Upload
hadiep
View
282
Download
6
Embed Size (px)
Citation preview
1
Chapter 4 Quadrilaterals
Level 1
1. In the figure, ABDE is a rectangle. The diagonals AD and BE intersect at
M. C is a point outside ABDE such that
△BMC is an equilateral triangle. If ∠BAD = 54o, find
(a) ∠BMD,
(b) ∠CDM.
(10 marks)
2. In the figure, ABCD is a rhombus. The side DA is produced
to a point E such that BE = BA. If ∠ADC = 68o, find
(a) ∠ABE,
(b) ∠BCE.
(10 marks)
3. In the figure, ABCD is a square, EB = FD.
(a) Prove that △BCE△DCF.
(b) Find ∠BEC.
(10 marks)
4. In the figure, ABCD is a parallelogram.
(a) Express z in terms of x and y.
(b) If DP is the angle bisector of ∠ADC and x = 58o,
find y and z.
(12 marks)
5. In the figure, ABCD is a parallelogram. P and Q are two
points on the diagonal AC such that BP ⊥ AC and DQ ⊥
AC.
(a) Prove that △APB△CQD.
(b) Prove that PBQD is a parallelogram.
(12 marks)
A
B
C
D
E
M
A D
CB
P
Q
A D
C B
P
x
y
z
A D
CB
E
A D
C B
E
F
64o
2
6. In the figure, A, B, C and D are four points on the four
sides of parallelogram PQRS respectively. If DS = QB
and PA = CR, prove that ABCD is a parallelogram.
(15 marks)
7. Find the unknown(s) in each of the following questions.
(a) (b) (c)
(16 marks)
8. In the figure, P, Q and R are the mid-points of the
three sides of △ABC respectively. Prove that APQR
is a parallelogram.
(15 marks)
Level 2
1. In the figure, ABCD is a parallelogram and ABEF is a square. AD and the diagonal BF of the
square intersect at G. If ∠BGD = 83o, find ∠BCD.
(6 marks)
F D CE
BA
G
P
Q R
SD
C
B
A
P
Q C B
A
R
B
A
C
D
E
F
G
H
7
7
x5
5
y
A
B E
C D
x cm
y cm
28 cm
A
55o M N
B C
xo
3
2. In the figure, PQRS is a rhombus. V is a point on SR such that QV and PR intersect at T, where
∠PTQ = 84o. If ∠PSR = 76o, find ∠RVQ.
(12 marks)
3. In the figure, ABCD is a rectangle. AC and BD intersect at E, AD = 3 and
CD = 7 . Find the length of BE and x.
(Give the answer correct to 3 significant figures if necessary.)
(12 marks)
4. In the figure, BA = BC, PR // BC and BP = BQ. Prove that PQCR is a parallelogram.
(12 marks)
5. In the figure, PQRS is a square. XY is perpendicular to the diagonal PR and RM = RS.
(a) Prove that △RSY△RMY.
(b) Using the result in (a), find ∠XRY.
(14 marks)
P
Q R
S
M
X
Y
C
P
A
B Q
R
S
84o
P
QR
V
T
7
A
B C
D
E
3
x
4
6.
In the figure, ABCD is a rectangle. The diagonals AC and BD intersect at E. BD and CD are
produced to F and G respectively such that ∠DCF = 30o and CF = GF. BA is produced to H
such that ∠AEH = 30o. If BH = GH and ∠BHE = 30o, prove that
(a) △CEF △EBH,
(b) EFGH is a rectangle.
(Hints: If a quadrilateral is a parallelogram and any one of the angles is a
right angle, then the quadrilateral is a rectangle.)
(16 marks)
7. In the figure, AB // CD // EF and BD = DF.
Prove that AB + EF = 2CD.
(12 marks)
8. In the figure, P and S are the mid-points of AD
and BC respectively. PQRS // DC.
(a) Prove that AB // DC.
(b) If AT = 18, CT = 45 and DT = 40, find QT.
(16 marks)
A B
C D
E F
D C
P
A B
S
T
Q R
30o
DC
E
BA
F
G
H
30o
5
Multiple-choice Questions
Answer ALL questions. Each question carries 5 marks.
