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8
(a)
AB=5 cm
BC=(30-2*5)/2=10 cm
EC^2+4^2=BC^2
EC^2+16=100
EC=9.17 cm (correct to 3 significant figures)
(b)
The area of ABE
=4*3/2
=6 cm^2
The area of BEC
=4*9.17/2
=18.34 cm^2
The area of ABCD
=2(6+18.34)
=48.7 cm^2
10
angle BCD=180-105=75
angle E
=180-angle BCD
=180-75
=105
angle F
=180-angle E
=180-105
=75
11
since angle P= angle QRS=118
a
=angle P/2
=118/2=59
angle S
=180-angle QRS
=180-118
=62
angle Q
=angle S
=62
a+b+118+angle Q=360
59+b+118+62=360
b=121

8.

AB=AD=SQRT(3^2+4^2)=5
AB+AD+BC+CD=30
BC+CD+5+5=30, BC+CD=20, BC=CD=10
BC^2=BE^2+EC^2, EC^2=10^2-4^2 ==> EC=SQRT(84)

EC=9.165cm

area of ABD=3x4=12cm^2
area of BCD=4x9.165=36.661cm^2

area of ABCD=12+36.661=48.661cm^2


10.
FGD=ABC=105
DEF=FGD=105
EFG=180-DEF ==> EFG=180-105=75

E=105
F=75

11.
QRS=SPQ ==> 2a=118 ==> a=118 / 2 = 59
PQR=RSP ==> PQR + RSP + 2 * 118 = 360
PQR = (360 - 236) / 2 = 124 / 2 = 62

a + b + 118 + 62 = 360
b = 360 - 59 - 118 - 62 = 121

a = 59
b=121

8a.
AD^2=AE^2+DE^2 <----畢氏定理
AD^2=3^2+4^2
AD^2=9+16
AD^2=25
AD=5
AE=AD <--從圖所知
AE=5

perimeter of ABCD=30
AD+AE+CB+CD=30
CB+CD=20
asCB=CD <--從圖所知
CB=CD=20/2=10

CB^2=EC^2+BE^2 <----畢氏定理
EC^2=100-16
EC^2=84
EC=9.16515139
EC=9.165cm <--3 significant figures

b.
area of ABD
=3*(4+4)/2
=12cm^2

area of BCD
=EC*(4+4)/2
=36.661cm^2

area of ABCD
=area of ABD+area of BCD
=12+36.661
=48.661cm^2 <--3 significant figures
--------------------------------------------------------------------
10.( *=角 )
as ABCD is parallelogam
*ADC=*ABC=105 <----對角相等

as CDE is a straight
*GDE+*ADC=180
*GDE=75

as ABCD is parallelogam
*F=*GDE=75 <----對角相等

as GF//DE
*E+*F=180 <---當一直線cut兩條平行線,兩夾角相加=180度
*E=105
------------------------------------------------------------------
11.
as PQRS is parallelogam
*SPQ=*QRS=118 <----對角相等

as PT bisects *SPQ
a=*SPQ/2
a=59

as SR//PQ
a+b=180 <---當一直線cut兩條平行線,兩夾角相加=180度
b=121

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8.

a)

AB = BD = (AE^2+BE^2)^0.5 = 5 cm [Pythagoras Th.]

AB + BC + CD + DA = 30
therefore, BC = CD = 10 cm

EC = (BC^2 - BE^2)^0.5 [Pythagoras Th.] (ANSWER)

b)

total area = area of triangle ABD + area of triangle BCD
= 0.5*BD*AE + 0.5*BD*EC (ANSWER)

10

Angle(CBA) + Angle(BAD) = Angle(CDA) + Angle(BAD) = 180

therefore Angle(CBA) = Angle(CDA) = 105

Angle(CDA) = Angle(DEF) = 105 (ANSWER)

Angle(GFE) + Angle(DEF) = 180
Angle(GFE) = 180 - Angle(DEF) (ANSWER)

11

Angle(R) + Angle(S) = 180
Angle(S) = 180 - 118 = 62

Angle(STP) = Angle(S) = 62 [Base angle of isosceles triangle]

b + Angle(STP) = 180
b = 180 - Angle(STP) (ANSWER)

a + b = 180
a = 180 - b (ANSWER)