Download Book 3A Chapter 05 (1)

Mathematics in Action (3rd Edition) 3A Full Solutions
5 Quadrilaterals
In △BCD,
BD 2  BC 2  CD 2
Quick Review
y  15  8
2
(Pyth. theorem)
2
 17
Let’s Try (p. 5.3)
1. (a) x  130  180
(int. s, CB // EF)
2.
x  50
37  ACD  75
ACD  38
(b) BCD  50  15
 65
∵ ABC  BCD  65
∴ AB // CD
(alt. s equal)
2.
(a) In △ABC and △CDA,
ABC  CDA
BAC  DCA
AC  CA
∴ △ABC  △CDA
given
alt. s, BA // CD
common side
AAS
(b) ∵ △ABC  △CDA
∴ BCA  DAC
∴ AD // BC
proved
corr. s, △s
alt. s equal
(b) ∵
∴
ca
 70
(corr. s, BA // CD)
b  70  180
b  110
(int. s, BA // CD)
∵ △BCE is an equilateral triangle.
∴ EBC  ECB  60
prop. of equil. △
ABC  BCD
 (20  60)  (40  60)
 80  100
 180
∴ AB // DC
int. s supp.
4.
(a) In △ABC and △ADC,
AB  AD
BAC  DAC
AC  AC
∴ △ABC  △ADC
b  20  a
 20  20
 40
(b) ∵
∴
∴
5.
(base ∠s, isos. △)
(ext. ∠ of △)
∵ CD  BD
c  b  40 (base ∠s, isos. △)
∴
BAC  ACD  38
BA // CD
alt. s equal
3.
Review Exercise 5 (p. 5.5)
1. (a) a  70
(alt. s, AD // BE)
(b) In △ABD,
∵ BD  AB
a  20
∴
BCD  ADE (corr. s, DA // CB)
(a)
given
given
common side
SAS
△ABC  △ADC
(proved in (a))
(corr. sides, △s)
BC  DC
△BCD is an isosceles triangle.
(a) In △ADB,
∵ AB  AD
∴ ABD  ADB (base s, isos. △)
DAB  ADB  ABD  180 ( sum of △)
56  2ADB  180
2ADB  124
ADB  62
In △ACD,
ACD  CAD  ADB (ext.  of △)
28  CAD  62
(c) In △ABC,
∵ BC  AC
∴ ABC  x
60  x  x  180
CAD  34
(base ∠s, isos. △)
(∠ sum of △)
(b)
x  60
BCE  ABC
(alt. ∠s, DE // AB)
yx
 60
EAB  EAD  DAB
 34  56
 90
AED  90
∵ EAB  AED  90  90
 180
∴ AB // ED
(int. s supp.)
(d) In △ABC,
BC 2  AB 2  AC 2
(Pyth. theorem)
x  122  9 2
 15
76
5 Quadrilaterals
6.
(a) In △BCD,
∴
BC 2  BD 2  (10 2  24 2 ) cm 2
 (100  576) cm 2
 676 cm 2
CD  26 cm 2  676 cm 2
2
2
∵ BC 2  BD 2  CD 2
∴ △BCD is a right-angled triangle, where
CBD  90 . (converse of Pyth. theorem)
ADB  CBD (alt. s, AD // BC)
Classwork
 90
Classwork (p. 5.9)
(a) ∵ ABCD is a parallelogram.
∴ AB  DC and BC  AD
∴ x  13 and y  22
(b) ∵ BAD  ABD
∴ AD  BD  24 cm (sides opp. equal s)
In △ABD,
AB 2  AD 2  BD 2
CG CF

(intercept theorem)
GD FB
y cm
7.2 cm

7.2 cm 12 cm
7.2
y
 7.2
12
 4.32
(Pyth. theorem)
AB  24 2 24 2 cm
 1152 cm (or 24 2 cm)
Activity
Activity 5.1 (p. 5.8)
1. (a) Yes (Reason: alt. s, BC // AD)
(opp. sides of // gram)
(b) ∵
∴
∴
ABCD is a parallelogram.
DO  BO and CO  AO (diags. of // gram)
a  9 and b  7
(c) ∵
∴
∴
ABCD is a parallelogram.
B  D and C  A (opp. s of // gram)
c  68 and d  112
Classwork (p. 5.18)
1. Yes (Reason: opp. sides equal)
(b) Yes (Reason: alt. s, BA // CD)
2.
Yes (Reason: opp. s equal)
(c) Yes (Reason: ASA)
2.
3.
No
(d) (i) They are equal.
(ii) They are equal.
4.
Yes (Reason: opp. sides equal and //)
(a) Yes (Reason: AAS (or ASA))
5.
Yes (Reason: diags. bisect each other)
6.
No
(b) OA = OC, OB = OD
(Reason: corr. sides, △s)
Classwork (p. 5.25)
(a) ∵ ABCD is a rhombus.
∴ BC  CD  AD (property of rhombus)
∴ x y7
Warm-Up Activity (p. 5.24)
1. A: Rhombus
B: Square
C: Rectangle
D: Trapezium
2.
rhombus, square, rectangle
(b) ∵
∴
Activity 5.2 (p. 5.47)
1. The length of MN is half of that of BC.
2.
ABCD is a rhombus.
x  90 (property of rhombus)
y  40 (property of rhombus)
Yes. It is because MEFN is a rectangle. (or any other
reasonable answers)
(c) ∵
∴
To Learn More
∴
To Learn More (p. 5.60)
In △ABC,
∵ EF // AC
FC EA

∴
(intercept theorem)
BF BE
x cm
6 cm

12 cm 10 cm
6
x   12
10
 7 .2
ABCD is a rhombus.
OA  OC and OB  OD
6
x   3 and y  4
2
(property of rhombus)
Classwork (p. 5.28)
(a) ∵ ABCD is a rectangle.
∴ ABC  90 (property of rectangle)
x  20  90
x  70
BC  AD
y6
In △BDC,
∵ FG // BD
77
(property of rectangle)
Mathematics in Action (3rd Edition) 3A Full Solutions
(b) ∵
∴
∴
∴
(b) In △ABC,
∵ AN  NB and AM  MC
1
∴ MN   BC and NM // BC
2
1
4 y
2
y 8
ABCD is a rectangle.
OB  OC  OA  5 cm (property of rectangle)
x5
y  25
 10
(c) ∵ ABCD is a rectangle.
∴ OA  OD
(property of rectangle)
∴
(base s, isos. △)
x y
In △AOD,
(ext.  of △)
x  y  46
2 x  46
x  23
∴
MNB  CBN  180
(b) ∵
∴
x  68
Quick Practice
y  23
Quick Practice 5.1 (p. 5.10)
(a) ∵ PQRS is a parallelogram.
∴ PQR  PSR
(opp. s of // gram)
20  3 x  4 x  1
20  1  4 x  3 x
x  19
ADC  90 (property of square)
y  90
(b) TQR  3x
ABCD is a square.
AOD  90 (property of square)
∵
∴
x  90
∵
∴
ABCD is a square.
ACD  45 (property of square)
ABCD is a square.
OA  OD  OB  7 cm (property of square)
x y7
2x  1  3
2x  4
Classwork (p. 5.34)
(a) ∵ ABCD is an isosceles trapezium.
∴ x  65
x  y  180
65  y  180
y  115
(b) ∵
∴
x2
∴
(int. s, AD // BC)
OD  OB
(diags. of // gram)
y34
y 1
Quick Practice 5.3 (p. 5.12)
∵ ABCD is a parallelogram.
∴ ABC  ADC (opp. s of // gram)
 70
∵ BEFG is a parallelogram.
∴ EBG  EFG (opp. s of // gram)
 55
ABG  ABC  EBG
ABCD is an isosceles trapezium.
BD  AC
x 42
6
Classwork (p. 5.48)
(a) In △ABC,
∵ AM  MB and AN  NC
1
∴ MN   BC and MN // BC
2
1
x   10
2
5
 3  19
 57
TQR  STQ  180 (int. s, QR // PS)
STQ  180  57
 123
Quick Practice 5.2 (p. 5.11)
∵ OCED is a parallelogram.
∴ OC  DE  ( 2 x  1) cm
(opp. sides of // gram)
and OD  CE  ( y  3) cm (opp. sides of // gram)
∵ ABCD is a parallelogram.
∴
(diags. of // gram)
OC  OA
y  45
(c) ∵
∴
∴
(int. s, NM // BC)
x  112  180
Classwork (p. 5.31)
(a) ∵ ABCD is a square.
∴ BC  AB
(property of square)
x6
∴
(mid-pt. theorem)
 70  55
 15
In △ABG,
BAG  ABG  AGB  180 ( sum of △)
BAG  15  140  180
BAG  25
∵ ABG  BAG
∴ △ABG is not an isosceles triangle.
(mid-pt. theorem)
y  55 (corr. s, MN // BC)
78
5 Quadrilaterals
Quick Practice 5.4 (p. 5.19)
(a) In EFGH,
(4 x  5)  115  (3x  20)  115  360
7 x  255  360
7 x  105
x  15
(b) FEH  4 15  5  65
FGH  3 15  20  65
∵ EFG  EHG
and FEH  FGH
∴ EFGH is a parallelogram.
Quick Practice 5.5 (p. 5.20)
∵ ABCD is a parallelogram.
∴ OA  OC
and OB  OD
∵ OM  2OA
 2OC
 ON
BCD  BDC  DBC  180 ( sum of △)
b  36  36  180
b  72  180
b  108
( sum of
polygon)
Quick Practice 5.8 (p. 5.28)
∵ PQRS is a rhombus.
∴ PQ  QR  RS  SP and PR  QS
(property of rhombus)
i.e. 4 PQ  68 cm
PQ  17 cm
In △OPQ,
1
OP  PR (property of rhombus)
2
30

cm
2
 15 cm
OP 2  OQ 2  PQ 2
(Pyth. theorem)
opp. s equal
diags. of // gram
diags. of // gram
given
OQ  17 2  152 cm
 8 cm
1
∴ Area of △OPQ   OQ  OP
2
1
  8  15 cm 2
2
 60 cm 2
∴ Area of PQRS = 4  area of △OPQ
 4  60 cm 2
given
∵ OM  ON and OB  OD
∴ BNDM is a parallelogram. diags. bisect each other
Quick Practice 5.6 (p. 5.21)
(a) ∵ ABCD is a parallelogram.
∴ AD  BC
and ADC  ABC
BG  BC  CG
 AD  CG
 (7  2) cm
 9 cm
∴ BG  EF
∵ ABC  ADC
and ADC  AEF
∴ ABC  AEF
∴ BG // EF
∴ BEFG is a parallelogram.
opp. sides of // gram
opp. s of // gram
 240 cm 2
Quick Practice 5.9 (p. 5.29)
∵ ABCD is a rectangle.
∴ KB  KA
(property of rectangle)
∴ KBA  x
(base ∠s, isos. △)
(vert. opp. s)
AKB  76
In △ABK,
BAK  ABK  AKB  180 (∠ sum of △)
x  x  76  180
proved
given
corr. s equal
opp. sides equal and //
(b) ∵ BEFG is a parallelogram.
∴ EBG  EFG (opp. s of // gram)
EBG  BCD
(alt. s, AB // DC)
 118
∴ EFG  118
2 x  104
x  52
∵ KA  KD
∴ KAD  KDA
y
BAD  90
Quick Practice 5.7 (p. 5.27)
(a) ∵ ABCD is a rhombus.
∴ AOD  90 (property of rhombus)
5a  90
(property of rectangle)
(base s, isos. △)
(property of rectangle)
x  y  90
52  y  90
y  38
a  18
Quick Practice 5.10 (p. 5.30)
∵ ABFE is a rhombus.
∴ BF  AE  13 cm
∵ BCDE is a rectangle.
∴ DF  BF  13 cm
i.e. BD  2 13 cm  26 cm
In △BDE,
BED  90
(b) ADB  2a
 2  18
 36
In △BCD,
BDC  DBC  ADB  36 (property of rhombus)
79
(property of rhombus)
(property of rectangle)
(property of rectangle)
Mathematics in Action (3rd Edition) 3A Full Solutions
BE 2  DE 2  BD 2
Construct a line TR such that PQ // TR.
∵ PQ // TR and PT // QR
∴ PQRT is a parallelogram.
∴ TR  PQ
(opp. sides of // gram)
(int. s, PS // QR)
PQR  QPS  180
(Pyth. theorem)
DE  26 2  10 2 cm
 24 cm
∴ Perimeter of BCDE  2  ( BE  DE )
 2  (10  24) cm
120  QPS  180
QPS  60
∵ PQRS is an isosceles trapezium.
∴ PQ  SR  9 cm and RSP  QPS
∴ TR  SR  9 cm and RSP  60
∵ △RST is an isosceles triangle.
i.e. RTS  RST  60
(base s, isos. △)
RST  RTS  TRS  180 ( sum of △)
 68 cm
Quick Practice 5.11 (p. 5.32)
(a) ∵ ABCD is a square.
∴
OA  OB  OC  OD  2 5 cm (property of square)
∴
BD  2  2 5 cm
 4 5 cm
(b) In △OAB,
AOB  90
(property of square)
AB  OA  OB
2
60  60  TRS  180
TRS  60
∴ △RST is an equilateral triangle.
i.e. TS  SR  9 cm
∴ QR  PT
 PS  TS
 (15  9) cm
 6 cm
2
2
(Pyth. theorem)
 [(2 5 )  (2 5 ) 2 ] cm 2
2
∴
 40 cm 2
Area of square ABCD  AB 2
 40 cm 2
Quick Practice 5.14 (p. 5.41)
In △ABE and △CDF,
AB  CD
opp. sides of // gram
given
AEB  CFD  90
alt. ∠s, AB // DC
ABE  CDF
∴ △ABE  △CDF
AAS
∴ BE  DF
corr. sides,  △s
Alternative Solution
∵ ABCD consists of four congruent right-angled
triangles △OAB, △ODA, △OBC and △OCD.
1