1. In the figure, ABCD is a trapezium.
Find x and y.
A. x = 85o, y = 115o
B. x = 85o, y = 125o
C. x = 90o, y = 115o
D. x = 90o, y = 125o 2. In the figure, PQRS is a square. Find a and b. A. a = 15o, b = 75o
B. a = 15o, b = 90o
C. a = 18o, b = 75o
D. a = 18o, b = 90o 3. In the figure, ABCD is a parallelogram. Find ∠ADB.
A. 20o
B. 25o
C. 30o
D. 35o
4. In the figure, KLMN is a rectangle. Find x.
A. 8
B. 8.94, cor. to 3 sig. fig.
C. 9.49, cor. to 3 sig. fig.
D. 10
5. Which of the following may NOT be true?
I. A parallelogram has two pairs of parallel opposite sides.
II. A quadrilateral with 4 equal sides must be a rhombus.
III. A quadrilateral with all its angles equal must be a square.
A. I only
B. II only
C. III only
D. II and III only
x – 35o
x y
A
B C
D
a + b
6a
P
Q R
S
K
L
N
M
x
2x
20
B
75o
A D
C
6
6. Which of the following is/are true?
I. The diagonals of a parallelogram are perpendicular to each other.
II. The diagonals of a rectangle are equal in length.
III. Each interior angle of a rhombus is bisected by the diagonal.
A. II only
B. III only
C. II and III only
D. I, II and III
7. In the figure, ABCD is a parallelogram. AC and DE intersect at F. AF = 8, CF = 6, DF = 4. Find
DE.
A. 3
B. 4
C. 6
D. 7
8. In the figure, PQRS is a parallelogram and SR = TR. Find ∠QRT.
A. 70o
B. 75o
C. 80o
D. 85o 9. In the figure, ABCD is a rectangle and AEC is a straight line. If ∠ABE : ∠DCE = 2 : 3, find
∠ABE.
A. 18o
B. 24o
C. 36o
D. 54o
10. In the figure, PQRS is a rhombus. PR = 10 cm and QS = 24 cm. Find the perimeter of PQRS.
A. 52 cm
B. 54 cm
C. 60 cm
D. 68 cm
B
8
A D
C
4
6 F
E
Q
P S
R
T
105o
Q
R
S
P
B
A D
C
E
7
11. In the figure, PQRS is a rectangle. PT = 2.5a, RS = 7 and QR = 5a – 1. Find a.
A. 4
B. 5
C. 6
D. 7
12. If ABCD is a rhombus, then the ratio of A : B : C : D can be
A. 2 : 1 : 1 : 2.
B. 1 : 2 : 3 : 4.
C. 2 : 1 : 3 : 1.
D. 1 : 3 : 1 : 3.
13. In the figure, ABCD is a rhombus. AD = AE = AF = EF.
Find ∠ECF.
A. 90o
B. 95o
C. 100o
D. 120o
14. Adding which of the following conditions is sufficient to prove that ABCD in the figure is a
parallelogram?
A. AB = DC
B. AD = BC
C. ∠BAD + ∠ADC = 180o
D. ∠ADC + ∠ABC = 180o 15. Which of the following conditions is sufficient to prove that PQRS in the figure is a
parallelogram?
A. ∠PSQ = ∠RQS
B. PQ = SR
C. PS // QR
D. PT = RT and ST = QT
16. Find x in the figure.
A. 70o
B. 65o
C. 50o
D. 40o
B
A D
C
A
70o
M N
B C
x
Q
P S
R
T
5a – 1
7 2.5a
Q
P S
R
T
A
D
C
E
B
F
8
17. Find x and y in the figure.
A. x = 2, y = 3
B. x = 2, y = 6
C. x = 2, y = 9
D. x = 3, y = 12
18. Find TU in the figure.
A. 14 cm
B. 12 cm
C. 10 cm
D. 8 cm
19. Find x in the figure.
A. 7
B. 7.5
C. 8
D. 8.5
20. In △ABC, D, E and F are the mid-points of AB, BC and CA respectively. Which of the
following must be true?
