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Quadrilateral
2 pairs of equal adjacent sides
1 pair of // opp. Sides
Kite
Trapezium
Sum of interior angles is 1800
One of the diagionals is axis of symmetry
2 diagionals are 
Parallelogram
2 pairs of opp.// sides
4 equal sides
Rhombus
Properties of trapesium
2 pairs of opposite sides are equal.(opp. sides of // gram)
2 pairs of opposite angles are equal (opp. s of // gram)
Diagonals bisect each other (diag. Of // gram)
Properties of // gram and kite
Diagonals bisects each interior angle
4 right angles
4 right angles and
4 equal sides
Rectangle
Properties of // gram
Diagonals are equal
Properties of rhombus/rectangle
Square
Angles between each diagional and each side is 45 0
450
Trapeziums
Definition : 1 pair of parallel sides
Properties:
Sum of interior angles is 1800
Parallelogram
Definition : 2 pairs of opp. parallel sides
Properties:
2 pairs of opposite sides are equal.
(opp. sides of // gram)
2 pairs of opposite angles are equal
(opp. s of // gram)
Diagonals bisect each other
(diag. Of // gram)
Conditions for Parallelogram
If 2 pairs of opposite sides are equal then
the quadrilateral is parallelogram.
(opp. sides eq.)
If 2 pairs of opposite angles are equal then
the quadrilateral is parallelogram.
(opp. s of eq.)
If diagonals bisect each other then
the quadrilateral is parallelogram
(diag. Bisect each other)
If 1 pair of opposite sides is equal and parallel then
the quadrilateral is parallelogram
(opp. sides eq. and //)
Rhombus
Definition : a // gram or a kite of 4 equal sides
Properties:
2 pairs of opposite sides are equal.
(opp. sides of // gram)
2 pairs of opposite angles are equal
(opp. s of // gram)
Diagonals bisect each other
(diag. Of // gram)
Diagonals bisects each interior angle
Diagonals are 
Rectangle
Definition : a parallelogram of 4 right angles
Properties:
2 pairs of opposite sides are equal.
(opp. sides of // gram)
2 pairs of opposite angles are equal
(opp. s of // gram)
Diagonals bisect each other
(diag. Of // gram)
Diagonals are equal
Square
Definition : a // gram of 4 right angles and 4 equal sides
Properties:
2 pairs of opposite sides are equal.
(opp. sides of // gram)
2 pairs of opposite angles are equal
(opp. s of // gram)
Diagonals bisect each other
(diag. Of // gram)
450
Diagonals are equal
Diagonals are 
Angles between each diagonal and each side is 450
Example 1: In the figure, PQRS is a kite
(a) Find x and y.
(b) Find the perimeter of the kite PQRS
(a) PQ = PS (given)
x+1 = y+3
x-y=2
P
QR=SR (given)
x+y=8
y+3
x+1
S
Q
x+y
(1)+(2),
8
R
(2)
2x=10
x=5
Put x=5 into (1),
(b)
(1)
5-y=2
y=3
PQ = x+1=5+1=6
 PQ+PS+SR+QR = 6 + 6 + 8 + 8 =28
Example 2: In the figure, ABCD is a kite. E is a point of intersection of
diagonals AC and BD, AE=9 cm, EC=16 cm and DE=EB=12 cm
(a) Find the area of ABCD.
(a) ABC= ADC (axis of symmetry AC)
(b) Find the perimeter of ABCD
AED=900
Area of ADC = 1  AC  DE
2
1
 (9  16)  12
2
 150cm 2
Area of kite ABCD=Area of ABC+Area of ADC
= 150+150 =300 cm2
(b)
In ADE,
AD2=AE2+DE2=92+122=225 cm2 (Pyth theorem)
AD=15 cm
In CDE,
DC2=DE2+EC2=122+162=400 cm2 (Pyth theorem)
DC=20 cm
 Perimeter of ABCD=AD+AB+ DC+CB = 15 + 15 + 20 + 20 =70 cm
Example 3: In the figure, ABCD is a parallelogram. Find x and y.
A
D
x
680
B
1500-y
2y
C
AD//BC (Given)
x+680=1800 (prop. Of trapezium)
 x=1120
(1500-y)+2y=1800 (prop. Of trapezium)
1500+y=1800
 y=1800 -1500=300
Example 4: In the figure, ABCD is a parallelogram. Find x and y.
D
C
3x+100
x+200
A
y
B
DAB=DCB (opp. s of // gram)
x+200=3x-100
2x=300
x=150
DAB+CBA=1800 (int.s , AD//BC)
x+200+y=1800
150+200+y=1800
y=1450
Example 5: In the figure, ABCD is a isosceles trapezium with AB=DC.
Find x , y and z
A
D
z
y
B
AD//BC (Given)
x+1260=1800 (prop. Of trapezium)
 x=540
1260
x
a
E
C
Construct AE // DC
AD//EC and AE//DC
 ADCE is a parallelogram (Definition of // gram)
ADCE is a parallelogram (proof)
 AE=DC (opp.sides of // gram)
In ABE,
AE=DC (proof)
AB=AC (given)
AB=AE
 y=a (base s. isos  )
a= x (corr. s. AE//DC)
 y=x
=540
y+z=1800 (prop. Of trapesium)
z= 1800-540
= 1260
MID-POINT THEOREM
A
IF AM = MB and AN =NC then
(a) MN // BC
1
(b) MN = BC
2
M
N
(Abbreviation: Mid-point theorem)
B
C
Example 13: In the figure, ABC is a triangle, find x and y.
CE=BE (given)
AD=DB (given)
DE//AC
x=
(mid-point theorem)
EDB
=420
1
DE  BC
2
1
6 y
2
(corr. s , DE//AC)
C
(mid-point theorem)
y
y  2  6  12
E
6
420
x
A
D
B
Example 14: Prove that BPQR is a parallelgram
AR=RB
(given)
AQ=QC
(given)
1
 RQ // BC and RQ  BC
2
 BP  PC
(given)
1
 BC
2
(mid-point theorem)
A
 RQ  BP
R
Q
 BPQR is a paral log ram
(opp-sides eq. And //)
B
P
C
Ex 11D
(b)
AM=AC (given)
BN=NC (given)
BM=MD (given)
BN=NC (given)
1
 MN  AB (mid-point theorem)
2
1
5 x
2
x  10
1
 MN  CD
2
1
5 y
2
y  10
D
A
M
x cm
B
y cm
(mid-point theorem)
Ex 11D
2(b)
AP=BP (given)
AQ=CQ (given)
 PQ // BC
(mid-point theorem)
 APQ  PBC  46 (corr.s. PQ//BC)
0
In APQ,
APQ+ PAQ+ a = 1800 (adj s. on a st line)
460+1100+a=1800
A
a=240
1100
P
460
B
a
Q
C
A
3(a)
10
F
B
E 8
D
9
C
A
3(b)
60
F
B
E
70
50
D
9
C
4. BP=PA (given)
CR=RB (given)
1
 PR  AB
2
1
6   AB
2
AB  12cm
C
(mid-point theorem)
P
R
8
AQ=QB (given)
AP=PC (given)
1
 PQ  BC
2
1
8   BC
2
BC  16cm
6
A
(mid-point theorem)
Q
B
AB  BC
Area of ABC 
2
12  16
5
2
x  96 cm 2
INTERCEPT THEOREM
transversal
X
B
P
A
D
C
Q
Y
INTERCEPT THEOREM
If AB//CD//EF then
A
C
E
BD AC

