Q. 23 B4.2( 6 Votes )

# In an equilateral

Answer :

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 cm^{2}

Area of circle = π × (radius)^{2}

= 3.14 × (3.46)^{2}

= 37.59 cm^{2}

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

= Area of Δ ABC – Area of circle

= (62.28 – 37.59) cm^{2}

= 24.69 cm^{2}

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