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earnstyle [38]
3 years ago
9

each side of triangle xyz has length 9 .Find the area of the region inside the circumcircle of the triangle but outside the tria

ngle. PLEASE HELP QUICK!

Mathematics
1 answer:
mote1985 [20]3 years ago
6 0

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

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