Question 827493
<pre>
{{{drawing(400,4400/21,-1,20,-1,10,

line(0,0,sqrt(356),0),arc(sqrt(356)/2,0,sqrt(356),-sqrt(356),0,180),
line(0,0,13.56797286,8.47998304),line(13.56797286,8.47998304,sqrt(356),0),
locate(0,0,A), locate(13.5,9.4,C),locate(18.86,0,B)


)}}}

An angle inscribed in a semicircle is always a right angle.
Therefore &#916;ABC is a right triangle, and we can take its two legs
as its base and height.

So the area of &#916;ABC = {{{expr(1/2)AC*CB}}} = {{{expr(1/2)(16)(10)}}} = 80

Since &#916;ABR is a right triangle, we can find the hypotenuse AB using
the Pythagorean theorem:

ABē=ACē+CBē
ABē=16ē+10ē
ABē=256+100
ABē=356
 AB={{{sqrt(356)}}} = {{{sqrt(4*89)}}} = {{{2sqrt(89)}}}

AB is the diameter of the semicircle.  Since the radius is
half the diameter, the radius is {{{sqrt(89)}}}.

Area of a whole circle is {{{pi*r^2}}} so 
Area of a semicircle = {{{expr(1/2)pi*r^2}}}

So the area of this semicircle is 

{{{expr(1/2)pi*(sqrt(89))^2}}} = {{{expr(1/2)pi*89}}} = 139.8008731

So the area of the portion of the semicircle which is outside of the 
inscribed right triangle &#916;ABC is 

area of semicircle - area of triangle =

139.8008731 - 80 = 59.80087308

Edwin</pre>