f2 maths on area 4 qus

Q13.
Let area of the unshaded region = m.
Let radius of the quadrant = r.
area of quadrant = pr^2/4 = p + q + 2m..............(1)
q + m = area of semi-circle with diameter r = p(r/2)^2/2 = pr^2/8.............(2)
From (2)
2q + 2m = pr^2/4 = eqt.(1)
so 2q + 2m = p + q + 2m
so p = q.
Q14.
Let centre of C4 be O4. Radius = r.
Length of O1O2 = 24/4 = 6.
Length of O1O4 = 12 - r.
Length of O2O4 = 6 + r.
By Pythagoras theorem,
(12 - r)^2 + 6^2 = (6 + r)^2
144 - 24r + r^2 + 36 = 36 + 12r + r^2
144 = 36r
r = 144/36 = 4.
So area of C4 = 16pi.
Q15.
Let M be the mid-point of PQ. So PM = 4/2 = 2. BAMC is a straight line.
Let radius of the circle with centre A = r, so AP = r.
Let radius of circle with centre C = s.
So AM = r - 2s.
By Pythagoras theorem,
AP^2 = PM^2 + AM^2
r^2 = 2^2 + ( r - 2s)^2
r^2 = 4 + r^2 + 4s^2 - 4rs
rs - s^2 = 4...................(1)
Diameter of circle with centre B = 2r - 2s , so radius = r - s.
So area of shaded region
= area of A circle - area of B circle - area of C circle
= pr^2 - p(r-s)^2 - ps^2
= pr^2 - pr^2 - ps^2 + 2prs - ps^2
= 2prs - 2ps^2
= 2p(rs - s^2). From the result (1), This is = 2p(4) = 8pi = area of shaded region.







2009-05-25 08:18:47 補充：
For the remaining question, area of shaded region = area of the 2 smaller semicircles + area of the right angled triangle - area of the big semicircle.