Question 1094682
Question 1094682
<pre>
{{{drawing(300,300,-9,9,-9,9,
green(arc(0,0,16,-16,90,390)),circle(0,0,.1), 
red(line(0,8,4sqrt(3),4),
red(line(0,8,4sqrt(3)-.1,4-.1))


,arc(0,0,16,-16,30,90),
arc(0,0,16.2,-16.2,30,90),arc(0,0,16.1,-16.1,30,90)),
locate(3.3,2,8),locate(-.6,4,8),
green(line(0,0,0,8),line(0,0,4sqrt(3),4)),
locate(3.3,6,8) )}}}

We want the area of the SEGMENT, which is the region
inside the red boundary.

The SECTOR consists of both the triangle and the SECTOR.
The triangle is equilateral because all its sides are 8.
Therefore the angle is 60° or {{{pi/3}}} radians.

The area of the whole SECTOR is given by the formula:

{{{A}}}{{{""=""}}}{{{expr(1/2)r*theta}}} and {{{theta}}}{{{""=""}}}{{{pi/3}}},

{{{A}}}{{{""=""}}}{{{expr(1/2)8*expr(pi/3)}}}{{{""=""}}}{{{4pi/3}}}

There is a formula for the area of the equilateral triangle 
in terms of its sides which all equal s.  It is

{{{A}}}{{{""=""}}}{{{expr(sqrt(3)/4)*s^2}}}

Since the sides are 8, s=8, the area of the equilateral
triangle is 

{{{A}}}{{{""=""}}}{{{expr(sqrt(3)/4)*8^2}}}{{{""=""}}}{{{16sqrt(3)}}}

{{{matrix(1,7,
(matrix(3,1,Area,of,SEGMENT)),
"",
""="",
"",
(matrix(3,1,Area,of,SECTOR)),
""-"",
(matrix(3,1,Area,of,TRIANGLE))
)}}}

{{{matrix(1,7,
(matrix(3,1,Area,of,SEGMENT)),
"",
""="",
"",
4pi/3,
""-"",
16sqrt(3)
)}}}

Edwin</pre>