To enjoy the geometra 2

from the figure let O be the centre of the semicircle . connect OC .length Of AC = 15√3 , BC = 15 . Angle COB = 75 Degrees .

let the coordinates of A be ( 0,0) and B = (30,0) then C = ( 22.5,7.5√3). and let the coordinates of O = (a,0).

Then the slope of CO = tan75 = 2 + √3 = $\frac{ 7.5√3 }{22.5 - a }$ , then a = $\frac{ 45(3 - √3) }{3}$ = 15(3 - √3).

The equations of the lines AC and HO are :

Y = $\frac{x }{ √3 }$ , y = -√3(x-a) respectively , Solving simultaneously the coordinates of H = (x,y) in terms of a can be ;

The coordinats of H = ( $\frac{3a }{ 4 }$ , $\frac{ √3a }{ 4 }$ )

and O = (a,0).

Then OH = R = √ ( $\frac{a^2}{16}$ + $\frac{3a^2}{16}$ ) = $\frac{a }{2}$ = $\frac{ 15(3 - √3) }{2}$

Then :

A = (15²√3)/2 – π/2 ( 15/2)² (3 - √3)² , 4A = (15)²(2√3 - 6π + 3√3).

ANSWER = 15² = 225.