1.
Let O be the center of the semi-circle
Draw CE perp to AB
Connect OC = r
OE = 8.5
EB = r - 8.5
CB = 3
Using the Pythagorean Theorem
Height of triangle OCE = Height of triangle ECB
CE^2 = CE^2
r^2 - (OE)^2 = (CB)^2 - ( EB)^2
r^2 - (8.5)^2 = (3)^2 - (r - 8.5)^2
r^2 - 72.25 = 9 - r^2 + 17r - 72.25
r^2 - 72.25 = -r^2 + 17r - 63.25
2r^2 - 17r -9 = 0
(2r + 1) ( r - 9) = 0
r = 9
2.
Area of Sector = pi *r^2 * (100 / 360)
250 pi = pi * r^2 * (5/18)
r^2 = 250 (18 / 5)
r^2 = 50 * 18
r^2 = 100 * 9
r^2 = 900
r = 30
3.
2* area ABC = AD * BC 2* area ABC = BE * AC
2* area ABC = 12 * BC 2* area ABC = 16 * AC
area ABC = 6 BC area ABC = 8 AC
6 BC = 8 AC
AC = (8/6) BC = (4/3) BC
Triangle inequality
AB + BC > AC
AB + BC > (4/3)BC
AB > (4/3)BC - BC
AB > (1/3)BC
Then
2 [ ABC ] = BC * AD
2[ ABC ] = BC * 12
2 [ ABC ] = (1/3)BC * 36
Since AB > (1/3)BC
2[ ABC ] < AB * 36
2[ ABC ] < AB * CF
Then CF < 36
So
Max integer value for CF = 35