I. DE // AC
II. Perimeter of BEFD = AB + BC
III. △ABC ~ △EFD
A. I only
B. I and II only
C. II and III only
D. I, II and III
P Q
R S
T U
20 cm
16 cm
7 cm
Q S U
T
R
P
4 cm 4 cm
x cm
A 3
y
3
x2
6
4
3
B
C
D
E
F G
9
Solutions
Level 1 1. (a) BM = AM (property of rectangle)
∴ ∠ABM = ∠BAM (base ∠s, isos. △)
= 54o
∠BMD = ∠BAM + ∠ABM (ext. ∠ of △)
= 54o + 54o
= 108o
(b) ∠BMC = 60o (property of equil. △)
∴ ∠CMD = 108o – 60o
= 48o
DM = BM (property of rectangle)
= CM
∴ ∠CDM = ∠DCM (base ∠s, isos. △)
∠CDM + ∠DCM + ∠CMD = 180o (∠ sum of △)
2∠CDM + 48o = 180o
∠CDM = 66o
2. (a) ∠BAE = ∠ADC (corr. ∠s, AB // DC)
= 68o
BE = BA (given)
∴ ∠BEA = ∠BAE (base ∠s, isos. △)
= 68o
∠ABE + ∠BAE + ∠BEA = 180o (∠ sum of △)
∠ABE + 68o + 68o = 180o
∠ABE = 44o
(b) ∠ABC = ∠ADC (property of rhombus)
= 68o
∠CBE = 68o + 44o
= 112o
∵ BC = BA (property of rhombus)
= BE (given)
∴ ∠BEC = ∠BCE (base ∠s, isos. △)
∠BCE + ∠BEC + ∠CBE = 180o (∠ sum of △)
2∠BCE + 112o = 180o
∠BCE = 34o
10
3. (a) EB = FD given
∠CBE = ∠CDF = 90o property of square
BC = DC
∴ △BCE△DCF SAS
(b) ∠BCE = ∠DCF (corr. ∠s, △s)
∠BCD = 90o (property of square)
∠BCE + ∠DCF + 64o = 90o
2∠BCE = 26o
∠BCE = 13o
∠BEC + ∠CBE + ∠BCE = 180o (∠ sum of △)
∠BEC + 90o + 13o = 180o
∠BEC = 77o
4. (a) ∠BAD = ∠BCD (opp. ∠s of // gram)
= y
∠BPD = ∠ADP + ∠DAP (ext. ∠ of △)
z = x + y
(b) ∠PDC = ∠ADP (given)
= 58o
∠ADC + ∠BCD= 180o (int. ∠s, AD // BC)
58o + 58o + y = 180o
y = 64o
z = x + y (proved)
= 58o + 64o
= 122o
5. (a) ∠APB = ∠CQD = 90o given
∠BAP = ∠DCQ alt. ∠s, AB // DC
AB = CD opp. sides of // gram
∴ △APB△CQD AAS
(b) PB = QD corr. sides, △s
∠BPQ = ∠DQP = 90o given
∴ PB // DQ alt. ∠s equal
∴ PBQD is a parallelogram. 2 sides equal and //
11
6. PS = QR opp. sides of // gram
PD + DS = QB + BR
PD + DS = DS + BR
PD = BR
In △APD and △CRB,
AP = CR given
∠APD = ∠CRB opp. ∠s of // gram
PD = BR proved
∴ △APD △CRB SAS
∴ AD = CB corr. sides, △s
PQ = SR opp. sides of //gram
PA + AQ = SC + CR
PA + AQ = SC + PA
AQ = SC
In △AQB and △CSD,
QB = SD given
∠AQB = ∠CSD opp. ∠s of // gram
AQ = CS proved
∴ △AQB △CSD SAS
∴ AB = CD corr. sides, △s
∴ ABCD is a parallelogram. opp. sides equal
7. (a) ∵ AM = MB and AN = NC (given)
∴ MN // BC (mid-pt. theorem)
∴ x = 55 (corr. ∠s, MN // BC)
(b) BF // CG // DH (given)
BC = CD (given)
∴ FG = GH (intercept theorem)
x = 7
AE // BF // CG (given)
