中四數學問題

7. The answer is B.
angle DBC = 25, angle CDB = 90 (angle in semi-circle)
Hence angle DCB = 180-90-25 = 65 (angle sum of triangle)
angle BAD = 180 - 65 = 115 (opp angle, cyclic quadrilaterial).
12. The answer is A.
Let the radius be r cm.
OP =OA = r
OM = 40 - r
MP = 20.
Note that triangle OPM is a right-angled triangle.
Hence (40-r)^2 + 20^2 = r^2
1600-80r + r^2 + 400 = r^2
80r = 2000
r = 25
15. The answer is B.
Let angle A = 5k, angle B = 8k and angle C = 7k, where k is non-zero number.
angle A + angle C = 180 (opp. angles, cyclic quadrilateral)
12k = 180
k = 15
angle B = 8k = 8(15) = 120
angle D = 180-120 = 60
Hence y = 2(60) = 120 (angle at centre = 2*angle at circumstance)

7) angle BDC = 90 , angle DCB = 180-90-25=65 , angle BAD = 180 - angle DCB = 115
ans:B
12) AM^2 + 20^2 = (40- AM)^2 AM=15 r=40-AM =25
ans:A
15) angle d = 180 - angle b, angle a + angle b + angle c + angle d =360 ,
so by calculation angle d = 60
angle y = 2 time of angle d, so angle y = 120
ans:B

7.
角bdc=90 (半圓上的圓周角)
角bcd=90-25=65
角bad=180-65=115 (圓內接四邊形對角)
12.
20^2 +(40-r)^2 =r^2 <==20為pm 40-r為om pq為r
400+1600-80r+r^2=r^2
80r=2000
r=25
15.
set
角a=5k
角b=8k
角c=7k
5k+7k=180 (圓內接四邊形對角)
k=15
b=8*15=120
角d+120=180 (圓內接四邊形對角)
角d=60
2角d=y (圓心角=圓周角*2)
y=120