Download Mathematics department 2015 -2016 1st Prep Time: 2 hours [1] a

Year Exam
Geometry
[3] a) In the opposite figure:
If AB
‫محافظة القاهرة‬
‫إدارة القاهرة الجديدة‬
[1] a) Choose the correct answer:
1] The obtuse angle supplements ………… angle
i) acute
ii) obtuse
iii) right
FA = FB and m (B) = 125°,
then
Find: m (A)
(Show the reason)
Solution:
In ∆ ADF, ∆ BCF:
iv) reflex
2] In the opposite figure:
If AB CD = {C}, then x = …….º
i) 120
ii) 90
iii) 60
iv) 20
D
B
6x
C
3x
A
3] If X complements Y and X ≡ Y, then mX = ……..
i) 90°
ii) 45°
iii) 70°
iv) 180°
A

120º
C
B

D
[2] a) Complete each of the following:
1] The sum of the measure of the accumulative angles at a point =
……360….º
2] Two triangles are congruent if two angles and … and the side
drawn between their vertices …. are congruent to the
corresponding parts of the other triangle
3] If mA = 150º, then m(reflex A) = …210º……
4] If a straight line intersects two parallel straight lines, then every
two interior angles in the same side of the transversal are
…supplementary..
5] In the opposite figure:
L // M
x = …70º…..
M
110°
L
x
A
/
\\
F
/
125°B
D
V.O.A.
∴ ∆ ADF≡ ∆ BCF
b) In the opposite figure:
If AB = AD, CB = CD m BAC = 30°
A
m ACB = 110°.
(1) Show that ∆ ABC ≡ ∆ ADC
30°
(2) Find m D
Solution:
In ∆ ABC: mB = 180º - (30º + 110º) = 40º
110°
In ∆ABC & ∆ADC:
\ C /
1] BC = DC
given
D

given
2] AB = AD

3] AC is a common side
∆ABC ≡ ∆ADC
∴ mD = B = 40º
//
If AB//CD , then mC = …….º
i) 30
ii) 180
iii) 60
iv) 100
1] FD = FC given

2] FA = FB given
3] mAFD = mBFC

\\
∴ mA = mB = 125º
4] If ∆ ABC ≡ ∆ XYZ and mA + mB = 130º, then mZ = ……º
i) 40
ii) 50
iii) 60
iv) 80
5] In the opposite figure:
C
CD = {F}, FC = FD,
//
Mathematics department
2015 -2016
1st Prep
Time: 2 hours
Mid
B
[4] a) In the opposite figure:
CB // AD // EH , mE = 115°,
mBAD = 45°
Find: mB & mC and mCAB
(Show the reason)
E
115°
A
45°
>
[5] a) In the opposite figure:
If AB CE = {M}, MD  CE and MB
H
>
bisects DME find:
m(BME) , m(DME) , m(AMC) and
m(AME)
Solution
D
Solution:
∵ CB // AD & AB is a transversal
C
∴ mB = mBAD = 45º
(Alternate angles)
C
(1)
∴ m(DME) + m(DMC) = 180°
∴ m(DME) = 180° - 90° = 90°
(2)
∵ MB bisects DME
∴ m(BME) = m(BMD) = ½  90° = 45°
(3)
● 90°
● M
B
∵ CE is a straight line
B
∵ CB // EH & CE is a transversal
∴ mC + mE = 180º
∴ mC = 180º - 115º = 65º
In ∆ ABC:
mCAB = 180º - (45º + 65º) = 70º
D
A
E
(1)
(2)
∵ AB is a straight line
∴ m(AME) = 180 - 45° = 135°
b) In the opposite figure:
AB//CD//EF & CD bisects ACE
F
<
Find with: mACD and mE
(Show the reason)
C
AB//CD, AC is a transversal
∴ mA + mACD = 180º (interior angles)
∴ mACD = 180º - 130º = 50º (1)
∵ CD bisects ACE
D
>
mECD = mACD = 50º
EF//CD, CE is a transversal
∴ mE = mECD = 50º
b) Using the geometric instruments, draw an angle of measure 120°,
then bisect it.
(Don’t remove the arcs)
E
A
(Alternate angles)
(3)
130°
>
B