EF = FG (proved)
∴ AB = BC (intercept theorem)
AB = 5
y = AB + BC + CD
= 5 + 5 + 5
= 15
12
(c) BE // CD (given)
AB = BC (given)
∴ AE = ED (intercept theorem)
x = y
x + y = 28
2x = 28
x = 14
y = x
= 14
8. BP = PA given
BQ = QC given
∴ PQ // AC mid-pt. theorem i.e. PQ // AR
and ARACPQ 2
1
∴ APQR is a parallelogram. 2 sides equal and //
Level 2 1. ∠ABF = 45o (property of square)
∠BAG + ∠ABG = ∠BGD (ext. ∠ of △)
∠BAG + 45o = 83o
∠BAG = 38o
∠BCD = ∠BAG (opp. ∠s of // gram)
= 38o
2. ∠PSR + ∠QRS = 180o (int. ∠s, SP // RQ)
76o + ∠QRS = 180o
∠QRS = 104o
∠PRS = ∠PRQ (property of rhombus)
∠PRS + ∠PRQ = ∠QRS
2∠PRS = 104o
∠PRS = 52o
∠RTV = ∠PTQ (vert. opp. ∠s)
= 84o
∠RVT + ∠TRV + ∠RTV = 180o (∠ sum of △)
∠RVT + 52o + 84o = 180o
∠RVT = 44o
13
i.e. ∠RVQ = 44o
3. ∠ADC = 90o (property of rectangle)
AC2 = AD2 + CD2 (Pyth. theorem)
AC = 22 )7( + 3
= 4
AE = BE = CE = DE (property of rectangle)
∴ BE = AC2
1
= 42
1
= 2
tan ∠CAD = 3
7
∠CAD = 41.410, cor. to 5 sig. fig.
∠EDA = ∠EAD (base ∠s, isos. △)
= 41.410
∠CED = ∠EAD + ∠EDA (ext. ∠ of △)
x = 41.410o + 41.410o
= 82.8o, cor. to 3 sig. fig.
4. BA = BC given
BP + PA = BQ + QC
BP + PA = BP + QC
∴ PA = QC
BA = BC given
∴ ∠BAC = ∠ACB base ∠s, isos. △
∠ARP = ∠ACB corr. ∠s, PR // BC
∴ ∠ARP = ∠BAC
∴ PA = PR sides opp. equal ∠s
∴ QC = PR
QC // PR given
∴ PQCR is a parallelogram. 2 sides equal and //
5. (a) In △RSY and △RMY,
RY = RY common side
∠RMY = 90o given
∠RSY = 90o property of square
∴ ∠RSY = ∠RMY
RS = RM given
∴ △RSY△RMY RHS
14
(b) ∠SRP = 45o (property of square)
∠SRY = ∠MRY (corr. ∠s, △s)
∴ ∠MRY = SRP2
1
= 452
1
= 22.5o
QR = RS
= RM
Similarly, ∠MRX = 22.5o.
∠XRY = ∠MRX + ∠MRY
= 22.5o + 22.5o
= 45o
6. (a) ∠BAE = ∠AHE + ∠AEH ext. ∠ of △
= 30o + 30o
= 60o
AE = BE = CE = DE property of rectangle
∠ABE = ∠BAE base ∠s, isos. △
= 60o
∠ABE + ∠BAE + ∠AEB = 180o ∠ sum of △
60o + 60o + ∠AEB = 180o
∠AEB = 60o
∠FEC = ∠AEB = 60o vert. opp. ∠s
∠DCE = ∠BAE = 60o alt. ∠s, AB // DC
In △HEB and △FCE,
∠HBE = ∠FEC = 60o proved
∠HEB = ∠AEH + ∠AEB
= 30o + 60o
= 90o
∠FCE = ∠FCD + ∠ECD
= 30o + 60o
= 90o
∴ ∠HEB = ∠FCE
BE = EC property of rectangle
∴ △CEF △EBH ASA
(b) EF = BH corr. sides, △s
= GH
EH = CF corr. sides, △s
= GF
∴ EFGH is a parallelogram. opp. sides equal
∠HEB = 90o
∴ ∠HEF = 180 o – 90o = 90o adj. ∠s on st. line
15
∴ EFGH is a rectangle.