∴ Area of square ABCD  4    2 5  2 5  cm 2
2

 40 cm 2
Quick Practice 5.12 (p. 5.33)
∵ ABCD is a square.
∴ AB  AD
(property of square)
∵ △ADE is an equilateral triangle.
∴ AD  AE
∴ AE AB
∴ AEB  ABE
(base s, isos. △)
x
BAD  90
(property of square)
DAE  60
(prop. of equil. △)
In △ABE,
ABE  BAE  AEB  180 ( sum of △)
x  (90  60)  x  180
2 x  150  180
2 x  30
x  15
ABD  45
x  y  45
15  y  45
y  30
Quick Practice 5.15 (p. 5.41)
In △ABE and △DCE,
AB  DC
ABE  DCE  90
∵ E is the mid-point of BC.
∴ BE  CE
∴ △ABE  △DCE
∴ AE  DE
∵ FD  AE
and FA  DE
∴ AE  DE  FD  FA
∴ AEDF is a rhombus.
Quick Practice 5.16 (p. 5.42)
∵ ABCD is a rhombus.
∴ BA// CD and AD // BC
In △AEG and △DHG,
AEG  DHG
EG  HG
AGE  DGH
∴ △AEG  △DHG
AE  DH
∴ AG  DG
In △CHF and △DHG,
CFH  DGH
HG  HF
DHG  CHF
∴ △CHF  △DHG
CF  DG
(property of square)
Quick Practice 5.13 (p. 5.35)
T
80
property of rectangle
property of rectangle
SAS
corr. sides, △s
opp. sides of // gram
opp. sides of // gram
alt. ∠s, BA // CD
given
vert. opp. ∠s
ASA
corr. sides, △s 
corr. sides, △s
alt. ∠s, AD // BC
given
vert. opp. ∠s
ASA
corr. sides, △s 
5 Quadrilaterals
∴
∵
∴
∵
∴
∴
CH  DH
AD  DC
AE  DH
1
 DC
2
1
 AD
2
 DG
 CF
AB  BC
BE  AB  AE
 BC  CF
 BF
AEG  CFH
corr. sides, △s
property of rhombus
From 
Quick Practice 5.19 (p. 5.51)
In △ABC,
∵ M is the mid-point of BC and
AF  FC .
1
∴ BA// MF and FM   AB
2
In △DEF,
∵ N is the mid-point of EF and
FC  CD .
1
∴ DE // CN and CN   DE
2
∵ BA// DE , BA// MF and
DE // CN
∴ MF // CN
1
∵ FM   AB and
2
1
CN   DE
2
∴ AB  2FM and DE  2CN
∵ AB  DE
∴ 2 FM  2CN
FM  CN
∴ FMCN is a parallelogram.
From 
property of rhombus
base s, isos. △
Quick Practice 5.17 (p. 5.49)
(a) In △XYZ,
∵ P and Q are the mid-points of XY and XZ
respectively.
1
PQ   YZ
∴
(mid-pt. theorem)
2
1
4 cm   YZ
2
YZ  8 cm
mid-pt. theorem
mid-pt. theorem
opp. sides equal and //
Quick Practice 5.20 (p. 5.56)
∵ AB // CD // EF // GH and BD  DF  FH
∴ AC  CE  EG
(intercept theorem)
1
∴ AC   AG
3
1
  12 cm
3
 4 cm
(b) In △XYZ,
∵ P and Q are the mid-points of XY and XZ
respectively.
∴ PQ // YZ
(mid-pt. theorem)
∴ XPQ  XYZ
(corr. s, PQ // YZ)
 55
In △XPQ,
XPQ  XQP  PXQ  180 ( sum of △)
Quick Practice 5.21 (p. 5.58)
(a) ∵ AD // EF // BC and AE  EB
∴ DF  FC
(intercept theorem)
In △ACD,
∵ AD // GF and DF  FC
∴ AG  GC
(intercept theorem)
∵ DF  FC and AG  GC
1
∴ GF   AD
(mid-pt. theorem)
2
1
  10 cm
2
 5 cm
55  XQP  38  180
XQP  87
Quick Practice 5.18 (p. 5.50)
(a) In △RST,
∵ U and V are the mid-points of ST and RT respectively.
1
UV   SR
∴
(mid-pt. theorem)
2
1
5 cm   SR
2
SR  10 cm
(b) In △RST,
∵ U and V are the mid-points of ST and RT respectively.
∴ SR // UV
(mid-pt. theorem)
∴ TSR  TUV  68
(corr. s, SR // UV)
In △PQR,
∵ S and U are the mid-points of PR and QT
respectively.
∴ PQ // SU
(mid-pt. theorem)
∴ QPR  USR  68
(corr. s, PQ // SU)
PQR  PRQ  QPR  180 ( sum of △)
PQR  25  68  180
PQR  87
(b)
EG  EF  GF
 (14  5) cm
 9 cm
In △ABC,
∵ AE  EB and AG  GC
1
∴
(mid-pt. theorem)
EG   BC
2
1
9 cm   BC
2
BC  18 cm
81
Mathematics in Action (3rd Edition) 3A Full Solutions
Quick Practice 5.22 (p. 5.59)
(a) ∵ AMCN is a parallelogram.
∴ AN // MC
Consider △ABK.
∵ BM  MA and MH // AK
∴ BH  HK
Consider △CDH.
∵ DN  NC and KN // HC
∴ KD  HK
∴ BH  HK  KD
(b) In △CDH,
∵ DN  NC and DK  KH
1
∴ KN   HC
2
HC  2 KN
∵ ABCD is a parallelogram.
∴ AB  CD
In △ABK and △CDH,
AB  CD
ABK  CDH
BK  BH  HK
 KD  HK
 DH
∴ △ABK  △CDH
∴ AK  CH
∴ AK  2KN
4.
∵
∴
x6
OC  OA
intercept theorem
y 1
intercept theorem
5.
(a) ∵
∴
ABCD is a parallelogram.
(opp.∠s of // gram)
ADC  ABC
 134
ADE  180  ADC
(adj.∠s on st. line)
 180  134
 46
mid-pt. theorem
(b) ∵
∴
opp. sides of // gram
proved
alt. s, AB // DC
Consolidation Corner (p. 5.21)
1. ∵ AB  DC  6 cm and
AD  BC  5 cm
∴ ABCD is a parallelogram.
SAS
corr. sides, △s
2.
Consolidation Corner (p. 5.12)
1. m  55  180
(int. s, AD // BC)
∵
4.
(a) ∵
∴
n6
3.
∵
∴
Perimeter of PQRS  28 cm
PQ  QR  RS  SP  28 cm
8 x  4  28
(opp. sides of // gram)
8 x  32
b  28  b
2b  28
b  14
C  A
c  108
given
opp. sides equal and //
(2 x  2)  2 x  ( x  2)  (3 x  4)  28
a  24
AB  DC
given
opp. sides equal
BAG  BEG  30 and
(given)
AGE  ABE  150
∴ AGEB is a parallelogram.
(opp. s equal)
∴ GE  AB
(opp. sides of // gram)
 6 cm
∵
ABCD is a parallelogram.
(opp. sides of // gram)
DC  AB
ABCD is a parallelogram.
AD  BC
(opp. sides of // gram)
BC  EF  5 cm and
BC // EF
BEFC is a parallelogram.
3.
m  125
∵
∴
ADE  AED  46°
(sides opp. equal∠s)
AD  AE
 5 cm
(opp. sides of // gram)
BC  AD
 5 cm
∴
2.
(diags. of // gram)
y 1  2
Consolidation Corner
∵
∴
ABCD is a parallelogram.
OB  OD (diags. of // gram)
x24
x4
(b)
(opp.∠s of // gram)
ABCD is a parallelogram.
(opp.∠s of // gram)
B  D
2h  30  h
h  30
C  D  180
(int.∠s, BC // AD)
k  h  180
k  180  30
5.
 150
82
PQ  [2(4)  2] cm
 6 cm
QR  2(4) cm
 8 cm
SR  (4  2) cm
 6 cm
PS  [3(4)  4] cm
 8 cm
∵ PQ  SR and QR PS
∴ PQRS is a parallelogram.
opp. sides equal
(a) ∵ ABCD is a parallelogram.
∴ BAD  BCD  80 and
(opp. s of // gram)
ABC  ADC
5 Quadrilaterals
In △CDE,
BED  CDE  DCE
 30  80
 110
In △ABF,
BFD  ABF  BAF
 30  80
 110
5.
(ext.  of △)
 FDE
BFD  BED and
FBE  FDE
∴ BEDF is a parallelogram.
 (9  6) cm
 15 cm
i.e. AD  15 cm
6.
proved
opp. s equal
Consolidation Corner (p. 5.36)
2.
∵
∴
∴
ABCD is a rhombus.
ABD  ADB  CBD  35
(property of rhombus)
In △ABD,
BAD  ABD  ADB  180 ( sum of △)
BAD  110
BDE  ADE  ADB
 15  35
 50
∵ EB  ED
∴ DBE  BDE  50 (base s, isos. △)
In △BDE,
BED  DBE  BDE  180 ( sum of △)
BED  50  50  180
BED  100  180
BED  80
(property of rhombus)
In △PQR,
y  180  x  z
 180  54  54
 72
∵
∴
BAD  35  35  180
BAD  70  180
PQRS is a rhombus.
(property of rhombus)
PQ  PS
w 1  4
w3
x  z  54
(Pyth. theorem)
PC  10 2  8 2 cm
 6 cm
BC  BP  PC
(ext.  of △)
∵
∵
∴
ABCD is a rectangle.
DC  AB  8 cm , AD  BC and BCD  90
(property of rectangle)
In △CDP,
PC 2  CD 2  DP 2
(b) FBE  ABE  ABF
 FDC  CDE
1.
∵
∴
( sum of △)
PQRS is a rectangle.
KQ  KR
(property of rectangle)
(base s, isos.△)
a  KQR
 28
In △KQR,
b  a  KQR
 28  28
 56
(ext. of △)
In △PQR,
PQR  90
c  180  PQR  a
 180  90  28
 62
3.
4.
∵
∴
∵
∴
Consolidation Corner (p. 5.43)
1. ∵ AB  BC
∴ CAB  ACB
∵ ADEF is a parallelogram.
∴ DAF  DEF
∴ ACB  DEF
(property of rectangle)
( sum of △)
2.
(property of square)
c  90
(property of square)
PQ  PS
d 8
(property of square)
opp. s of // gram
property of rectangle
given
 DF  OD
 FO
PQRS is a square.
a  45
(property of square)
b  90
In △EBO and △FCO,
∵ OA  OD
and EA  DF
∴ EO  EA  OA
given
base s, isos. △
OB  OC
EOB  FOC
∴ △EBO  △FCO
3.
ABCD is an isosceles trapezium.
a  65
b3
83
In △BEF and △CED,
BFE  CDE
EBF  ECD
BE  CE
∴ △BEF  △CED
∴ FE  DE
∵ BE  CE and FE  DE
∴ BFCD is a parallelogram.
∴ BD  FC
property of rectangle
vert. opp. s
SAS
alt. s, FA // CD
alt. s, FA // CD
given
AAS
corr. sides, △s
diags. bisect each other
opp. sides of // gram
Mathematics in Action (3rd Edition) 3A Full Solutions
In △PST,
∵ PV  VS and VU // ST
∴ PU  UT
(intercept theorem)
∵ PV  VS and PU  UT
1
∴ VU   ST
(mid-pt. theorem)
2
1
q   10
2
5
Consolidation Corner (p. 5.52)
1. (a) In △ACD,
∵ AB  BC and AE  ED
1
∴ BE   CD
(mid-pt. theorem)
2
1
x   8.4
2
 4.2
∵
∴
(mid-pt. theorem)
(corr. s, BE // CD)
BE // CD
y  35
(b) In △ACD,
∵ ABE  ACD  55
∴ BE // CD
(corr. s equal)
∵ AB  BC and BE // CD
(intercept theorem)
∴ AE  ED
pq
∵ AE  ED  AD
p  q  12
∴
p  p  12
p6
(b) In △ABD,
∵ AF  FD and BC  CD
1
∴ FC   AB
(mid-pt. theorem)
2
1
4 x
2
x 8
In △BDE,
∵ BF  FE and BC  CD
1
∴ FC   ED
(mid-pt. theorem)
2
1
4 y
2
y 8
2.
3.
In △ABC,
∵ D, E and F are the mid-points of
AB, BC and CA respectively.
1
1
∴ FE   AB , DF   BC
2
2
1
and DE   AC
mid-pt. theorem
2
∴ AD  EF and AF  ED
common side
DF  FD
∴ △ADF  △EFD
SSS
Consolidation Corner (p. 5.60)
1. ∵ L1 // L2 // L3 and BD  DF
∴
CE  AC
(intercept theorem)
 3 cm
2.
∵
AC  CE and L1 // L2
∴
GE  BG
(intercept theorem)
7
 cm
2
 3.5 cm
q6
∴
In △ACF,
∵ AB  BC and BG // CF
∴ AG  GF
∵ AB  BC and AG  GF
1
∴ BG   CF
2
In △BDG,
∵ BC  CD and BG // CE
∴ GE  ED
∵ BC  CD and GE  ED
1
∴ CE   BG
2
1 1
   CF
2 2
1
 CF
4
∴ CF  4CE
EF  CF  CE
intercept theorem
mid-pt. theorem
intercept theorem
mid-pt. theorem
 4CE  CE
 3CE
Exercise
Exercise 5A (p. 5.12)
Level 1
1. ∵ ABCD is a parallelogram.
∴ AB  DC
(opp. sides of // gram)
x4
(a) In △PQS,
∵ QR  RS and RV // QP
∴ PV  VS
(intercept theorem)
∵ QR  RS and PV  VS
1
∴ RV   QP
(mid-pt. theorem)
2
1
p  8
2
4
AD  BC
(opp. sides of // gram)
y3
2.
84
∵
∴
ABCD is a parallelogram.
(diags. of // gram)
OC  OA
x2
∴
OD  OB  y cm (diags. of // gram)
5 Quadrilaterals
BD  3 cm
∴
AD  BC
5 z  3z  4
2z  4
z2
∵
∴
ABCD is a parallelogram.
(diags. of // gram)
OB  OD
y y 3
y  1 .5
3.
∵
∴
ABCD is a parallelogram.
(opp. s of // gram)
B  D
9.
x  30
∵
∴
4.
5.
ABCD is a parallelogram.
(opp. sides of // gram)
AB  DC
x4
∴
(opp. sides of // gram)
AD  BC
y  2x 1
 2(4)  1
9
∵
∴
∴
2( x  1)  3( x  1)
C  D  180 (int. s, BC // AD)
(3 y  30)  30  180
3 y  120
y  40
∵
∴
2 x  2  3x  3
x5
∴
(opp. s of // gram)
BCD  BAD
y  2 y  108
3 y  108
y  36
10. (a) ∵ ABCD is a parallelogram.
BAD  BCD
∴
(opp. s of // gram)
58  4 x  7 x  10
48  3x
x  16
ABCD is a parallelogram.
(opp. sides of // gram)
OB  OD
4 x  2  x  14
3 x  12
x4
OA  OC
(opp. sides of // gram)
(b) DAE  4x
 4 16
∵
∴
(opp. sides of // gram)
2 y  3y  2
 64
DAE  AEC  180 (int. s, AD // EC)
AEC  180  64
 116
y2
6.
∵
∴
∵
∴
7.
∵
∴
11. ∵
∴
ABCD is a parallelogram.
x  ADC (opp. s of // gram)
 40
∵
∴
ABC  BAD  180 (int. s, AD // BC)
x  [(2 x  30)  y ]  180
y  180  30  3x
 150  3(40)
 30
12. (a) ∵ AB  AC
∴ ABC  ACB (base s, isos. △)
In △ABC,
BAC  ABC  ACB  180 ( sum of △)
ABCD is a parallelogram.
AB  DC (opp. sides of // gram)
6  x  2x  3
70  2ABC  180
3x  3
x 1
∴
2ABC  110
ABC  55
OB  OD (opp. s of // gram)
13  y  10
(b) ∵ ABCD is a parallelogram.
∴ ADC  ABC (opp. s of // gram)
y3
8.
∵
∴
WXYZ is a parallelogram.
ZY  WX  8 cm (opp. sides of // gram)
and WZ  XY
(opp. sides of // gram)
Perimeter of parallelogram WXYZ  54 cm
WX  XY  YZ  ZW  54 cm
8 cm  XY  8 cm  XY  54 cm
2 XY  38 cm
XY  19 cm
 55
ABCD is a parallelogram.
(opp. s of // gram)
A  C
13. ∵
∴
60  x  5 x  6
