This one gets ugly, ACG....but....it is "do-able"
If we position the bottom left vertex of the square at (0,0) the circle has a center of (0,0) and a radius of 4.....so.....the equation is x^2 + y^2 = 16
Note....since AN = 3...the intersection of NM and the circle occurs at x = 3
So.... 3^2 + y^2 = 16
9+ y^2 = 16
y = √[16 - 9 ] = √7
Now...if you can picture this.....let this intersection be the point, Q
Draw AQ and QD
Now....we want to find the area of the sector AQD
Note that the sine of the central angle QAD is NQ / AQ = √7/4
So...the measure [in degrees] of this angle can be expressed as arcsin (√7/4)
So....the area of the sector is (1/2)r^2 *arcsin (√7/4) (/360° }=
(1/2) 4^2* [arcsin (√7/4) / 360°] = 8 [arcsin (√7/4) / 360° ] (1)
Don't worry....this looks ugly, but it will eventually disappear !!!!
And the area of the triangle AQD is given by
(1/2) r^2 sin QAD =
(1/2) 4^2 (√7/4) =
2√7 (2)
So....the area of the sector - the area of the triangle = the area between the chord QD and the edge of the circle = (1) - (2)
And this is part of the white area NQD
And the other part of this white area is the triangle with base ND and height NQ =
(1/2)ND * NQ = (1/2)(1)*√7 = √7/2 (3)
So..... the "white" area NQD is given by [ (1) - (2) ] + (3) =
8 [arcsin (√7/4) / 360° ] - 2√7 + √7/2 =
8 [arcsin (√7/4) / 360° ] - 3√7/2 (4)
So.....Area "y" = Area of rectangle MCDN - (4) =
MC * CD - (4) =
1 * 4 - (4) =
4 - [8 arcsin (√7/4) / 360° ] - 3√7/2 ] =
[ 4 + 3√7/2 - 8 [arcsin (√7/4) / 360° ] = (5)
Now ...area "x" is just the area of the quater circle - the white area, (4)
The area of the quarater circle is just (1/4)pi (4)^2) = 4pi
So "x" =
4pi - [8 [arcsin (√7/4) / 360° ] - 3√7/2] =
4pi + 3√7/2 - [8 arcsin (√7/4) / 360°] (6)
So..... "x" - "y" = (6) - (5) =
(4pi + 3√7/2 - [8 arcsin (√7/4) / 360°] ) - ( [ 4 + 3√7/2 - 8 [arcsin (√7/4) / 360° ] ) =
( 4pi - 4) units^2
See???...I told you it was ugly.... LOL!!!!
