Answer:
The area of the region inside the circumcircle of the triangle but outside the triangle is
[tex]A=\frac{27}{4}[\pi-3\sqrt{3}]\ units^2[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the area of triangle
we have an equilateral triangle
Applying the law of sines
[tex]A_t=\frac{1}{2}(b^2)sin(60^o)[/tex]
where b is the length side of the equilateral triangle
we have
[tex]b=9\ units[/tex]
[tex]A_t=\frac{1}{2}(81)sin(60^o)[/tex]
[tex]A_t=\frac{1}{2}(81)\frac{\sqrt{3}}{2}[/tex]
[tex]A_t=81\frac{\sqrt{3}}{4}\ units^2[/tex]
step 2
Find the area of circle
The area of the circle is equal to
[tex]A_c=\pi r^{2}[/tex]
The formula to calculate the radius of the circumcircle of the triangle equilateral is equal to
[tex]r=b\frac{\sqrt{3}}{6}[/tex]
where b is the length side of the equilateral triangle
we have
[tex]b=9\ units[/tex]
substitute
[tex]r=(9)\frac{\sqrt{3}}{6}[/tex]
[tex]r=3\frac{\sqrt{3}}{2}\ units[/tex]
Find the area
[tex]A_c=\pi (3\frac{\sqrt{3}}{2})^{2}[/tex]
[tex]A_c=\frac{27}{4} \pi\ units^2[/tex]
step 3
Find the area of the shaded region
we know that
The area of the region inside the circumcircle of the triangle but outside the triangle is equal to the area pf the circle minus the area of triangle
so
[tex]A=(\frac{27}{4} \pi-81\frac{\sqrt{3}}{4})\ units^2[/tex]
Simplify
[tex]A=\frac{27}{4}[\pi-3\sqrt{3}]\ units^2[/tex]