66  6 x
x  11
∵
∴
85
ABCD is a parallelogram.
EBC  AEB (alt. s, BC // AD)
 41
EB bisects ABC.
ABE  EBC  41
Mathematics in Action (3rd Edition) 3A Full Solutions
ABC  ABE  EBC
 41  41
 82
ADC  ABC
 82
18. ∵
∴
5x  6  2 x  3
3x  9
(opp. s of // gram)
x3
∴
Level 2
14. ∵ ABCD and BEFC are parallelograms.
∴ BCD  BAD
(opp. s of // gram)
 124
and BCF  BEF
(opp. s of // gram)
∵
∴
32  CDE  72
CDE  40
∵ CED  CDE
∴ △CDE is not an isosceles triangle.
 7 cm
16. ∵
∴
(opp. sides of // gram)
BCEF is a parallelogram.
CBF  CEF
 56
∵ BC  BG
∴ BCG  BGC
In △BCG,
CBG  BCG  BGC  180
56  2BCG  180
2BCG  124
BCG  62
∵ ABCD is a parallelogram.
∴ BAD  BCD
 62
(diags. of // gram)
19. CED  ADE
(alt. s, AD // BC)
 32
∵ ABCD is a parallelogram.
∴ ADC  ABC (opp. s of // gram)
 72
ADE  CDE  ADC
BEFG is a parallelogram.
(diags. of // gram)
AB  AF  7 cm
and AE  AG  6 cm (diags. of // gram)
ABCD is a parallelogram.
(opp. sides of // gram)
CD  BA
AD  BC
 9 cm
DE  AD  AE
 (9  6) cm
 3 cm
∴
OA  OC
x  3 y  3 x  18
3 y  2 x  18
2
y  x6
3
2
 (3)  6
3
8
 146
BCD  BCF  DCF  360 (s at a pt.)
124  146  DCF  360
270  DCF  360
DCF  90
∴ DC is perpendicular to CF.
15. ∵
∴
ABCD is a parallelogram.
AB  DC
(opp. sides of // gram)
20. ∵
∴
PXYS is a parallelogram.
(opp. s of // gram)
SPX  SYX
 40
In △PZS,
PSZ  SPZ  PZR
(ext.  of △)
PSZ  40  142
(opp. s of // gram)
∵
∴
(base s, isos. △)
PSZ  102
PQRS is a parallelogram.
(opp. s of // gram)
PQR  PSZ
 102
( sum of △)
21. ∵
∴
ABCD is a parallelogram.
BCD  BAD
 106
DCE  180  BCD
 180  106
 74
(opp. s of // gram)
∵
∴
17. ABF  FED
(alt. s, AB // EC)
 48
∵ BE bisects ABC.
∴ ABC  2ABF
 2  48
CEFG is a parallelogram.
EFG  DCE
 74
DGF  DCE
 74
(opp. s of // gram)
(adj. s on st. line)
(opp. s of // gram)
(corr. s, GF // CE)
FD  FG
(base. s, isos. △)
GDF  DGF
 74
In △DGF,
DFG  GDF  DGF  180 ( sum of △)
∵
∴
 96
∵ ABCD is a parallelogram.
∴ ADC  ABC (opp. s of // gram)
 96
ADE  ADC  180 (adj. s on st. line)
ADE  96  180
ADE  84
DFG  74  74  180
DFG  32
86
5 Quadrilaterals
DFE  DFG  EFG
∴ Perimeter of ACED  2( AC  AD)
 32  74
 2(12  10) cm
 106
 44 cm
22. (a) ∵ ABCD is a parallelogram.
∴ BCD  BAD (opp. s of // gram)
 60
(alt. s, BC // AD)
DEC  ADE
24. In △ABD,
AB 2  BD 2  AD 2
BD  AD  AB 2
 60
In △CDE,
CDE  DEC  ECD  180 ( sum of △)
∵
∴
∵
∴
CDE  60  60  180
CDE  60
∵ CDE  DEC  ECD  60
∴ CD  DE  CE
∴ △CDE is an equilateral triangle.
AE 2  AB 2  BE 2
AE  AB  BE
2
(Pyth. theorem)
(Pyth. theorem)
2
 193 cm
AC  AE  CE
 ( 193  193 ) cm
 2 193 cm
Exercise 5B (p. 5.21)
Level 1
1. ∵ OA  OC  3 cm and
OB  OD  4 cm
∴ ABCD is a parallelogram.
23. (a) ∵ ABCD is a parallelogram.
∴ GD  BG
(diags. of // gram)
 8 cm
and GC  AG
(diags. of // gram)
 6 cm
In △AGD,
(Pyth. theorem)
AD 2  AG 2  GD 2
AB 2  BG 2  AG 2
(diags. of // gram)
 7 2  12 2 cm
 (5  8) cm
 13 cm
Perimeter of the parallelogram  2( AB  BC )
 2(8  13) cm
 42 cm
(opp. sides of // gram)
 252  7 2 cm
 24 cm
ABCD is a parallelogram.
BE  DE and AE  CE
BD  24 cm
BE  DE  24 cm
2 BE  24 cm
BE  12 cm
In △AEB,
(b) ∵ ABCD is a parallelogram.
∴ AD = BC and AB = DC (opp. sides of // gram)
∵ △CDE is an equilateral triangle.
∴ EC  DC  DE  8 cm
BC  BE  EC
AD  6 2  8 2 cm
 10 cm
BC  AD
 10 cm
In △AGB,
(Pyth. theorem)
2
2.
3.
4.
AB  8 2  6 2 cm
 10 cm
DC  AB
(opp. sides of // gram)
 10 cm
∴ Perimeter of ABCD  2( AB  BC )
 2(10  10) cm
 40 cm
(b) ∵ ACED is a parallelogram.
∴ AC  DE
(opp. sides of // gram)
and AD  CE
(opp. sides of // gram)
AC  AG  GC
 (6  6) cm
 12 cm
87
given
diags. bisect each other
AB  DC  2.3 cm and
BA// CD
∴ ABCD is a parallelogram.
given
opp. sides equal and //
AB  DC and
AD  BC  3 cm
∴ ABCD is a parallelogram.
given
opp. sides equal
∵
∵
A  B  C  D  360
A  60  120  60  360
A  240  360
A  120
∵ A  C  120 and
B  D  60
∴ ABCD is a parallelogram.
ADB  CBD
AD// BC
AD  BC and AD// BC
ABCD is a parallelogram.
5.
∵
∴
∵
∴
6.
(a) ∵
∴
AD  BC
5a  2  3a  4
2a  6
a3
 sum of polygon
opp. s equal
given
alt. s equal
given
opp. sides equal and //
Mathematics in Action (3rd Edition) 3A Full Solutions
(b)
7.
AB  (3  1) cm  4 cm
DC  (7  3) cm  4 cm
∵ AD  BC and AB  DC
∴ ABCD is a parallelogram.
Method 2:
∵ △AOD  △COB
∴ OA  OC and OD  OB
∴ ABCD is a parallelogram.
opp. sides equal
Level 2
12. (a) In ABCD,
A  B  C  D  360
(a) In ABCD,
143  (3x  11)  143  (2 x  5)  360 ( sum of
280  5 x  360 polygon)
(5a  5)  (2a  10)  120 
5 x  80
x  16
(b) ABC  3(16)  11  37
ADC  2(16)  5  37
∵ BAD  BCD and
ABC  ADC
∴ ABCD is a parallelogram.
8.
(a) In △AGD,
GAD  ADG  BGD
CF  EF
ECF  CEF
 50
(3a  15)  360
10a  250
a  25
(b) A  5( 25)  5  120
B  2( 25)  10  60
D  3( 25)  15  60
∴ A  C and B  D
∴ ABCD is a parallelogram.
opp. s equal
(ext.  of △)
(base s, isos. △)
13. (a) ∵ ABCD is a parallelogram.
∴ BC  AD
opp. sides of // gram
∵ PADQ is a parallelogram.
∴ AD  PQ
opp. sides of // gram
∴ BC  PQ
BAD  ABC  50  120
 170
 180
∴ AD is not parallel to BC.
∴ ABCD is not a parallelogram.
P  S  180
S  180  P
Q  R  180
(b) ∵ ABCD is a parallelogram.
∴ BC // AD
∵ PADQ is a parallelogram.
∴ AD // PQ
∴ BC // PQ
Also, BC  PQ
proved in (a)
∴ PBCQ is a parallelogram. opp. sides equal and //
int. s, QP // RS
int. s, QP // RS
Q  180  R
∵
∴
∴
10. ∵
∴
∵
∴
∵
∴
∴
P  R
given
S  Q
PQRS is a parallelogram.
opp. s equal
14. ∵
ABCD is a parallelogram.
(opp. sides of // gram)
BC  AD
(given)
FC  AE
BF  BC  FC
 AD  AE
 ED
BC // AD
BF // ED
BFDE is a parallelogram.
11. Method 1:
∵ △AOD  △COB
∴ AD  CB
and OAD  OCB
∴ AD // BC
∴ ABCD is a parallelogram.
opp. s equal
(c) ∵ ABCD is a parallelogram.
∴
(opp. sides of // gram)
AD  BC
18  3b  b  2
4b  20
b5
(b) ∵
9.
( sum of
polygon)
10a  110  360
GAD  15  65
GAD  50
i.e. BAD  50
∵
∴
given
corr. sides, △s
diags. bisect each other
∴
∵
∴
∴
15. ∵
∴
(opp. sides equal and //)
∵
given
corr. sides, △s
corr. s, △s
alt. s equal
opp. sides equal and //
∴
∵
∴
88
AB  FC
EF  FC
AB  EF
AE  FD
BF  FD
AE  BF
ABFE is a parallelogram.
PQRS is a parallelogram.
PO  OR
and QO  OS
X is the mid-point of PO.
1
XO  PO
2
Y is the mid-point of OR.
1
OY  OR
2
1
 PO
2
 XO
given
diags. of // gram
given
diags. of // gram
opp. sides equal
diags. of // gram
diags. of // gram
5 Quadrilaterals
∵
∴
XO  OY and QO  OS
XQYS is a parallelogram.
PQR  MNO
 66
diags. bisect each other
16. (a) ∵ △ABC  △CDE
∴ BC  DE
and ACB  CED
∵ BC is the angle bisector
of ACE.
∴ ACB  BCE
∴ BCE  CED
∴ CB // DE
∴ BCDE is a parallelogram.
given
corr. sides, △s
corr. s, △s
19. (a) ∵ ABCD is a parallelogram.
∴ AD  BC
∵ △ABE and △CDF are
equilateral triangles.
∴ AB  AE  BE and
DC  DF  FC
∵ AB  DC
∴ DF  BE
∵ ADC  CBA
FDC  ABE  60
∴ ADF  ADC  60
alt. s equal
opp. sides equal
and //
(b) ∵ △ABC  △CDE
∴ ABC  CDE (corr. s, △s)
 54
∵ BCDE is a parallelogram.
∴ CBE  CDE (opp. s of // gram)
 54
ABE  ABC  CBE
 54  54
 108
17. (a) In △ABO and △CDO,
AOB  COD
OAB  OCD  90
BO  DO
∴ △ABO △CDO
 CBA  60
 CBE
∴ △AFD  △CEB
(b) ∵ △AFD  △CEB
∴ AF  CE
AE  FC
∴ AECF is a parallelogram.
Exercise 5C (p. 5.37)
Level 1
1. ∵ ABCD is a rhombus.
∴ AB  BC  CD  DA
∴ x4
vert. opp. s
given
given
AAS
(b) ∵ △ABO △CDO
∴ AO  CO
BO  DO
∴ ABCD is a parallelogram.
(proved in (a))
opp. sides of // gram
opp. sides of // gram
opp. s of // gram
prop. of equil.△
SAS
proved in (a)
corr. sides, △s
from (a)
opp. sides equal
(property of rhombus)
2y  2  4
2y  2
proved in (a)
corr. sides, △s
given
diags. bisect each other
y 1
3z  5  4
3z  9
z3
18. (a)
2.
ABCD is a rhombus.
DAB  2  BAC
(property of rhombus)
 2  30
 60
S
With the notation in the figure,
∵ △MNO  △PQR
∴ MN  PQ
and MNO  PQR
PSO  PQR
∴ MNO  PSO
∴ MN // PQ
∴ MNQP is a parallelogram.
∵
∴
p  DAB
(alt. s, AD // EC)
 60
given
corr. sides, △s
corr. s, △s
corr. s, NO // QR
(int. s, AB // DC)
DAB  q  180
60  q  180
q  120
3.
corr. s equal
opp. sides equal
and //
∵
∴
ABCD is a rhombus.
a  55
(property of rhombus)
b  90
(property of rhombus)
In △DOC,
OCD  CDO  AOD
(b) ∵ MNQP is a parallelogram.
∴ MNQ  MPQ (opp. s of // gram)
 112
MNO  MNQ  ONQ
∵
∴
 112  46
 66
89
(ext.  of △)
CDO  90  55
 35
ADO  CDO (property of rhombus)
ADC  ADO  CDO
c  35  35
 70
Mathematics in Action (3rd Edition) 3A Full Solutions
4.
∵
∴
ABCD is a rectangle.
(property of rectangle)
OD  OB
2x  2
x 1
∵
∴
AC  BD
(property of rectangle)
y  2  2x
 2  2(1)
4
UQR  PQR  PQT
 90  35
 55
In △UQR,
x  UQR  PRQ
 55  45
 100
y  TQR  180
(ext.  of △)
(int. s, PS // QR)
y  55  180
5.
∵
∴
y  125
ABCD is a rectangle.
AD  BC (property of rectangle)
2x  1  7
2x  6
10. BAD  ABC  180 (int. s, AD // BC)
2 x  62  180
x3
2 x  118
x  59
∴ BAD  90 (property of rectangle)
In △ABD,
ABD  ADB  BAD  180 ( sum of △)
∵
∴
64  y  90  180
3 y  10  62
3 y  72
y  24
154  y  180
y  26
6.
∵
∴
∵
∴
ABCD is a rectangle.
(property of rectangle)
ABC  90
x  70  90
x  20
OA  OB
OAB  OBA
11. ∵
∴
∵
∴
∴
(property of rectangle)
(base s, isos.△)
ADB  DBC
y  30
12. ∵
∴
EFGH is a rhombus of side 10 cm.
EH  10 cm and EOH  90
(property of rhombus)
∵ EO  OG
(property of rhombus)
1
∴ EO  EG
2
1
  12 cm
2
 6 cm
In △EOH,
(property of square)
2b  7  13
2b  6
b3
8.
∵
∴
EO 2  OH 2  EH 2
PQRS is a square.
(property of square)
POS  90
2 y  30  90
2 y  120
∵
∴
y  60
OR  OQ
(property of square)
z  1  11
(Pyth. theorem)
OH  10 2  6 2 cm
 8 cm
FO  OH
(property of rhombus)
FH  2  OH
 2  8 cm
 16 cm
13. (a) ∵ PQRS is a rhombus.
∴ PQS  RQS (property of rhombus)
3 x  4  4 x  9
x  13
z  12
9.
(alt. s, AD // BC)
(ext.  of △)
PQRS is a square.
a  45
(property of square)
SR  PS
ABCD is an isosceles trapezium.
BD  AC
2  (2 x  1)  7
2x  3  7
2x  4
x2
y  70
In △OAB,
z  y  70
 70  70
 140
7.
ABCD is an isosceles trapezium.
DCB  ABC
∵ PQRS is a square.
∴ QRP  45
(property of square)
and PQR  90
(property of square)
(b) ∵ PQRS is a rhombus.
∴ PRQ  PRS and PR  QS
(property of rhombus)
90
5 Quadrilaterals
RQT  4  13  9  43
In △QRT,
RQT  QRT  QTR  180 ( sum of △)
43  QRT  90  180
In △ABC,
AC 2  AB 2  BC 2
URS  2  QRT
 2  47
 94
 90
∴ △SUR is an obtuse-angled triangle.
∴ Peter’s claim is incorrect.
∵
ABCD is a rectangle.
OB  OA and ABC  90
(property of rectangle)
(base s, isos. △)
OBA  OAB
 34
In △OAB,
BOC  OAB  OBA
m  34  34
 68
In △ABC,
ACE  BAC  ABC
n  34  90
 124

