Answer:
4π
Step-by-step explanation:
As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:
A = [tex]\frac{1}{2}[/tex] x b x h -------------(i)
Where;
A = 18[tex]\sqrt{3}[/tex]
b = XZ = base of the triangle
h = YZ = height of the triangle = 6
Substitute these values into equation(i) and solve as follows:
18[tex]\sqrt{3}[/tex] = [tex]\frac{1}{2}[/tex] x b x 6
18[tex]\sqrt{3}[/tex] = 3b
Divide through by 3
6[tex]\sqrt{3}[/tex] = b
Therefore, b = XZ = 6[tex]\sqrt{3}[/tex]
Now, assume that the circle is centered at O;
Triangle XOZ is isosceles, therefore the following are true;
(i) |OZ| = |OX|
(ii) XZO = ZXO = 30°
(iii) XOZ + XZO + ZXO = 180° [sum of angles in a triangle]
=> XOZ + 30° + 30° = 180°
=> XOZ + 60° = 180°
=> XOZ = 180° - 60°
=> XOZ = 120°
Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;
[tex]\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}[/tex]
[tex]\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}[/tex]
[tex]\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}[/tex]
[tex]\frac{1}{12} = \frac{1}{2|OZ|}[/tex]
[tex]\frac{1}{6} = \frac{1}{|OZ|}[/tex]
|OZ| = 6
The radius of the circle is therefore 6.
Now, let's calculate the length of the arc XZ
The length(L) of an arc is given by;
L = θ / 360 x 2 π r ------------------(ii)
Where;
θ = angle subtended by the arc at the center.
r = radius of the circle.
In our case,
θ = ZOX = 120°
r = |OZ| = 6
Substitute these values into equation (ii) as follows;
L = 120/360 x 2π x 6
L = 4π
Therefore the length of the arc XZ is 4π