(intercept theorem)
DF CE
B
D
F
INTERCEPT THEOREM
Proved: AD AC
DF

CE
Construct GB through A such that BG//CD//EF
GB//CD//EF (given)
A
G
C
E
B
D
F
AD AC


(intercept theorem)
DF CE
Example 15. AP//BQ//CR, AB=BC, AP=11 and CR=5. Find BQ.
A
B
Join AR to cut BQ at S
AB  BC
AP//BQ//CR
(given)
(given)
11
S
AS PQ AB



 1 (intercept theorem)
SR QR BC
 AS  SR and PQ  QR
In ARC ,
P
C
5
Q
R
In APR,
AB  BC
(given)
AS  SR
(proved)
AS  SR
PQ  QR
(proved)
(proved)
CR 5
 BS 
  2.5 (mid-pt theorem)  SQ  AP  11  5.5(mid-pt theorem
2
2
2
2
BQ=BS+SQ = 2.5+5.5=8
Example 16. AB and DC are straight lined. Find x and y.
(a) Proved: AB AC

A
D
PB QC
E
Join DE through A and // BC
DE//PQ//BC
(given)
AP AQ


PB QC
(intercept theorem)
AB  PB AC  QC

PB
QC
AB
AC
1 
1
PB
QC
AB AC

PB QC
P
Q
B
(b) AB=6, PB=2 and AQ=9. Find QC

AB AC

(proved)
PB QC
6 9  QC
 
2
QC
3QC  9  QC
2QC  9
QC  4.5
C
Example 16. Find QR and CD.
A
B 3
C
6
P
2Q
R
8 S
D
AP//BQ//CR (given)

AB PQ

BC QR
3
2

6 QR
26
3
4
QR 
(intercept theorem)
BQ//CR//DS (given)

BC QR

(intercept theorem)
CD RS
6
4

CD 8
86
CD 
4
 12