(ext.  of △)
1
72

cm
2
2

72
cm
4

 3 2

18
cm  or
cm 


2
2


17. ∵ PQRS is a square.
∴ TPK  45
(property of square)
∵ TP  KP
∴ PTK  PKT (base s, isos. △)
In △PTK,
PTK  PKT  TPK  180
( sum of △)
2PKT  45  180
2PKT  135
PKT  67.5
∵ PKQ  90
(property of square)
∴ TKQ  PKQ  PKT
 90  67.5
 22.5
(ext.  of △)
15. (a) ∵ ABCD is a rectangle.
∴ OA  OB  OC  OD and BCD  90
(property of rectangle)
∴ OBC  OCB
(base s, isos. △)
In △OBC,
BOC  OBC  OCB  180 ( sum of △)
60  2OBC  180
2OBC  120
OBC  60
∵ OBC  OCB  BOC  60
∴ △OBC is an equilateral triangle.
i.e. OB  BC  5 cm
BD  OB  OD
 (5  5) cm
 10 cm
18.
Construct ST such that ST  QR , where T is a point on
QR.
QT  PS  3 cm
TR  QR  QT
 (8  3) cm
 5 cm
ST  PQ  12 cm
In △RST,
(b) In △BCD,
BC 2  CD 2  BD 2
72
cm
2
OE  EC
1
OE  OC
2

∴
∴
2
 72 cm
1
OC  AC
2
1
  72 cm
2
133  QRT  180
QRT  47
14. ∵
∴
(Pyth. theorem)
AC  6  6 cm
2
(Pyth. theorem)
CD  10  5 2 cm
2
 75 cm (or 5 3 cm)
16. ∵
∴
ABCD is a square.
OA  OB  OC  OD and ABC  90
(property of square)
AB  BC
24

