First: notice that triangle(DCE) is similar to triangle(BCF) by AA.
Since BC = 3·DC: if DE = x, then BF = 3x
and if CE = y, then CF = 3y.
Since triangle(ABE) is a right triangle: (21 + 3x)2 = (27)2 + (30 + y)2
---> 441 + 126x + 9x2 = 729 + 900 + 60y + y2
---> 9x2 - y2 + 126x - 60y = 1188 (Equation #1)
Since triangle(ADF) is a right triangle: (27 + x)2 = (21)2 + (3y + 10)2
---> 729 + 54x + x2 = 441 + 9y2 + 60y + 100
---> x2 - 9y2 + 54x - 60y = -188 (Equation #2)
Subtracting Equation #2 from Equation #1: 8x2 + 8y2 + 72x = 1376 (Equation #3)
We need to get rid of the y2-term; let's look at triangle(CDE): x2 + y2 = 100 ---> y2 = 100 - x2
Substituting this value into Equation #3: 8x2 + 8y2 + 72x = 1376 ---> 8x2 + 8(100 - x2) + 72x = 1376
---> 8x2 + 800 - 8x2 + 72x = 1376 ---> 800 + 72x = 1376 ---> 72x = 576 ---> x = 8
Since x2 + y2 = 100 and x = 8 ---> y = 6.
You can now determine all the length of all the sides; divide the region into non-overlapping triangles and find the appropriate areas.