chapter nine quadrilaterals

8/18/2019 Chapter Nine Quadrilaterals

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Chapter Nine Quadrilaterals

1. In the figure, ABCD is a parallelogram and ABEF is a square. AD and the

diagonal BF of the square meet at G. If ∠ BGD = 83o, find ∠ BCD.

(6 marks)

2. In the figure, PQRS is a rhombus. V is a point on SR such that QV and PR meet atT where ∠PTQ = 84

o. If ∠PSR = 76

o, find ∠ RVQ.

(12 marks)

S

84o

P

Q R

V

T

F D C E

B A

G


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3. In the figure, ABCD is a rectangle. AC cuts BD at E , AD = 3 and CD = 7 . Find

the length of BE and x. (Give the answer correct to 3 significant figures when

necessary.)

(12 marks)

4. In the figure, BA = BC , PR // BC and BP = BQ. Prove that PQCR is a

parallelogram.

(12 marks)

7

A

B C

D

E

3

x

C

P

A

B Q

R


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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)

6. In the figure, ABCD is a rectangle. The diagonals AC and BD meet 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.

(If a quadrilateral is a parallelogram and any one of the angles is a right angle,

then the quadrilateral is a rectangle.)

(16 marks)

P

Q R

S

M

X

Y

30o

DC

E

B A

F

G

H

30o


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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)

− END OF PAPER −

A B

C D

E F

D C

P

A B

S

T

Q R


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Solutions:

1. ∠ ABF = 45o (property of square)

∠ BAG + ∠ ABG = ∠ BGD (ext.∠ of △)

∠ BAG + 45o = 83o

∠ BAG = 38o

∠ BCD = ∠ BAG (property of parallelogram)

= 38o

2. ∠PSR + ∠QRS = 180o (int. ∠s, PS // QR)

76o + ∠QRS = 180

o

∠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 + 84

o = 180

o

∠ RVT = 44

o

i.e. ∠ RVQ = 44o

3. ∠ ADC = 90o (property of rectangle)

AC 2 = AD

2 + CD

2 (Pyth. theorem)

AC 2 = 3

2 + ( 7 )

2

AC 2 = 16

AC = 4

AE = BE = CE = DE (property of rectangle)

∴ BE = AC 2

1

= )4(2

1

= 2

tan ∠CAD =3

7

∠CAD = 41.410o (cor. to 5 sig. fig.)

∠ EDA=∠ EAD

(base∠

s, isos.△

)= 41.410

o


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∠CED = ∠ EAD + ∠ EDA

(ext. ∠ of △)

x = 41.410o + 41.410

o

= 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, PQ // 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 = 90

o

property of square∴ ∠ RSY = ∠ RMY

RS = RM given

∴ △ RSY ≅△ RMY RHS

(b) ∠SRP = 45o

(property of square)

∠SRY = ∠ MRY (corr. ∠s, ≅△s)

∴ ∠ MRY = SRP∠2

1

= 22.5o

QR = RS

= RM

Similarly, ∠ MRX = 22.5o.

∠ XRY = ∠ MRX + ∠ MRY

= 22.5o + 22.5

o

= 45o


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6. (a) ∠ BAE = ∠ AHE + ∠ AEH ext.∠ of △

= 30o + 30

o

= 60o

AE = BE = CE = DE property of rectangle

∠ ABE = ∠ BAE base ∠s, isos. △

= 60o

∠ ABE + ∠ BAE + ∠ AEB = 180o ∠ sum of △

60o + 60

o + ∠ AEB = 180

o

∠ AEB = 60o

∠FEC = ∠ AEB vert. opp. ∠s

= 60o

∠ DCE = ∠ BAE alt ∠s, AB // DC

= 60o

In △ HEB and △FCE ,

∠ HBE = ∠FEC = 60o proved

∠ HEB = ∠ AEH + ∠ AEB

= 30o + 60

o = 90o

∠FCE = ∠FCD + ∠ ECD

= 30o + 60

o = 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 = 90o adj. ∠s on st. line

∴ EFGH is a rectangle.


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7. Join AF . Suppose AF and CD meet at G.

AB // CD // EF and FD = DB given

∴ FG = AG intercept theorem

and EC = AC intercept theorem

∴ DG = AB2

1 mid-pt. theorem

and CG = EF 2

1 mid-pt. theorem

CG + DG = CD

EF 2

1 + AB

2

1 = CD

2

1( AB + EF ) = CD

AB + EF = 2CD

8. (a) In △ ACD,

AP =

DP given

PR // DC given

∴ AR = RC intercept theorem

BS = CS given

AB // RS

i.e. AB // DC . mid-pt. theorem

(b) DP = AP (given)

PQ // AB (proved)

∴ DQ = BQ (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

A B

C D

E F

G


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BQ = BT + QT

∴ DQ = 16 + QT

DQ + QT = DT

16 + QT + QT = 40

2QT = 24

QT = 12

chapter nine quadrilaterals

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