cm
4
 6 cm
RS 2  ST 2  RT 2 (Pyth. theorem)
RS  12 2  52 cm
 13 cm
Perimeter of trapezium PQRS  PQ  QR  RS  PS
 (12  8  13  3) cm
 36 cm
91
Mathematics in Action (3rd Edition) 3A Full Solutions
Level 2
19. ∵ PQRS is a rhombus.
∴ PRS  PRQ
(property of rhombus)
SRQ  PQR  180 (int. s, PQ // SR)
22. ∵
∴
∵
∴
∴
SRQ  76  180
SRQ  104
1
PRS  SRQ
2
1
 104
2
 52
∵ TR is the angle bisector of PRS.
1
∴ PRT  PRS
2
1
  52
2
 26
SPR  PRS  52 (property of rhombus)
In △TPR,
PTR  TPR  PRT  180 ( sum of △)
ABCD is a square.
BCD  ADC  90
BCE  ADE
(property of square)
 90  60
 30
Similarly, ADE  30
∵ △CDE is an equilateral triangle.
∴ CE  DE  CD
AB  BC  CD  DA (property of square)
∴ CE  BC and DE  DA
∴ CEB  CBE (base s, isos.△)
and DEA  DAE (base s, isos.△)
In △BCE,
CEB  CBE  BCE  180
( sum of △)
2CEB  30  180
2CEB  150
CEB  75
PTR  52  26  180
Similarly, DEA  75
AEB  360  CEB  DEA  CED (s at a pt.)
 360  75  75  60
 150
PTR  78  180
PTR  102
20. ABC  ADC
(opp. s of // gram)
 132
ABK  ABC  KBC
23. (a) ∵ DBCE is a parallelogram.
∴ BD  CE
(opp. sides of // gram)
 8 cm
∵ ABCD is a rectangle.
∴ AC  BD
(property of rectangle)
 8 cm
 132  54
 78
∵ ABCD is a rhombus.
∴ DAC  BAC (property of rhombus)
(int. s, AB // DC)
ADC  (DAC  BAC )  180
132  2BAC  180
2BAC  48
BAC  24
AC  CE
CAE  CED (base s, isos. △)
 60
In △ECA,
ECA  CAE  CED  180 ( sum of △)
ECA  60  60  180
(b) ∵
∴
In △ABK,
BAK  ABK  AKB  180 ( sum of △)
24  78  AKB  180
102  AKB  180
AKB  78
∵ ABK  AKB
∴ AB  AK
(sides opp. equal s)
i.e. △ABK is an isosceles triangle.
21. ∵
∴
∵
∴
∴
△CDE is an equilateral triangle.
CED  ECD  EDC  60 (prop. of equil.△)
ECA  60
∴ △ECA is an equilateral triangle.
i.e. AE  CE  8 cm
∵ ABCD is a rectangle.
∴ ADC  90
(property of rectangle)
∵ AC  CE and CD  AE
∴ AD  DE
(prop. of isos. △)
1
  AE
2
1
  8 cm
2
 4 cm
PQRS is a square.
PRQ  45
(property of square)
UVRP is a rhombus.
UVP  PVR (property of rhombus)
(UVP  PVR)  PRQ  180 (int.s, UV // PR)
2PVR  45  180
2PVR  135
PVR  67.5
24.
Construct a line FG such that DC // FG.
92
5 Quadrilaterals
∵ BA // CD // GF and AF // BG
∴ ABGF is a parallelogram.
(int. s, AF // BE)
ABC  120  180
ABC  60
FGE  ABC  60 (corr. s, AB // FG)
∵ CEFD is an isosceles trapezium.
∴ FEG  60
and FE  DC
Exercise 5D (p. 5.43)
Level 1
1. ∵ ABCD is a parallelogram.
∴ ABC  ADC
∵ DEFG is a parallelogram.
∴ EFG  EDG
∴ ABC  EFG
2.
(property of rhombus)
 AB
 15 cm
In △EFG,
EFG  FGE  FEG  180 ( sum of △)
EFG  60  60  180
EFG  120  180
EFG  60
∴ △EFG is an equilateral triangle.
i.e. GE  FE  15 cm
∵ DF // CG and DC // FG
∴ CGFD is a parallelogram.
∴ DF  CG
(opp. sides of // gram)
 CE  GE
 (23  15) cm
 8 cm
ABCD is a parallelogram.
BC  AD
AFCE is a square.
AF  FC  CE  AE
 BC  FC
opp. s of // gram
opp. sides of // gram
property of square
 AD  AE
 ED
3.
BC  BE  EF  CF
AD  BC and AB  DC
∵ CDF  CFD
∴ CD  CF
∴ AB  BC  CD  DA
∴ ABCD is a rhombus.
4.
In △EAF and △EDG,
∵ ABCD is a rectangle.
∴ AB  DC
1
∵ AF  AB and
2
1
DG  DC
2
∴ AF  DG
EAF  EDG  90
EA  ED
∴ △EAF  △EDG
∴ EF  EG
i.e. △EFG is an isosceles triangle.
25. (a) ∵ ABCD is a square.
∴ AB  BC , ABC  90 and BAC  45
(property of square)
∵ BCE is an equilateral triangle.
∴ BC  CE  BE and EBC  60
(prop. equil. △)
∵ BA  BE
∴ BAE  BEA (base s, isos. △)
ABE  ABC  EBC
 90  60
 30
In △ABE,
ABE  BEA  BAE  180
30  2BAE  180
2BAE  150
BAE  75
CAE  BAE  BAC
∵
∴
∵
∴
BF
opp. s of // gram
5.
( sum of △)
 75  45
 30
∵ ACFE is a parallelogram.
∴ CFE  CAE (opp. s of // gram)
 30
6.
(b) CAE  AEF  180 (int. s, AC // EF)
30  AEF  180
AEF  150
BEF  AEF  BEA
7.
 150  75
 75
∵ BEA  BEF  75
∴ BE bisects AEF.
93
In △AED and △CFD,
DAE  DCF
DEC  DFA
AED  180  DEC
CFD  180  DFA
∴ AED  CFD
AD  CD
∴ △AED  △CFD
∴ AE  FC
∵ AB  BC
∴ BAC  BCA
DFA  BCA
∴ DAF  DFA
∴ AD  DF
(a)
property of square
opp. sides of // gram
given
sides opp. equal s
property of rectangle
given
given
property of rectangle
given
SAS
corr. sides, △s
property of rhombus
given
adj. s on st. line
adj. s on st. line
property of rhombus
AAS
corr. sides, △s
given
base s, isos. △
corr. s, DF // BC
sides opp. equal s
ADC  90
DCE  ADC
ACD  CDE
ACE  ACD  DCE
 CDE  ADC
 ADE
property of rectangle
alt. s, AD // BE
alt. s, AC // DE
Mathematics in Action (3rd Edition) 3A Full Solutions
(b)
8.
property of rectangle
BC  AD
∵ AC // DE and AD // BE
∴ ACED is a parallelogram.
∴ AD  CE
opp. sides of // gram
∴ BC  CE
i.e. C is the mid-point of BE.
∵ ABCD is a parallelogram.
∴ BCD  BAD
∵ BQCP is a rhombus.
∴ CBP  BCP
In △BCP,
BPD  CBP  BCP
In △CFE,
CFE  CEF  ECF  180
45  CEF  45  180
90  CEF  180
CEF  90
∴ FE  EC
Level 2
11. (a) ∵
∴
opp. s of // gram
property of rhombus
∵
∴
ext.  of △
 2BCP
 2BAD
9.
∵ ABCD is a square.
∴ AB  BC  CD  AD
Also, BP  BQ
In △APD and △CQD,
AP  AB  BP
∴
∴
property of square
given
 CQ
DAP  DCQ  90
AD  CD
∴ △APD  △CQD
∴ DP  DQ
∴ △PDQ is an isosceles triangle.
property of square
proved
SAS
corr. sides, △s
10. ∵
∴
∵
∴
DP  DQ
△PDQ is an isosceles triangle.
given
base s, isos.△
property of rectangle
 EDA  ADC
 EDC
In △ABE and △DCE,
AB  DC
EAB  EDC
EA  ED
∴ △ABE  △DCE
∴ BE  CE
∴ △EBC is an isosceles triangle.
13. (a)
∴
∴
opp. sides of // gram
opp. sides of // gram
opp. sides of // gram
SSS
12. ∵ EA  ED
∴ EAD  EDA
BAD  ADC  90
∴ EAB  EAD  BAD
Alternative Solution
Join BD.
In △BPD and △BQD,
BP  BQ
∵ ABCD is a square.
∴ PBD  QBD  45
BD  BD
∴ △ BPD  △ BQD
ABCD is a parallelogram.
(opp. sides of // gram)
AD  BC
and AD // BC
AEFD is a parallelogram.
(opp. sides of // gram)
AD  EF
and AD // EF
BC  EF and BC // EF
BCFE is a parallelogram.
(opp. sides equal and //)
(b) In △ABE and △DCF,
AB  DC
AE  DF
BE  CF
∴ △ABE  △DCF
 BC  BQ
 sum of △
given
property of square
common side
SAS
corr. sides, △s
ABCD is a rectangle.
BCD  90
property of
rectangle
(b)
CE and CF are the angle bisectors of
ACB and ACD respectively.
ACE  BCE and
ACF  DCF
BCD  90
BAC  ACD
∵ AX and CY bisect
BAE and DCE
respectively.
∴ BAX  EAX
and DCY  ECY
∴ EAX  ECY
In △AXE and △CYE,
EAX  ECY
AE  CE
AEX  CEY
∴ △AXE  △CYE
alt. s, BA // CD
BE  DE
∵ △AXE  △CYE
∴ XE  YE
BX  BE  XE
diags. of // gram
proved in (a)
corr. sides, △s
 DE  YE
 DY
ACE  BCE  ACF  DCF  90
2(ACE  ACF )  90
ACE  ACF  45
ECF  45
94
property of rectangle
proved
given
SAS
corr. sides, △s
proved
diags. of // gram
vert. opp. s
ASA
5 Quadrilaterals
14. (a) ∵ ABCD is a square.
∴ AB  CB
and ABC  90
ABF  ABC  180
ABF  90  180
ABF  90
In △ABF and △CBE,
AB  CB
ABF  CBE  90
BF  BE
∴ △ABF  △CBE
(b) ∵ △ABF  △CBE
∴ BAF  BCE
In △AGE,
AGE  GAE  AEC
In △BCE,
CBE  BCE  AEC
△BAX  △DCY (proved in (a)(i))
(corr. sides, △s)
BX  DY
and
BX
// YD
BX  DY
BYDX is a parallelogram.
(opp. sides equal and //)
In △BAX and △DAX,
(common side)
AX  AX
DAX  DAC  180 (adj. s on st. line)
(b) ∵
∴
∵
∴
property of square
adj. s on st. line
proved
proved
given
SAS
DAX  45  180
DAX  135
(proved)
BAX  DAX
(property of square)
AB  AD
∴ △BAX  △DAX (SAS)
∴ BX  DX
(corr. sides, △s)
Also, BX  DY
(opp. sides of // gram)
and BY  DX
(opp. sides of // gram)
∴ BX  DY  BY  DX
∴ BYDX is a parallelogram with four equal sides.
∴ BYDX is a rhombus.
(proved in (a))
(corr. s, △s)
(ext.  of △)
(ext.  of △)
90  BCE  AGE  GAE
AGE  90
∴ CG  AF
15. (a) ∵ ABCD is a square.
∴ BAD  90
∵ AEFG is a square.
∴ EAG  90
∵ EAB  EAG  GAB
 90  GAB
GAD  GAB  BAD
 GAB  90
∴ EAB  GAD
(b) In △AEB and △AGD,
AB  AD
AE  AG
EAB  GAD
∴ △AEB  △AGD
∴ BE  DG
16. (a) (i)
17. (a) In △BPC and △DQA,
BPC  DQA
PBC  QDA
BC  DA
∴ △BPC  △DQA
property of square
property of square
(b) ∵ △BPC  △DQA
∴ BP  DQ
and PC  QA
BA  DC
AP  BP  BA
 DQ  DC
 CQ
∴ AQCP is a parallelogram.
property of square
property of square
proved in (a)
SAS
corr. sides, △s
(ii) ∵ △BAX  △DCY
∴ AXB  CYD
∴ BX // YD
proved in (a)
corr. sides, △s
corr. sides, △s
opp. sides of // gram
opp. sides equal
18. (a) ∵ ABCD is a square.
∴ BAD  90 (property of square)
BAO  BAE  OAE
 90  OAE
∵ ABCD is a square.
∴ AB  BC  CD  DA
BAC  BCA 
and
property of square
DAC  DCA  45
In △BAX,
adj. s on st. line
BAX  BAC  180
BAX  45  180
BAX  135
In △DCY,
DCY  DCA  180
DCY  45  180
DCY  135
In △BAX and △DCY,
AX  CY
BAX  DCY
AB  CD
∴ △BAX  △DCY
given
opp. s of // gram
opp. sides of // gram
AAS
In △ABO,
ABO  BAO  AOE
(ext.  of △)
ABO  (90  OAE)  90
OAE  ABO
(b) In △ADF and △BAE,
∵ OAE  OBA
i.e. DAF  ABE
DA  AB
ADF  BAE  90
∴ △ADF  △BAE
∴ AF  BE
adj. s on st. line
given
proved
property of square
SAS
19. ∵ AB  AC
∴ ABC  ACB
DGB  ACB
∴ DBG  DGB
∴ DB  DG
∵ DG  DB  CE and
DG // CE
proved in (a)(i)
corr. s, △s
alt. s equal
95
property of square
property of square
ASA
corr. sides, △s
given
base s, isos. △
corr. s, DG // AE
sides opp. equal s
given
Mathematics in Action (3rd Edition) 3A Full Solutions
∴
∴
∴
CDGE is a parallelogram.
GF  FC
F is the mid-point of CG.
(b) In △FBG,
∵ BD  DF and BE  EG
1
∴ DE   FG
(mid-pt. theorem)
2
1
  6 cm
2
 3 cm
opp. sides equal and //
diags. bisect each other
Exercise 5E (p. 5.52)
Level 1
1. In △ABC,
∵ AD  DB and AE  EC
1
∴ DE   BC
(mid-pt. theorem)
2
1
3 x
2
x6
2.
6.
In △ABC,
∵ AD  DB and AE  EC
∴ DE // BC
(mid-pt. theorem)
∴ ACB  AED (corr. s, DE // BC)
(opp. s of // gram)
ABC  108
In △ACE,
∵ AB  BE and F is the mid-point of CE.
∴ BF // AC
(mid-pt. theorem)
∴ ACB  CBF (alt. s, BF // AC)
 32
In △ACB,
BAC  ACB  ABC  180 ( sum of △)
BAC  32  108  180
BAC  40
x  77
7.
3.
In △ABC,
∵ AD  DB and AE  EC
∴ DE // BC
(mid-pt. theorem)
∴ AED  ACB (corr. s, DE // BC)
x  50
∴
4.
∴
∴
(mid-pt. theorem)
∴
∴
In △ABC,
∵ AD  DB and AE  EC
∴ DE // BC
(mid-pt. theorem)
∴ AED  ACB (corr. s, DE // BC)
 45
In △ADE,
ADE  AED  DAE  180 ( sum of △)
a  45  60  180
a  75
∴
5.
1
DE   BC
2
1
14   y
2
y  28
1
DE   DB
2
1
b   11
2
 5.5
∵
∴
∵
(mid-pt. theorem)
(a) In △ABC,
∵ BF  (3  3) cm  6 cm  FA
and BG  ( 4  4) cm  8 cm  GC
1
∴ FG   AC
(mid-pt. theorem)
2
1
  12 cm
2
 6 cm
96
△ABC is an equilateral triangle.
AB  BC  AC
P, Q and R are the mid-points
of AB, BC and CA respectively.
AP  PB  BQ  QC  AR  RC
1
1
PQ   AC , QR   AB
2
2
1
and PR   BC
mid-pt. theorem
2
PQ  QR  PR
△PQR is an equilateral triangle.
8.
In △ABC,
∵ E and F are the mid-points of AC
and BC respectively.
∴ AB // EF
mid-pt. theorem
∵ E and F are the mid-points of BD
and BC respectively.
∴ EF // DC
mid-pt. theorem
∵ AB // EF and EF // DC
∴ AB // DC
9.
In △ABC,
∵ M and N are the mid-points of AC and BC
respectively.
1
∴ AB // MN and MN   AB (mid-pt. theorem)
2
1
MN   12 cm
2
 6 cm
∵ AB // CD and AB // MN
∴ MN // CD
∵ MN  CD  6 cm and MN // CD
∴ MNDC is a parallelogram. (opp. sides equal and //)
∴ Cathy’s claim is correct.
5 Quadrilaterals
(b) In △AFG,
∵ E and C are the mid-points of AF and AG
respectively.
∴ EC // FG
(mid-pt. theorem)
(alt. s, EC // FG)
EFG  BEF
Level 2
10. In △DEF,
∵ FG  GD and FC  CE
1
∴ GC   DE
(mid-pt. theorem)
2
1
a  8
2
4
 62
FAG  AFG  AGF  180 ( sum of △)
52  62  AGF  180
AGF  66
In △ABC,
∵ CF  FA and CG  4 cm  GB
1
∴ FG   AB
(mid-pt. theorem)
2
1
5  b
2
b  10
14. In △ABD,
∵ E and F are the mid-points
of AB and BD respectively.
∴ EF // AD
∴ BFE  BDA
In △BCD,
∵ F and G are the mid-points
of BD and BC respectively.
∴ FG // DC
∴ BFG  BDC
EFG  BFE  BFG
 BDA  BDC
 ADC
11. In △BCD,
∵ DG  GB and DF  FC
1
∴ GF   BC
(mid-pt. theorem)
2
1
  8 cm
2
 4 cm
EG  EF  GF
 (6  4) cm
 2 cm
In △ABD,
∵ BE  EA and BG  GD
1
∴ EG   AD
(mid-pt. theorem)
2
1
2 x
2
x4
mid-pt. theorem
corr. s, FG // DC
15. (a) In △ABC,
∵ D and F are the mid-points of AB and CA
respectively.
1
∴ DF // BC and DF   BC
2
(mid-pt. theorem)
∵ E and F are the mid-points of BC and CA
respectively.
1
∴ FE // AB and FE   AB
2
(mid-pt. theorem)
∴ BEFD is a parallelogram.
∴ DFE  DBE
(opp. s of // gram)
 90
∴ △DEF is a right-angled triangle.
12. In △ACD,
∵ AF  FC and AG  GD
∴ FG // CD
(mid-pt. theorem)
∴ ADC  AGF (corr. s, CD // FG)
 35
Reflex BCD  360  x
(s at a pt.)
∵ ABCD is a quadrilateral.
∴ BAD  ABC  reflex BCD  ADC  360
( sum of polygon)
50  60  (360  x)  35  360
(b)
x  145
13. (a) In △ABC,
∵ D and E are the mid-points
of AB and BC respectively.
1
∴
DE   AC
2
1
4 cm   AC
2
AC  8 cm
∵ AC  CG  8 cm
∴ C is the mid-point of AG.
mid-pt. theorem
corr. s, EF // AD
1
 BC
2
1
  8 cm
2
 4 cm
1
FE   AB
2
1
  6 cm
2
 3 cm
DF 
1
 FE  DF
2
1
  3  4 cm 2
2
 6 cm 2
Area of △DEF 
mid-pt. theorem
97
Mathematics in Action (3rd Edition) 3A Full Solutions
18. In △ADC,
∵ AB  BD and N is the
mid-point of CD.
1
∴ BN // AC and BN   AC
2
∵ AB  AC
1
∴ BN   AB
2
∵ M is the mid-point of AB.
1
∴ BM   AB  BN
2
∵ AB  AC
∴ ABC  ACB
∵ NBC  ACB
∴ ABC  NBC
In △CMB and △CNB,
BM  BN
MBC  NBC
BC  BC
∴ △CMB  △CNB
CM  CN
∵ N is the mid-point of CD.
1
∴ CM  CD
2
i.e. CD  2CM
16. (a) In △BFC,
∵ BD  DF and CH  HF
1
∴ DH   BC
(mid-pt. theorem)
2
In △ADE,
∵ AB  BD and AC  CE
1
∴ BC   DE
(mid-pt. theorem)
2
DE  2 BC
HE  DE  DH
 2 BC 

