數學一題 要步驟

Let centre of circle with AB as diameter be X.
Let Y be the point on AB such that OY is perpendicular to AB.
Let radius of semi-circle AP be b and radius of semi-circle PB be a.
So radius of semi-circle AB = (a + b).
Since area of shaded region = 39p, therefore
p(a + b)^2/2 - pa^2/2 - pb^2/2 - 9p = 39p
a^2 + b^2 + 2ab - a^2 - b^2 - 18 = 78
2ab = 78 + 18 = 96
ab = 48 or b = 48/a...................................(1)

Since area of circle with centre O = 9p ( p stands for pi), so its radius = 3.
Let centre of semi-circle PB be Z.
OZ = a + 3.
PY = radius of circle O = 3, so YZ = a - 3.
so OY^2 = OZ^2 - YZ^2 = (a + 3)^2 - ( a - 3)^ = 12a...........(2)
OX = (a + b) - 3 = (a + b -3)
PX = AP - AX = 2b - ( a + b) = b - a
XY = PX + PY = (b - a + 3)
OY^2 = OX^2 - XY^2 = ( a + b - 3)^2 - (b - a + 3)^2 = 4b(a -3)......(3)
Eqt.(2) = (3), so we get
12a = 4b(a -3)
3a = b( a - 3)......................(4)
Sub. (1) into (4),
3a = 48(a - 3)/a
a^2 = 16(a - 3)
a^2 - 16a + 48 = 0
(a - 4)(a - 12) = 0
a = 4, so b = 48/a = 12. Or
a = 12, so b = 48/12 = 4.
therefore, AB = 2(a + b) = 2 x 16 = 32.