F.2數 10分

4.
PRS = P + Q (D外角)
54o = 27o + h
h = 27o


P = Q = 27o
所以 DRPQ 是等腰 D。
PR = QR (等腰 D 的兩腰)
PR = 4.8 cm
k = 4.8


5.
ABD + DBE + EBC = 180o (直線同側鄰角)
58o + m + 79o = 180o
m = 43o

DBE = DEB = 43o
所以 DDBE 是等腰 D。
DB = DE (等腰 D 的兩腰)
n = 3n - 6
2n = 6
n = 3


7.
AB = AC = 4cm (己知)
AD = AD (公共邊)
BD = CD (己知)
DABD DACD (SSS)
ADB = ADC = a (全等三邊形對應角)
BAD = CAD = b (全等三邊形對應角)
CD = BD = 2 cm (全等三角形對應邊)


ADB + ADC = 180o (直線同側鄰角)
a + a = 180o
a = 90o

BC = (2 + 2) cm = 4 cm
所以，AB = BC = CA = 4 cm
DABC 是等邊D。
BAC = 60o
b + b = 60o
b = 30o


8.
右上角的三角形是等腰三角形，兩底角均等於 x。
100o + x + x = 180o (D內角和)
x = 40o

下方的三角形為等邊三角形。
每一個內角 = 60o
y = 60o + x
y = 100o
=

4.
h + 27 degrees = 54 degrees (ext. ∠ of triangle)
h = 54 degrees - 27 degrees
h = 27 degrees

∠Q = ∠P (from above)
so, QR = PR (opp. sides, = ∠s)
k cm = 4.8 cm
k = 4.8

5.
∠ABD + ∠DBE + ∠CBE = 180 degrees (adj. ∠s on st. line)
58 degrees + m + 79 degrees = 180 degrees
m = 180 degrees - 58 degrees - 79 degrees
m = 43 degrees

∠DBE = ∠BED (from above)
so, DB = DE (opp. sides, = ∠s)
n = 3n-6
6 = 3n-n
6 = 2n
n = 3

7.
because AB = AC and BD = DC
so triangle ABC is a isos. triangle
a = 90 degrees (prop. of isos. triangle)

BD = DC
so,
BC
= (2+2)cm
= 4cm
= AB
= CA
so triangle ABC is also an equil. triangle
2b = 60 degrees (prop. of equil. triangle)
b = 30 degrees

8.
Let B & C be the pts on the left corner and the right one such that triangle ABC is an equil. triangle & triangle ACD is an isos. triangle.

x = ∠ACD (base. ∠s, isos. triangle)
2x + 100 degrees = 180 degrees (∠ sum of triangle)
2x = 80 degrees
x = 40 degrees

y = ∠ACD + ∠ACB
& ∠ACD = x = 40 degrees (from above)
∠ACB = 60 degrees (prop. of equil. triangle)
so,
y
= 40 degrees + 60 degrees
= 100 degrees