∴
∴
In
∵
∴
3
BC
2
1
BC
DH 2

HE 3 BC
2
1

3
DH : HE  1 : 3
∴
(b) ∵
1
BC
2
DH : HE  1: 3
5 cm 1

HE 3
HE  15 cm
△CFG,
CH  HF and CE  EG
1
HE   FG (mid-pt. theorem)
2
1
15 cm   FG
2
FG  30 cm
In △BCD,
∵ F and G are the mid-points
of BC and DC respectively.
1
∴ FG // BD and FG   BD
2
∴ EH // FG and EH  FG
∴ EFGH is a parallelogram.
given
given
base s, isos. △
alt. s, BN // AC
proved
proved
common side
SAS
corr. sides, △s
Exercise 5F (p. 5.61)
Level 1
17. Join BD.
In △ABD,
∵ E and H are the mid-points
of AB and AD respectively.
1
∴ EH // BD and EH   BD
2
mid-pt. theorem
AB // CD // EF and BD  DF
CE  AC
(intercept theorem)
x3
1.
∵
∴
2.
In △ACD,
∵ AB  BC and BE // CD
∴ AE  ED
(intercept theorem)
y4
3.
∵
∴
∵
∴
∴
mid-pt. theorem
∵
∴
mid-pt. theorem
QXY  QPR  50
(corr. s equal)
XY // PR
QX  XP and XY // PR
QY  YR
(intercept theorem)
1
QY   QR
2
1
a   18
2
9
YR  QY
ba
9
opp. sides equal and //
98
4.
∵ AB // CD // EF // GH and AC  CE  EG
∴ BD  DF  FH  5 cm
(intercept theorem)
DH  DF  FH
x 55
 10
5 Quadrilaterals
5.
In △BFG,
∵ BE  EG and DE // FG
∴ DF  BD
(intercept theorem)
p2
∵
∴
6.
(b) In △ABE,
∵ AG  GE and DG // BE
∴ AD  BD
(intercept theorem)
18 cm
BD 
2
 9 cm
DE // FG // CA and EG  GA
(intercept theorem)
FC  DF
q p
2
∵
∴
In △ACF,
∵ AB  BC and BG // CF
∴ AG  GF
(intercept theorem)
AG  GE and AD  DB
1
DG   BE
(mid-pt. theorem)
2
1
  8 cm
2
 4 cm
y3
10. (a) In △PQR,
∵ QC  CP and RN  NP
∴ CN // QR
In △ACB,
∵ AQ  QC and MQ // BC
∴ AM  BM
EF  AE  AG  GF
z  12  y  3
 12  3  3
6
In △ADE,
∵ AF  (3  3) cm  6 cm  FE and CF // DE
∴ CD  AC
(intercept theorem)
x  22
4
7.
8.
intercept theorem
(b) In △ACB,
∵ AQ  QC and AM  MB
1
∴ MQ  BC
mid-pt. theorem
2
In △ABC,
∵ AM  MB and MN // BC
∴ AN  NC
(intercept theorem)
∵ AC  6 cm
6
∴ AN  cm
2
 3 cm
In △AND,
∵ DP  PN and AD // QP
∴ AQ  QN
(intercept theorem)
∵ AN  3 cm
3
∴ QN  cm
2
 1.5 cm
11. In △ABF,
∵ BD  DF and BA// DC
∴ AC  CF
∵ BD  DF and AC  CF
1
∴ CD   AB
2
AB  2CD
In △BFE,
∵ BD  DF and DC // FE
∴ BC  CE
∵ BD  DF and BC  CE
1
∴ CD   EF
2
EF  2CD
∴ AB  EF
In △ACE,
∵ AF  FE and BF // CE
∴ AB  BC
(intercept theorem)
In △ADE,
∵ AF  FE and CF // DE
∴ AC  CD
(intercept theorem)
∵ AB  BC
∴ CD  2 AB
∵
BD  12 cm
∴
BC  CD  12 cm
intercept theorem
mid-pt. theorem

intercept theorem
mid-pt. theorem

from  and 
Level 2
12. In △ACE,
∵ AB  BC and BD // CE
∴ DE  AD
(intercept theorem)
a3
In △AGJ,
∵ AH  HJ and FH // GJ
∴ AF  FG
(intercept theorem)
b4
AB  2 AB  12 cm
In △AEG,
∵ AD  DE and AF  FG
1
∴ DF   EG
(mid-pt. theorem)
2
1
2  c
2
c4
3 AB  12 cm
AB  4 cm
9.
mid-pt. theorem
(a) In △AEC,
∵ F and G are the mid-points of AC and AE
respectively.
∴ GF // EC
(mid-pt. theorem)
i.e. DF // BC
99
Mathematics in Action (3rd Edition) 3A Full Solutions
13. ∵
∴
BA // CF // DE and BC  CD
AF  FE
(intercept theorem)
x5
∴
In △ABE,
∵ EF  FA and GF // BA
∴ EG  GB
(intercept theorem)
∵ EF  FA and EG  GB
1
∴ FG   AB
(mid-pt. theorem)
2
1
y   14
2
7
16. (a) In △AMN and △BMN,
AMN  ABC
 90
BMN  180  AMN
 90
∴ AMN  BMN
AM  BM
MN  MN
∴ △AMN  △BMN
In △BDE,
∵ BC  CD and BG  GE
1
∴ CG   DE
(mid-pt. theorem)
2
1
z  8
2
4
∴
AG  GE and AB  BC
1
BG   CE
(mid-pt. theorem)
2
1
2  c
2
c4
∴
corr. s, MN //BC
adj. s on st. line
given
common side
SAS
17. (a) ∵ ABD  BDE
∴ AB // DF
In △ABC,
∵ BE  EC and AB // DF
∴ AD  CD
In △BDF,
∵ DE  EF and EC // FB
∴ DC  CB
(intercept theorem)
ba
3
∵
(mid-pt. theorem)
(b) In △ABC,
∵ AM  MB and MN // BC
∴ AN  NC
(intercept theorem)
∵ △AMN  △BMN (proved in (a))
∴ NA  NB
(corr. sides,  △s)
∴ NB  NC
i.e. △BNC is an isosceles triangle.
14. In △ACE,
∵ AG  GE and GB // EC
∴ BC  AB
(intercept theorem)
a3
∵
1
 PA
2
1
QD  4 cm   12 cm
2
QD  4 cm  6 cm
QD  2 cm
QE 
given
alt. s equal
intercept theorem
(b) In △ABC,
∵ BE  EC and
AD  DC
1
∴ DE   AB
mid-pt. theorem
2
AB  2 DE
∵ DE  EF
given
∴ AB  DF
∴ ABFD is a parallelogram. opp. sides equal and //
DE  EF and DC  CB
1
CE   BF
(mid-pt. theorem)
2
1
4   (2  d )
2
82d
d 6
18. (a) (i)
15. In △ABC,
∵ D and E are the mid-points of AB and AC
respectively.
1
∴ DE // BC and DE   BC
(mid-pt. theorem)
2
1
DE   8 cm
2
 4 cm
In △APC,
∵ AE  EC and PA // QE
∴ PQ  QC
(intercept theorem)
∵ AE  EC and PQ  QC
∵ ABCD is a parallelogram.
∴ AD // BC and
(diags. of // gram)
AE  EC
∵ BCEF is a parallelogram.
∴ FE // BC
In △ABC,
∵ AE  EC and GE // BC
∴ AG  GB
(intercept theorem)
∴ AG : GB  1 : 1
(ii) In △ABC,
∵ AE  EC and AG  GB
1
∴ GE   BC (mid-pt. theorem)
2
∵ FE  BC
(opp. sides of // gram)
∴ FG  GE

100
1
 BC
2
5 Quadrilaterals
∴
∴
3.
FG 1

BC 2
FG : BC  1 : 2
AG  GB and
FG  GE
∴ AFBE is a
parallelogram.
x  12  50 (property of rhombus)
x  38
y  50
(b) ∵
4.
proved in (a)
diags. bisect each other
19. Join BC.
Produce EF such that it meets BC at a point G.
(property of rectangle)
OA  OB
OAB  OBA (base s, isos. △)
 65
In △ABO,
OAB  OBA  AOB  180 ( sum of △)
65  65  x  180
x  50
∵
∴
AD  BC
y  11
5.
(property of rhombus)
(property of rectangle)
y  90
(property of square)
OC  OB
(property of square)
z 5
In △ACB,
∵ CF  FA and FG // AB
∴ CG  GB
∵ CF  FA and CG  GB
1
∴ FG   AB
2
In △BDC,
∵ BG  GC and EG // DC
∴ BE  ED
∵ BG  GC and BE  ED
1
∴ EG   DC
2
EF  EG  FG
1
1
  DC   AB
2
2
1
  ( DC  AB)
2
6.
ABC  DCB
 79
BAD  ABC  180 (int. s, AD // BC)
x  79  180
x  101
7.
(a) ∵ ABCD is a parallelogram.
∴ OB  OD and OA  OC
(opp. sides of // gram)
x  4 and 9  y  y  3
2 y  12
y6
intercept theorem
mid-pt. theorem
intercept theorem
mid-pt. theorem
(b) ∵ ABCD is a rhombus.
∴ AOB  90 and AB  BC  CD  DA
(property of rhombus)
In △ABO,
AB 2  OA2  OB 2 (Pyth. theorem)
Enrichment Topic (p. 5.64)
1. By the property of kite,
x  130
AB  (9  6) 2  4 2 cm
AB  AD
y  3 .5
2.
 32  4 2 cm
 5 cm
Perimeter of ABCD  4  AB
 4  5 cm
 20 cm
By the property of kite,
x  20
y  90
8.
Check Yourself (p. 5.69)
1. (a) 
2.
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
∵ AD  BE and BE  BC given
∴ AD  BC
∵ BC  BE
given
∴ BCE  BEC  65
In △BCE,
BCE  BEC  EBC  180  sum of △
65  65  EBC  180
EBC  50
∵ ADB  DBC  50
∴ AD // BC
alt. s equal
∴ ABCD is a parallelogram.
opp. sides equal and //
4 x  x  90 (opp. s of // gram)
3 x  90
x  30
101
Mathematics in Action (3rd Edition) 3A Full Solutions
9.
∵ ABCD is a square.
∴ CAD  45
(property of square)
∵ AFDE is a rhombus.
∴ ADF  67
(property of rhombus)
In △AGD,
AGD  ADG  GAD  180 ( sum of △)
AGD  67  45  180
AGD  68
10. ∵
∴
∵
∴
4.
EPN  PDC  90
(corr. s equal)
NP // CD
EP  PD and NP // CD
AN  NC
(intercept theorem)
 7 cm
5.
In △ABC,
∵ AN  NC and AM  MB
1
∴ MN   BC
(mid-pt. theorem)
2
1
 12 cm
2
 6 cm
∵ ABCD is a rhombus.
∴ AED  90
(property of rhombus)
In △AED,
( sum of △)
2 x  3 x  90  180
5 x  90
x  18
y  2x
 2  18
 36
(property of rhombus)
z  3x
 3  18
 54
(property of rhombus)
∵
∴
∴
ABCD is a rectangle.
(property of rectangle)
ED  EC
(base s, isos.△)
EDC  ECD
b
In △CDE,
ECD  EDC  CED  180 ( sum of △)
b  b  80  180
2b  100
b  50
Revision Exercise 5 (p. 5.71)
Level 1
1. ∵ ABCD is a parallelogram.
∴ BCD  BAD (opp. s of // gram)
x  140
In △ACD,
(property of rectangle)
ADC  90
CAD  ACD  ADC  180 ( sum of △)
a  b  ADC  180
a  50  90  180
a  40
BAD  ADC  180 (int. s, BA // CD)
140  y  180
y  40
∴
AD  BC
6.
(opp. sides of // gram)
a 3  5
a 8
2.
∵
∴
ABCD is a parallelogram.
OB  OD (opp. sides of // gram)
BCE  ECD  90 (property of rectangle)
60  x  90
y  1  11  y
2 y  10
y5
∴
OC  OA
BCE  y
(alt. s, BC // AD)
In △BCF,
BCF  BFC  CBF  180 ( sum of △)
y  80  40  180
y  60
x  30
7.
(opp. sides of // gram)
∵
∴
ABCD is a square.
(property of square)
CED  90
6 x  90
x  15
∴
AD  CD
5y  6  3y  2
2y  4
y2
3x  1  y  3
3x  1  5  3
3x  1  8
3x  9
x3
3.
∵
∴
ABCD is a rhombus.
AD  CD (property of rhombus)
a37
a4
∴
AED  90 (property of rhombus)
8.
b  30  90
b  60
102
(property of square)
∵ ABCD is a square.
∴ ACD  45
(property of square)
ACD  DCE  180 (adj. s on st. line)
45  x  180
x  135
5 Quadrilaterals
∴
OA  OB
14. ∵ ABCD is a square.
∴ ABD  45
(property of square)
∵ △AEB is an equilateral triangle.
∴ ABE  60
(prop. of equil. △)
DBE  ABD  ABE
 45  60
 105
(property of square)
y4 2
y6
9.
In △ACF,
∵ AB  BC and AE  EF
1
∴ BE   CF
(mid-pt. theorem)
2
1
x  9
2
 4.5
(prop. of equil.△)
EAB  60
(property of square)
BAD  90
EAD  EAB  BAD
 60  90
 150
∵ △AEB is an equilateral triangle.
∴ EA  AB
Also, AB  AD
(property of square)
∴ EA  AD
∴ ADE  AED (base s, isos.△)
In △AED,
EAD  ADE  AED  180 ( sum of △)
In △ADG,
∵ AC  ( 4  4) cm  8 cm  CD and
AF  (3  3) cm  6 cm  FG
1
∴ CF   DG
(mid-pt. theorem)
2
1
9 y
2
y  18
10. ∵
∴
150  2ADE  180
ADE  15
(property of square)
ADB  45
BDE  ADB  ADE
 45  15
 30
AE // BF // CG and AB  BC
(intercept theorem)
EF  FG
2x  3  4x  5
2x  8
x4
FG  (4 x  5) cm
 (4  4  5) cm
 11 cm
∵ BF // CG // DH and FG  11 cm  GH
∴ CD  BC
(intercept theorem)
15. ∵ AEDF is a rhombus.
∴ AF  DF
(property of rhombus)
∴ FAD  FDA (base s, isos. △)
In △ADF,
AFD  FAD  FDA  180 ( sum of △)
122  2FAD  180
2FAD  58
FAD  29
EAD  FAD
(property of rhombus)
 29
(property of rectangle)
ADC  90
In △ACD,
CAD  ACD  ADC  180 ( sum of △)
29  ACD  90  180
ACD  61
y5
11. In △BCD,
∵ DF  FC and EF // BC
∴ DE  EB
(intercept theorem)
x3
In △ACD,
∵ DF  FC and AD// EF
∴ AE  EC
(intercept theorem)
y2
12. ∵ AB  AE
∴ ABE  AEB (base s, isos. △)
In △ABE,
BAE  ABE  AEB  180 ( sum of △)
30  2ABE  180
2ABE  150
ABE  75
∵ ABCD is a parallelogram.
∴ ADC  ABC (opp. s of // gram)
 75
16.
13. By the property of kite,
CED  90
With the notations in the figure,
OA  OC and OB  OD (property of rhombus)
x  90
BAE  DAE
y  35
103
Mathematics in Action (3rd Edition) 3A Full Solutions
19. (a) ∵ ABCD is a parallelogram.
∴ BAD  DCB ,
ABC  ADC ,
AB  CD and AD  BC
∵ BE and FD are the angle
bisectors of ABC and
ADC respectively.
∴ ABE  CBE  ADF
 CDF
In △AEB and △CFD,
ABE  CDF
AB  CD
BAE  DCF
∴ △AEB  △CFD
24
cm
2
 12 cm
In △AOD,
OA 
OA2  OD 2  AD 2
(Pyth. theorem)
OD  20  122 cm
 16 cm
BD  2  OD
2
 2  16 cm
 32 cm
∴ The length of another diagonal of the rhombus is
32 cm.
17. (a) ∵ AQDR is a rectangle.
∴ ARD  90
(property of rectangle)
In △ADR,
(Pyth. theorem)
AD 2  AR 2  DR 2
AD 
AR  DR
2
(b) ∵ ABCD is a parallelogram.
∴ AD  BC and AD  BC
∵ △AEB  △CFD
∴ AE  CF
ED  AD  AE
2
 3 2  4 2 cm
 5 cm
∵ ABCD is a rhombus.
∴ AB  BC = CD  AD  5 cm
∴ Perimeter of ABCD  AB  BC  CD  AD
 (5  5  5  5) cm
 20 cm

