Question

Express answer in exact form.

An equilateral triangle is inscribed in a circle with a radius of 6". Find the area of the segment cut off by one side of the triangle.


HELP ME!!!

Answer 1

can someone help me

Answer 2

\[12 \pi - 9 \sqrt {3}\]

Answer 3

Give me a minute to get the explanation together.

Answer 4

The area of the arc of the circle is 1/3 of the circle which is:
\[(1/3) \pi (6)^{2} = 12 \pi\]
The area of the triangle is a bit trickier. If you draw the triangle you can divide it into 3 smaller triangles using the radius. These are 30-120-30 triangles. We're going to break them in two to get the area exactly. So we have 2 triangles or 2*(1/2)Base*height.

The hypotenuse is 6 in each case. We need to calculate the opposite:
\[sin(30) = x/6 = 1/2\]
\[6(1/2) = x = 3\]
Now to calculate the adjacent:
\[cos(30) = x/6 = \sqrt {3} /2\]
\[6 \sqrt {3} / 2 = x = 3 \sqrt {3}\]
The area of the 2 smaller triangles is:
\[3*3 \sqrt {3} = 9 \sqrt {3} \]
So we now subtract this from the arc area:
\[12 \pi - 9 \sqrt {3} \]

Answer 5

Keep in mind that
\[sin(30) = 1/2\]
is a unit circle value.
\[cos(30) = \sqrt {3}/2\]
is also a unit circle value.
You have a triangle with a hypotenuse of 2, adjacent of sqrt 3 and an opposite of 1.