δ ABC – Area of circle


= (62.28 – 37.59) cm2


= 24.69 cm2


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