 BC  CF
 BF
∴ BFDE is a parallelogram.
20. (a) In △ABP and △CDQ,
ABP  CDQ
∵ APQ  CQP
∴ APB  180  APQ
(b) ∵ AQDR is a rectangle.
∴ DQ  AR  3 cm and
AQ  DR  4 cm (property of rectangle)
∵ ABCD is a rhombus.
∴ CQ  AQ  4 cm and
BQ  DQ  3 cm (property of rhombus)
∵ BPCQ is a rectangle.
∴ CP  BQ  3 cm and
BP  CQ  4 cm (property of rectangle)


Area of ABPCDR = 6  area of △ADR
1
 6   AR  DR
2
1
 6   3  4 cm 2
2
 36 cm 2
proved
proved
proved
ASA
proved in (a)
corr. sides, △s
opp. sides equal
and //
alt. s, AB // DC
given
adj. s on st. line
 180  CQP
adj. s on st. line
 CQD
AB  CD
∴ △ABP  △CDQ
opp. sides of // gram
AAS
(b) ∵ APQ  CQP
∴ AP // QC
∵ △ABP  △CDQ
∴ AP  CQ
∴ APCQ is a parallelogram.
21. (a) In △AED and CFB,
AD  CB
BAD  BCD  90
∵ △ABE  △CDF
∴ BAE  DCF
EAD  BAE  BAD
18. (a) ∵ △ABE is an equilateral triangle.
∴ BAE  ABE  AEB  60
(prop. of equil. △)
In quadrilateral ABCD,
BAD  ABC  BCD  ADC  360
( sum of polygon)
(60  40)  (60  20)  100  ADC  360
100  80  100  ADC  360
ADC  80
(b) BAD  60  40  100
ABC  60  20  80
∵ BAD  BCD and
ABC  ADC
∴ ABCD is a parallelogram.
opp. s of // gram
opp. sides of // gram
 DCF  BCD
 FCB
AE  CF
∴ △AED  △CFB
∴ BF  ED
property of square
property of square
given
corr. s, △s
corr. sides, △s
SAS
corr. sides, △s
(b) ∵ △ABE  △CDF
∴ BE  FD
∴ BFDE is a parallelogram.
∴ Daniel’s claim is incorrect.
opp. s equal
104
given
alt. s equal
proved in (a)
corr. sides, △s
opp. sides equal and //
(given)
(corr. sides, △s)
(opp. sides equal)
5 Quadrilaterals
BDF  FDE
1
∴ BDF   BDE
2
1
  144
2
 72
DOA  90
In △DOF,
AFD  DOA  BDF
 90  72
 18
22. (a) ∵ BCHG is a parallelogram.
∴ GB  HC  7 cm (opp. sides of // gram)
In △AFB,
∵ AE  EF and EG // FB
∴ AG  GB
(intercept theorem)
AB  AG  GB
 (7  7) cm
 14 cm
∵
(b) In △AFB,
∵ AE  EF and AG  GB
1
EG   FB
∴
(mid-pt. theorem)
2
1
2 cm   FB
2
FB  4 cm
∵ ABCD is a parallelogram.
∴ BC  AD  6 cm (opp. sides of // gram)
FC  FB  BC
 ( 4  6) cm
 10 cm
(property of rhombus)
(ext.  of △)
25. ∵
∴
∵
∴
ABCD is a square.
(property of square)
DBC  45
BF // CE
(property of rhombus)
BCE  180  DBC (int.∠s, BF // CE)
 180  45
 135
∵ AB  BC  CD  DA (property of square)
DC  CE  EF  FD (property of rhombus)
∴ BC  CE
∴ EBC  BEC
(base s, isos. △)
In △BCE,
BCE  EBC  BEC  180 ( sum of △)
23. (a) ∵ ABCD is a parallelogram.
∴ AB  DC  10 cm and AD  BC  12 cm
(opp. sides of // gram)
MB  AB  AM
135  2EBC  180
EBC  22.5
DBG  DBC  EBC
 45  22.5
 22.5
 (10  5) cm
 5 cm
AN  AD  ND
 (12  6) cm
 6 cm
∵ AM  MB and AN  ND
1
∴ MN // BD and MN   BD
2
(mid-pt. theorem)
∴ BDNM is a trapezium.
(b)
(given)
26. ∵ △BCG is an equilateral triangle.
∴ CG  GB  BC
Also, BC  CD
(property of square)
∴ CD  CG
∴ CDG  CGD (base s, isos.△)
∵ BCG  60 (prop. of equil.△)
DCB  90 (property of square)
∴ DCG  DCB  BCG
 90  60
 30
In △CDG,
DCG  CDG  CGD  180 ( sum of △)
1
 BD
2
BD  2MN
∵
Perimeter of BDNM = 38 cm
∴
MN  MB  BD  ND  38 cm
MN  5 cm  2MN  6 cm  38 cm
3MN  11 cm  38 cm
3MN  27 cm
MN  9 cm
MN 
30  2CDG  180
CDG  75
(property of square)
CDB  45
GDB  CDG  CDB
 75  45
 30
Level 2
24. ∵ ABCD is a rhombus.
∴ ADB  CBD  ABD (property of rhombus)
∴ ADB  ABD
1
  ABC
2
1
  72
2
 36
(adj. s on st. line)
BDE  180  ADB
 180  36
 144
(property of rectangle)
KB  KC
KBC  KCB (base s, isos.△)
 24
In △BCK,
BKC  180  KBC  KCB ( sum of△)
 180  24  24
 132
AKD  BKC  132 (vert. opp. s)
(prop. of equil.△)
AKE  60
27. (a) ∵
∴
105
Mathematics in Action (3rd Edition) 3A Full Solutions
EKD  AKD  AKE
 132  60
 72
Perimeter of parallelogram ABCD
 AB  BC  CD  AD
 (7.5  15  7.5  15) cm
 45 cm
(b) ∵ △AKE is an equilateral triangle.
∴ EAK  60
(prop. of equil. △)
(alt. s, AD // BC)
DAK  ACB
 24
EAD  EAK  DAK
29. In △ADQ and △CBP,
ADB  CBD
ADQ  180  ADB
 180  CBD
 CBP
AD  CB
DQ  BP
∴ △ADQ  △CBP
∴ AQ  CP
AQD  CPB
∴ AQ // PC
∴ APCQ is a parallelogram.
 60  24
 36
∵ △AKE is an equilateral triangle.
∴ KA  KE
Also, KA  KD
(property of rectangle)
∴ KE  KD
∴ KED  KDE (base s, isos.△)
In △DEK,
EKD  KDE  KED  180 ( sum of △)
72  2KDE  180
KDE  54
KDA  KBC
(alt. s, AD // BC)
 24
EDA  KDE  KDA
 54  24
 30
∵ EAD  EDA
∴ △AED is not an isosceles triangle.
∴ Philip’s claim is incorrect.
28. (a) Let CBE  a and BCE  b.
ABE  CBE  a
DCE  BCE  b
ABC  BCD  180
2a  2b  180
a  b  90
In △BCE,
BEC  BCE  CBE  180
BEC  a  b  180
BEC  90  180
BEC  90
∴ △BEC is a right-angled
triangle.
BC  BE  CE
2
opp. sides of // gram
given
SAS
corr. sides,  △s
corr. s,  △s
alt. s equal
opp. sides equal and //
ABCDE is a regular pentagon.
ABC  BCD
(5  2)  180

5
 108
∵ AB  CB
∴ BAC  BCA
base s, isos. △
In △ABC,
 sum of △
ABC  BCA  BAC  180
108  2BCA  180
2BCA  72
BCA  36
∵ BCDF is a rhombus.
∴ BCF  DCF
property of rhombus
108

2
 54
FCG  BCF  BCA
given
given
int. s, AB // DC
 sum of △
 54  36
 18
GCD  GCF  DCF
 18  54
 72
BGC  GCD
 72
FCG  BGC  18  72
 90
(Pyth. theorem)
2
 12 2  9 2 cm
 15 cm
AD  BC  15 cm
∵ DEC  BCE
and DCE  BCE
∴ DCE  DEC
∴ DC  DE
 7.5 cm
AB  DC  7.5 cm
adj. s on st. line
30. ∵
∴
(b) In △BCE,
BC 2  BE 2  CE 2
alt. s, AD // BC
adj. s on st. line
31. (a) In △BCE and △BGA,
GBE  CBA  90
CBE  CBG  GBE
(opp. sides of // gram)
(alt. s, DE // CB)
 CBG  CBA
 GBA
∵ ABCD and BEFG are
two identical rectangles.
∴ BA  BE and BG  BC
∴ △BCE  △BGA
(sides opp. equal s)
(opp. sides of // gram)
106
alt. s, BF // CD
property of rectangle
SAS
5 Quadrilaterals
(b) ∵ △BCE  △BGA (proved in (a))
∴ BCE  BGA (corr. s,  △s)
BPG  180  PBQ  BGA ( sum of △ )
in △PED and △QEB,
∵ PD // BQ
∴ DPE  BQE
PED  QEB
ED  EB
∴ △PED  △QEB
∴ PE  QE
and ED  EB
∴ BQDP is a parallelogram.
In △PED and △PEB,
PED  PEB  90
ED  EB
PE  PE
∴ △PED  △PEB
∴ PD  PB
Also, PD  BQ and PB  QD
∴ PD  QD  BQ  PB
∴ BQDP is a parallelogram
with four equal sides.
∴ BQDP is a rhombus.
 180  PBQ  BCE
∴
32. (a) (i)
 BQC
BQC  107
In △ABC and △AFE,
BAC  FAE
ABC  90
 AFE
AC  AE
∴ △ABC  △AFE
(ii) In △ABG and △AFG,
ABG  AFG  90
∵ △ABC  △AFE
∴ AB  AF
AG  AG
∴ △ABG  △AFG
( sum of △ )
common angle
property of square
given
given
AAS
proved in (i)
proved in (i)
corr. sides,  △s
common side
RHS
35. (a) In △OCD and △OQP,
COD  QOP
∵ CO  DO
and CQ  DP
∴ QO  CO  CQ
(b) BAC  45
(property of square)
∵ △ABG  △AFG
(proved in (a)(ii))
∴ BAG  FAG
(corr. s, △s)
1
  BAC
2
1
  45
2
 22.5
In △ABG,
AGB  BAG  ABG  180 ( sum of △)
AGB  22.5  90  180
AGB  67.5
33. (a) In △ACE and △DBF,
AE  DF
∵ OA  OD
∴ CAE  BDF
AC  DB
∴ △ACE  △DBF
∴ ACE  DBF
(b) In △ACE,
NEF  CAE  ACE
In △DBF,
NFE  BDF  DBF
∴ NEF  NFE
∴ NE  NF
i.e. △NEF is an isosceles
triangle.
∴
given
given
common side
SAS
corr. sides, △s
opp. sides of // gram
common angle
property of rectangle
given
 DO  DP
 PO
CO DO