7. Join AF. Suppose AF and CD intersect at G.
AB // CD // EF and FD = DB given
∴ FG = GA intercept theorem
and EC = CA intercept theorem
∴ GD = AB2
1 mid-pt. theorem
and CG = EF2
1 mid-pt. theorem
CG + GD = CD
EF2
1 + AB
2
1 = CD
2
1(AB + EF) = CD
AB + EF = 2CD
8. (a) In △ACD,
AP = PD given
PR // DC given
∴ AR = RC intercept theorem
and BS = CS given
∴ AB // RS mid-pt. theorem
i.e. AB // DC
(b) DP = PA (given)
PQ // AB (proved)
∴ DQ = QB (intercept theorem)
In △ATB and △CTD,
∠BAT = ∠DCT (alt. ∠s, AB // DC)
∠ABT = ∠CDT (alt. ∠s, AB // DC)
∠ATB = ∠CTD (vert. opp. ∠s)
∴ △ATB ~ △CTD (AAA)
∴ CT
AT =
DT
BT (corr. sides, ~△s)
45
18 =
40
BT
BT = 16
BQ = BT + QT
∴ DQ = 16 + QT
DQ + QT = DT
16 + QT + QT = 40
2QT = 24
A B
C D
E F
G
16
QT = 12
MC
1. D 6. C 11. B 16. A 2. A 7. D 12. D 17. B 3. C 8. B 13. C 18. B 4. B 9. C 14. A 19. A 5. C 10. A 15. D 20. D
1. The answer is D.
∠BCD + ∠ADC = 180o (int. ∠s, AD // BC)
x + 90o = 180o
x = 90o
∠ABC + ∠BAD = 180o (int. ∠s, AD // BC)
35xy = 180o
3590y = 180o
y = 125o
2. The answer is A.
6a = 90o (property of square)
a = 15o
a + b = 90o (property of square)
15o + b = 90o
b = 75o
3. The answer is C.
BC = BD (given)
∴ ∠BDC = ∠BCD (base ∠s, isos. △)
= 75o
∠ADC + ∠BCD = 180o (int. ∠s, AD // BC)
∠ADB + 75o + 75o = 180o
∠ADB = 30o
4. The answer is B.
∠LMN = 90o (property of rectangle)
MN 2 + LM 2 = LN 2
x2 + (2x)2 = 202
5x2 = 400
x2 = 80
17
x = 8.94, cor. to 3 sig. fig.
5. The answer is C.
6. The answer is C.
7. The answer is D.
In △AFD and △CFE,
∠AFD = ∠CFE (vert. opp. ∠s)
∠DAF = ∠ECF (alt. ∠s, AD // BC)
∠ADF = ∠CEF (alt. ∠s, AD // BC)
∴ △AFD ~ △CFE (AAA)
∴ CF
AF
EF
DF (corr. sides, ~△s)
6
8
EF
4
EF = 3
∴ DE = DF + EF
= 4 + 3
= 7
8. The answer is B.
∠QPS + ∠PSR = 180o (int. ∠s, PQ // SR)
105o + ∠PSR = 180o
∠PSR = 75o
SR = TR (given)
∴ ∠STR = ∠TSR (base ∠s, isos. △)
= 75o
∠QRT = ∠STR (alt. ∠s, PS // QR)
= 75o
9. The answer is C.
Let ∠ABE = 2x,
then ∠DCE = 3x
∠BAE + ∠ABE = ∠BEC (ext. ∠ of △)
∠BAE + 2x = 90o
∠BAE = 90o – 2x
∠BAE = ∠DCE (alt. ∠s, AB // DC)
90o – 2x = 3x
90o = 5x
x = 18o
∠ABE = o182
= 36o
18
19
10. The answer is A.
Suppose the diagonals PR and QS intersect at T.
