I'll try this one......!!!
We need to find the radius of the large circle.BC = BD = BE
Now....DAC is a right triangle with DC = √[30^2 + 10^2] = 10√10 cm
And by the Law of Sines .... sin90 / 10√10 = sin DCA /30 .....sin-1 (3/√10) = DCA = DCB = about 71.565°......and in triangle DBC, DB = CB....so angle DBC = 180 - (2)71.565 = about 36.87°
And by the Law of Sines, again ........ 10√10 / sin 36.87 = DB / sin 71.565
So DB = (10√10) sin71.565 / sin 36.87 = 50 cm
So the area of the big circle is 2500pi cm^2
And 1/2 of the area of the small circle = (1/2)(30)^2 pi = 450pi cm^2
And to find the area bound by DCEA, we have area of sector BDE - area of triangle BDE = (1/2)(50)^2 (1.287002217586568 ) - (1/2)(60)(40) = 1608.75277198321 -1200 = 408.75277198321 cm^2
So....the shaded area = area of the big circle - (1/2) area of the small circle - area of DCEA =
[ 2500pi - 450pi - 408.75277198321 ) = about 6031.51 cm^2
Don't know if it's correct or not.......