QO PO
∴ △OCD ~ △OQP
∴ OCD  OQP
∴ DC // PQ
Also, AB // DC
∴ AB // PQ
given
property of rectangle
base s, isos. △
property of rectangle
SAS
corr. s, △s
property of rectangle
alt. s, PD // BQ
vert. opp. s
given
AAS
corr. sides, △s
given
diags. bisect each other
ratio of 2 sides, inc. 
corr. s, ~△s
corr. s equal
property of rectangle
(b) In △AOP and △BOQ,
PO  QO
(proved in (a))
AOP  BOQ
(vert. opp. s)
AO  BO
(property of rectangle)
∴ △AOP  △BOQ (SAS)
∴ AP  BQ
(corr. sides, △s)
∵ ABQP is a quadrilateral with AB // PQ and
AP  BQ.
∴ ABQP is an isosceles trapezium.
ext.  of △
ext.  of △
36. In △ABC,
∵ M and N are the mid-points
of AB and AC respectively.
∴ MN // BC and
1
MN   BC
2
In △BCD,
∵ P and Q are the mid-points
of CD and BD respectively.
∴ QP // BC and
1
QP   BC
2
∵ QP  MN and QP // MN
∴ MNPQ is a parallelogram.
∴ MQ  NP
sides opp. equal s
34. With the notation in the figure,
107
mid-pt. theorem
mid-pt. theorem
opp. sides equal and //
opp. sides of // gram
Mathematics in Action (3rd Edition) 3A Full Solutions
37. (a) (i)
In △APR and△CQR,
AR  CR
ARP  CRQ
PAR  QCR
∴ △APR  △CQR
(ii) ∵ △APR  △CQR
∴ PA  QC
and BQ  QC
∴ PA  BQ
Also, AP // BQ
∴ ABQP is a
parallelogram.
1
 AQ  h
Area of △ APQ 2

Area of △ QCD 1  RQ  h
2
AQ

RQ
AR  RQ

RQ
2 RQ  RQ

RQ
3RQ

RQ
3
∴ Area of △APQ : area of △QCD  3 : 1
given
vert. opp. s
alt. s, AP // QC
ASA
proved in (i)
corr. sides, △s
given
given
opp. sides equal and //
(b) In △ADE,
∵ B and C are the mid-points of AD and AE
respectively.
1
∴ BC   DE
(mid-pt. theorem)
2
DE  2 BC
∵ BQ  QC
∴ BC  2 BQ
 2 AP
∴ DE  2 BC
 4 AP
AP 1

DE 4
∴ AP : DE  1 : 4
39. (a) In △ABD,
∵ AE  EB and
∴ DG  GB
∵ AE  EB and
1
∴ EG   AD
2
In △BCD,
∵ DG  GB and
∴ DF  FC
∵ DG  GB and
1
∴ GF   BC
2
∴ EF  EG  GF
38. (a) In △PQR and △DQC,
QPR  QDC
alt. s, PR // CD
PQ  DQ
given
PQR  DQC
vert. opp. s
∴ △PQR  △DQC
ASA
(b) (i) ∵ AP  PB and PR // BC
∴ AR  RC
(intercept theorem)
∵ △PQR  △DQC (proved in (a))
∴ RQ  CQ
(corr. sides, △s)
1
  RC
2
AR
RC

RQ 1  RC
2
2
∴ AR : RQ  2 : 1
AD// EG
intercept theorem
DG  GB
mid-pt. theorem
GF // BC
intercept theorem
DF  FC
mid-pt. theorem
1
1
 AD   BC
2
2
1
 ( AD  BC )
2

(b) In △ABC,
∵ AE  EB and EH // BC
∴ AH  HC
∵ AE  EB and AH  HC
1
∴
EH   BC
2
1
EG  GH   BC
2
1
GH   BC  EG
2
In △ABD,
∵ AE  EB and EG // AD
∴ DG  GB
∵ AE  EB and DG  GB
1
∴ EG   AD
2
1
1
∴ GH   BC   AD
2
2
1
  ( BC  AD)
2
(ii)
Let h be the height of △APQ and △PQR.
Area of △QCD = area of △PQR
108
intercept theorem
mid-pt. theorem
intercept theorem
mid-pt. theorem
5 Quadrilaterals
∴
Multiple Choice Questions (p. 5.77)
1. Answer: C
(opp. sides of // gram)
BC  AD
∵ AD  BE
∴ BC  BE
∴ BCE  BEC
(base s, isos. △)
 35
In △BCF,
FBC  BCF  AFC
 180  100
 80
BCD  BCE  ECD
 80  60
 140
∵ △CDE is an equilateral triangle.
∴ EC  CD  DE
Also, AB  BC  CE  EA (property of rhombus)
∴ DE  EA and CD  CB
∴ EDA  EAD and
(base s, isos.△)
CDB  CBD
In △AED,
AED  EDA  EAD  180 ( sum of △)
(ext.  of △)
FBC  35  100
FBC  65
i.e. ABC  65
2.
Answer: D
BCD  x (opp. s of // gram)
(opp. s of // gram)
CBF  z
BGC  y (vert. opp. s)
In △BCG,
BCG  BGC  CBG  180
160  2EDA  180
EDA  10
In △BCD,
BCD  CBD  CDB  180 ( sum of △)
140  2CDB  180
CDB  20
ADB  EDC  EDA  CDB
 60  10  20
 30
( sum of △)
x  y  z  180
y  180  x  z
3.
Answer: B
ADC  90
In △ACD,
(property of rectangle)
AC  AD  CD
2
BCE  180  ABC (int. s, EC // AB)
2
2
Alternative Solution
Construct BF such that BF // AD.
(Pyth. theorem)
AC  2  7 cm
2
2
 53 cm
ACF  90 and
AC  CF  53 cm
In △ACF,
AF 2  AC 2  CF 2
(property of square)
(Pyth. theorem)
Let EAD  a and CBD  b.
∵ AE // BC
(property of rhombus)
∴ a  DAB  100  180
(int. s, AE // BC)
DAB  80  a
DAB  ABF  180 (int. s, AD // BF)
AF  ( 53 )  ( 53 ) 2 cm
2
 10.3 cm (cor. to 3 sig. fig.)
4.
Answer: C
AEC  ECD  180 (int. s, EA // CD)
(80  a)  (100  CBF )  180
CBF  a
(alt. s, AD // BF)
ADB  CBD  CBF
ba
∵ △CDE is an equilateral triangle.
∴ EC  CD  DE
Also, AB  BC  CE  EA
(property of rhombus)
∴ DE  EA and CD  CB
∴ EDA  EAD  a and
(base s, isos. △)
CDB  CBD  b
∵ CDE  60
(prop. of equil.△)
∴ EDA  ADB  CDB  60
a  (b  a )  b  60
2b  2a  60
b  a  30
∴ ADB  b  a
 30
125  ECD  180
ECD  55
(property of square)
BDC  45
In △CDF,
CFD  FCD  FDC  180 ( sum of △)
CFD  55  45  180
CFD  80
5.
Answer: C
∵ △CDE is an equilateral triangle.
∴ CED  ECD  EDC  60 (prop. of equil.△)
∵ ABCE is a rhombus.
∴ AEC  ABC
(property of rhombus)
 100
AED  AEC  CED
 100  60
 160
∵ EC // AB
(property of rhombus)
109
Mathematics in Action (3rd Edition) 3A Full Solutions
6.
(common side)
DE  DE
∴ △ADE  △CDE (SSS)
∴ III must be true.
∴ The answer is D.
Answer: A
OA  OC , OB  OD
and AOB  90 (property of rhombus)
1
OA   AC
2
1
  16 cm
2
 8 cm
In △AOB,
OA2  OB 2  AB 2
10. Answer: A
For I:
AEB  BED  180 (adj. s on st. line)
AEB  120  180
AEB  60
AB  BE
BAE  AEB (base s, isos. △)
 60
In △ABE,
ABE  AEB  BAE  180 ( sum of △)
ABE  60  60  180
ABE  60
∵ ABE  AEB  BAE  60
∴ AB  BE  AE
∴ △ABE is an equilateral triangle.
∴ I must be true.
For II:
It is not necessary that BCDE is a parallelogram because
BE may not be parallel to CD.
∴ II may be false.
For III:
It is not necessary that ABCE is a rhombus because BC
and EC may not equal to AB.
∴ III may be false.
∴ The answer is A.
(Pyth. theorem)
∵
∴
OB  17 2  82 cm
 15 cm
Area of ABCD = 4  area of △AOB
1
 4   OA  OB
2
1
 4   8  15 cm2
2
 240 cm2
7.
Answer: B
∵ The interior angles of a rhombus may not be all
equal.
∴ It is incorrect that any rhombus must be a regular
polygon.
8.
Answer: D
11. Answer: B
In △ADE,
∵ AB  BD and AC  CE
1
BC   DE (mid-pt. theorem)
∴ BC // DE and
2
1
2.5 cm   DE
2
DE  5 cm
In △AHI,
∵ AD  DH and AE  EI
1
∴ DE // HI and DE   HI
(mid-pt. theorem)
2
1
5 cm   HI
2
HI  10 cm
With the notation in the figure,
∵ AD  AB
∴ ADB  ABD
(base s, isos. △)
 23
By the property of kite,
AED  90
In △AED,
a  AED  ADE  180 ( sum of △)
a  90  23  180
a  67
9.
12. Answer: A
∵ P and R are the mid-points of AB and CA
respectively.
∴ PR// BC
(mid-pt. theorem)
APR  x and ARP  y (corr. s, PR // BC)
∵ P and Q are the mid-points of AB and BC
respectively.
∴ PQ // AC
(mid-pt. theorem)
RPQ  ARP  y
(alt. s, AR // PQ)
APQ  APR  RPQ
 x y
Answer: D
For I,
by the property of kite,
AC  BD
∴ I must be true.
For II,
by the property of kite,
BAE  BCE
∴ II must be true.
For III,
in △ADE and △CDE,
(given)
AE  CE
(property of rhombus)
AD  CD
110
5 Quadrilaterals
BAD  ABC  180 (int. s, AD // BC)
(85  CAD)  (25  CAD)  180
2CAD  70
13. Answer: A
In △ABC,
∵ AQ  QC and PQ // BC
∴ AP  PB
(intercept theorem)
∴ I must be true.
∵ AQ  QC and AP  PB
1
∴ PQ   BC (intercept theorem)
2
BC  2 PQ
∴ II must be true.
For III,
in △ACD,
∵ AR  RD and AQ  QC
∴ QR // CD
(mid-pt. theorem)
AQR  ACD (corr. s, QR // CD)
∵ It is not necessary that AC  AD .
∴ ACD  ADC may not be true.
∴ AQR  ADC may not be true.
∴ The answer is A.
CAD  35
2.
64  2DAE  180
DAE  58
(alt. s, AD // BC)
ACB  DAC
 58
In △BCE,
(ext.  of △)
CBE  ECB  AEB
CBE  58  89
CBE  31
3.
14. Answer: D
∵ BG : GC  3 :1
BG
3
∴
GC
BG  3GC
In △BGE,
∵ EA  AB and AF // BG
∴ EF  FG
(intercept theorem)
∵ EA  AB and EF  FG
1
∴ AF   BG
(mid-pt. theorem)
2
1
  3GC
2
3GC

2
(opp. sides of // gram)
BC  AD
BG  GC  AF  FD
3GC
3GC  GC 
 FD
2
5GC
FD 
2
FD 5

GC 2
∴ FD : GC  5 : 2
4.
Exam Corner
Exam-type Questions (p. 5.80)
1.
∵ AD  DE
∴ DAE  DEA
(base s, isos. △)
In △AED,
ADE  DAE  DEA  180 ( sum of △)
In △BCG and △DCH,
∵ CE and CF divide BCD
into three equal parts.
∴ BCG  DCH
∵ ABCD is a rhombus.
∴ BC  DC
∴ CBG  CDH
∴ △BCG  △DCH
∴ CG  CH
CHG  CGH
x
CHG  FHG  180
x  y  180
111
adj. s on st. line
∵ EB : BC  3 : 2
∴ Let EB  3x and BC  2 x .
In △BEF and △CED,
(corr. s, AB // DC)
EBF  ECD
(corr. s, AB // DC)
EFB  EDC
(common angle)
BEF  CED
∴ △BEF ~ △CED (AAA)
FB BE

∴
(corr. sides, ~△s)
DC CE
FB
3x

DC 3 x  2 x
FB 3

DC 5
3
FB  DC
5
(opp. sides of // gram)
AB  DC
AF  FB  DC
3
AF  DC  DC
5
2
AF  DC
5
AF 2

DC 5
∴ AF : DC  2 : 5
ACB  CAD (alt. s, AD // BC)
DCB  25  ACB
 25  CAD
∵ ABCD is a trapezium with AD // BC .
∴ ABC  DCB
 25  CAD
property of rhombus
property of rhombus
ASA
corr. sides, △s
base s, isos. △
Mathematics in Action (3rd Edition) 3A Full Solutions
5.
Do and Investigate (p. 5.82)
(a) In △BCF and △DCF,
(property of square)
BC  DC
BCF  DCF  45 (property of square)
(common side)
CF  CF
∴ △BCF  △DCF (SAS)
1.
(b) CDF  CBF (corr. s, △s)
 77
(property of square)
DCF  45
In △CDF,
AFD  CDF  DCF (ext.  of △)
 77  45
 122
6.
7.
∵ AD is a median of △ABC.
∴ BD  DC
∵ ADG  CED
∴ GD// FC
alt. s equal
In △AGD,
∵ AE  ED and FE // GD
∴ AF  FG
intercept theorem
In △BCF,
∵ CD  DB and FC // GD
∴ FG  GB
intercept theorem
∴ AF  FG  GB
2.
Join BD.
∵ AE  EB and AH  HD
1
∴ EH  BD and EH // BD
2
∵ CF  FB and CG  GD
1
∴ FG  BD and FG // BD
2
∴ EH  FG and EH // FG
∴ EFGH is a parallelogram.
(a) In △ACD,
∵ DN  NC and AD // ON
∴ AO  OC
intercept theorem
In △ABC,
∵ AO  OC and BM  MC
∴ BA// MO
mid-pt. theorem
∵ AD // BO and BA// OD
∴ ABOD is a parallelogram.
3.
(a) yes, rectangle
(b) yes, rhombus
(b) In △ACD,
∵ DN  NC and AO  OC
1
∴ ON   AD
(mid-pt. theorem)
2
∵ BO  AD
(opp. sides of // gram)
1
∴ ON   BO
2
BO
2
ON
∴ BO : ON  2 : 1
(c) yes, square
(d) yes, rhombus
112
mid-pt. theorem

mid-pt. theorem

from  and 
opp. sides equal and //