Question

each side of triangle xyz has length 9 .Find the area of the region inside the circumcircle of the triangle but outside the triangle. PLEASE HELP QUICK!

Answer 1

Answer:

The area of the region inside the circumcircle of the triangle but outside the triangle is

$A=\frac{27}{4}[\pi-3\sqrt{3}]\ units^2$

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

$A_t=\frac{1}{2}(b^2)sin(60^o)$

where b is the length side of the equilateral triangle

we have

$b=9\ units$

$A_t=\frac{1}{2}(81)sin(60^o)$

$A_t=\frac{1}{2}(81)\frac{\sqrt{3}}{2}$

$A_t=81\frac{\sqrt{3}}{4}\ units^2$

step 2

Find the area of circle

The area of the circle is equal to

$A_c=\pi r^{2}$

The formula to calculate the radius of the circumcircle of the triangle equilateral is equal to

$r=b\frac{\sqrt{3}}{6}$

where b is the length side of the equilateral triangle

we have

$b=9\ units$

substitute

$r=(9)\frac{\sqrt{3}}{6}$

$r=3\frac{\sqrt{3}}{2}\ units$

Find the area

$A_c=\pi (3\frac{\sqrt{3}}{2})^{2}$

$A_c=\frac{27}{4} \pi\ units^2$

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

$A=(\frac{27}{4} \pi-81\frac{\sqrt{3}}{4})\ units^2$

Simplify

$A=\frac{27}{4}[\pi-3\sqrt{3}]\ units^2$