Question 1209969
<pre>

We want to know the area of the two segments marked "XXX" below

{{{drawing(400,400,-10,10,-10,10,circle(0,0,.2), 

circle(0,0,9), line(9sqrt(3)/2,-4.5,0,-9), line(9sqrt(3)/2,-4.5,0,9),
line(0,-9,0,9), locate(.2,6.8,30^o), locate(4,3,9sqrt(3)),
locate(.2,-7,60^o), locate(6,-3.5,90^o),

locate(4.5,4.5,XXX), locate(3.3,-7,XXX)

  )}}}

The triangle is inscribed in a semicircle, i.e., its hypotenuse is a diameter,
so we know it is a right triangle.

Since it has a 30<sup>o</sup> angle, we know it is a 30-60-90 right triangle.

We know that the longer leg is {{{9sqrt(3)}}}

You should learn the following information about the special 30-60-90
special right triangle.  It occurs very often.

The sides of the standard 30-60-90 right triangle:
The shorter leg is 1 unit long. 
The longer leg is {{{sqrt(3)}}} units long.
The hypotenuse is 2 units long.

Let the shorter leg of your 30-60-90 right triangle be x.

You are given the longer leg of the 30-60-90 triangle as {{{9sqrt(3)}}}

So make the proportion:

{{{matrix(1,6,shorter,leg, of,my,30-60-90,triangle)/
matrix(1,6,longer,leg, of,my,30-60-90,triangle)}}}{{{""=""}}}{{{matrix(1,7,shorter,leg, of,the,standard,30-60-90,triangle)/
matrix(1,7,longer,leg, of,the,standard,30-60-90,triangle)}}}

{{{x/9sqrt(3)}}}{{{""=""}}}{{{1/sqrt(3)}}}

cross-multiply:

{{{x*sqrt(3)}}}{{{""=""}}}{{{9sqrt(3)}}}

{{{x}}}{{{""=""}}}{{{9sqrt(3)/sqrt(3)}}}

{{{x}}}{{{""=""}}}{{{9}}}

So the shorter leg of the 30-60-90 special right triangle is 9 cm.

Therefore the area of the 30-60-90 special right triangle is

{{{A = expr(1/2)base*height=expr(1/2)*(matrix(1,2,shorter,leg))*(matrix(1,2,longer,leg))}}}{{{""=""}}}
{{{expr(1/2)(9)(9sqrt(3))}}}{{{""=""}}}{{{expr(81/2)*sqrt(3)}}} cm<sup>2</sup>

To get the area of the two segments marked XXX, we must find the radius
of the circle.  Since the hypotenuse is twice the shorter side, the
hypotenuse is (9)(2) = 18 cm. The hypotenuse is a diameter, so the radius is
half that or 9.

The area of the whole circle = {{{pi*radius^2=pi*9^2=81pi}}}
The area of the semicircle the triangle is inscribed in is half that, or
{{{expr(81/2)pi}}}

So we subtract the area of the triangle from the area of the semicircle
and get

{{{expr(81/2)pi}}}{{{""-""}}}{{{expr(81/2)*sqrt(3)}}} cm<sup>2</sup>

Factor out {{{81/2}}},

{{{expr(81/2)(pi-sqrt(3))}}} cm<sup>2</sup>, about 57.1 cm<sup>2</sup>
 
Edwin</pre>