PT = RT = PR2
1 (property of rhombus)
cm 5
cm 102
1
QT = ST = QS2
1 (property of rhombus)
cm 12
cm 242
1
∠PTS = 90o (property of rhombus)
In △PTS,
PS 2 = PT 2 + ST 2
PS = 22 STPT
= 22 125 cm
= 13 cm
Perimeter of PQRS = 4PS
= 413 cm
= 52 cm 11. The answer is B. ST = QT = PT (property of rectangle)
= 2.5a
QS = 2 ST
= a2.52
= 5a
∠QRS = 90o (property of rectangle)
In △QRS,
QS 2 = SR 2 + QR 2 (5a)2 = 72 + (5a – 1)2 25a2 = 49 + 25a2 – 10a + 1 10a = 50
a = 5 12. The answer is D.
The opposite angles of a rhombus are equal.
∴ A = C and B = D.
∴ The ratio cannot be 2 : 1 : 1 : 2, 1 : 2 : 3 : 4 or 2 : 1 : 3 : 1.
13. The answer is C.
In △ABE and △ADF,
ABE = ADF (property of rhombus)
Q
R
S
P
T
20
AB = AD (property of rhombus)
= AE (given)
AEB = ABE (base s, isos. △)
= ADF
= AFD (base s, isos. △)
AE = AF (given)
△ABE △ADF (AAS)
∴ BE = DF (corr. sides, △s)
BC = DC (property of rhombus)
EC = BC BE
= DC DF
= FC
i.e. FEC = EFC (base s, isos. △)
In △CEF,
ECF + FEC + EFC = 180 ( sum of △)
ECF + 2FEC = 180
FEC = 2
180 ECF
AEF = 60 (property of equil. △)
ABE + ECF = 180 (int. s, AB // DC)
ABE = 180 ECF
∴ AEB = 180 ECF (AEB = ABE)
AEB + AEF + FEC = 180 (adj. s on st. line)
180 ECF + 60 + 2
180 ECF = 180
ECF + 2
ECF = 150
ECF = 100
14. The answer is A.
∵ AB // DC and AB = DC.
∴ ABCD is a parallelogram. (2 sides equal and //)
15. The answer is D.
∵ PT = RT and ST = QT.
∴ PQRS is a parallelogram. (diags. bisect each other)
16. The answer is A.
AM = MB and AN = NC (given)
∴ MN // BC (mid-pt. theorem)
∠AMN = ∠ABC (corr. ∠s, MN // BC)
= 70o
AM = AN (given)
21
∴ ∠ANM = ∠AMN (base s, isos. △)
x = 70o
17. The answer is B.
AC = CE (given)
AB // CD // EF (given)
∴ DF = BD (intercept theorem)
x = 2
AE = EG = 6 (given)
BF = FG = 4
∴ ABEF2
1 (mid-pt. theorem)
y = 23
= 6
18. The answer is B.
Join TQ.
Suppose TQ and RS intersect at V.
PQ // RS // TU and PR = RT (given)
∴ QS = SU (intercept theorem)
and QV = VT (intercept theorem)
∴ VS = TU2
1 (mid-pt. theorem)
and RV = PQ2
1 (mid-pt. theorem)
RV + VS = RS
PQ2
1 + TU
2
1 = RS
PQ + TU = 2RS
TU = PQRS 2
= )2016(2 cm
= 12 cm
P Q
R S
T U
20 cm
16 cm
V
22
19. The answer is A.
QS = SU and PR = RU (given)
∴ RS // PQ and PQRS2
1 (mid-pt. theorem)
QS = SU (given)
RS // TU
∴ QR = RT (intercept theorem)
QS = SU and QR = RT
∴ TURS2
1 (mid-pt. theorem)
∴ PQ2
1 = TU
2
1
PQ = TU
x = 7
20. The answer is D.
According to the information, draw △ABC as shown.
By the mid-point theorem, we have:
EF // AB and ABEF2
1
DF // BC and BCDF2
1
DE // AC and ACDE2
1
DE // AC
∴ I is true.
EF = AB2
1= AD;DF = BC
2
1= EC
Perimeter of BEFD = BE + EF + FD + DB
= BE + AD + EC + DB
= (AD + DB) + (BE + EC)
= AB + BC
∴ II is true.
2
1
AC
ED
BC
FD
AB
EF
∴ △ABC ~ △EFD (3 sides proportional)
∴ III is true i.e. I, II and III are true.
A
D
B C E
F