chapter 4 quadrilateralssyf@tanghin.edu.hk/f.3/ch. 4...8. in the figure, pqrs is a parallelogram and...

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