Question 1202027
In the figure below, AXC and BXD are arcs with centers B and A respectively.
ABCD is a square of side 14 cm.  Calculate the area of the region CXD.


{{{drawing(200,180,-5,19,-6,19,
locate(7,16,14),
line(0,0,14,0), line(14,14,14,0),line(0,0,0,14),line(0,14,14,14),
locate(0,0,A), locate(14.1,8.7,14), locate(7,0,14), locate(-2,8.7,14),
locate(0,16,D), locate(14,16,C),locate(14,0,B),locate(6.5,11.6,X),


arc(0,0,28,-28,0,90), arc(14,0,28,-28,90,180) )}}} 
<pre>
Draw AX and BX.

{{{drawing(200,180,-5,19,-6,19, locate(7,16,14),

line(0,0,14,0), line(14,14,14,0),line(0,0,0,14),line(0,14,14,14),
locate(0,0,A), locate(14.1,8.7,14), locate(7,0,14), locate(-2,8.7,14),
locate(0,16,D), locate(14,16,C),locate(14,0,B),locate(6.5,11.6,X),

green(line(0,0,7,7sqrt(3)), line(14,0,7,7sqrt(3))),
arc(0,0,28,-28,0,90), arc(14,0,28,-28,90,180) )}}} 

Triangle ABX is an equilateral triangle because AX = AD = AB and
BX = BD = AB, so AX = BX = AB.  So angle BAX = 60<sup>o</sup>.

Area of triangle ABX is {{{expr(1/2)AX*AB*sin(60^o)}}}{{{""=""}}}{{{expr(1/2)14*14*expr(srt(3)/2)}}}{{{""=""}}}{{{49sqrt(3)}}}

Angle DAX = 90<sup>o</sup> - angle XAB =  90<sup>o</sup>-30<sup>o</sup> =  60<sup>o</sup>

Area of sector ADX is {{{expr(30^o/360^o)*pi*14^2}}}{{{49pi/3}}}
Area of sector BCX is also {{{49pi/3}}}

Area of square ABCD is {{{14^2}}}{{{""=""}}}{{{196}}}.

Area of region CXD = Area of square - Area of sector ADX - Area of sector BDX - Area of triangle ABX

Area of region CXD = {{{196-49pi/3-49pi/3-49sqrt(3)}}}{{{""=""}}}{{{196-98pi/3-49sqrt(3)}}}

That's about 8.594159412 cm<sup>2</sup>.

Edwin</pre>