SlideShare a Scribd company logo
9011041155 / 9011031155

• Live Webinars (online lectures) with
recordings.
• Online Query Solving
• Online MCQ tests with detailed
solutions
• Online Notes and Solved Exercises
• Career Counseling

1
9011041155 / 9011031155

Quadrilaterals
We know that quadrilateral is a closed
figure with four sides. But by definition
suppose P, Q, R and S are four points
in a plane such that
i.

3 of these points are not collinear

ii. If sides PQ, QR, RS, SP are such that any of the
two such segments have a common point, it is an
endpoint only.
iii. If we draw a line containing any one of these
segments the remaining two points lie on the
same side of this line and
iv. If seg PR and seg QS intersect in the points other
than P, Q, R, S then the union of 4 segments
PQ, QR, RS, SP is called a quadrilateral.
is a symbol of ‘quadrilateral’.

2
9011041155 / 9011031155

(And Interior of the quadrilateral is a convex set
but quadrilateral is not a convex set)
Each quadrilateral has 4 sides, 4 angles, 4
vertices, 2 diagonals, 4 pairs of adjacent angles,
2 pairs of opposite sides, 4 pairs of adjacent or
consecutive sides and 2 pairs of opposite angles.
There are main 8 types of quadrilaterals
(as shown in the figures they are
i.

Parallelogram

ii. Rectangle
iii. Rhombus
iv. Square
v. Trapezium (non parallel sides are not
congruent)
vi. Isosceles trapezium
vii. Kite
viii. Kite (with all side, congruent)
3
9011041155 / 9011031155

Table
S. No. Quadrilateral

Type

1.

Parallelogram

2.

Rectangle

3.

Rhombus

4.

Square

4
9011041155 / 9011031155

5.

Trapezium

6.

Isosceles Trapezium

7.

Kite I

8.

Kite II

5
9011041155 / 9011031155

Property of a quadrilateral
Theorem 1
[Angle sum property of a quadrilateral]
The sum of the measures of all angles of a
quadrilateral is 360º
Given

:

To prove

:

Construction :

ABCD is a given quadrilateral

∠A + ∠B + ∠C + ∠D = 360º
Join the points A and C

6
9011041155 / 9011031155

Proof

:

In

ABC,

∠BAC + ∠ ABC + ∠BCA = 180º
.....sum of angles of
ABC .... (1)
and, In

ADC,

∠DAC + ∠ADC + ∠DCA = 180º
.....sum of angles of
ADC .... (2)

7
9011041155 / 9011031155

Adding (1) and (2) we get
∠BAC + ∠ABC + ∠BCA + ∠DAC
+ ∠ADC + ∠DCA = 180º + 180º
∴ ∠BAC + ∠DAC + ∠ABC + ∠BCA
+ ∠DCA + ∠ADC = 360º

......(3)

But ∠BAC + ∠DAC = ∠BAD
....Angle addition property....(4)
and ∠BCA + ∠DCA = ∠BCD
....Angle addition property....(5)
∴ from (3), (4) and (5) we get

8
9011041155 / 9011031155

∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º
i.e. ∠A + ∠B + ∠C + ∠D = 360º
Hence, sum of the measures of all
angles of a quadrilateral is 360º

9
9011041155 / 9011031155

Theorem 2
The opposite sides of a parallelogram are congruent.
Given

ABCD is a parallelogram in which

:

side AB || side CD and side BC || side
DA

:

Construction :
Proof

:

side AB

side CD and

side BC

To prove

side DA

Draw diagonal AC.
ABCD is a parallelogram in
which side AB || side CD and
seg AC is a transversals.
10
9011041155 / 9011031155

∴∠BAC

∠DCA ...(alternate angle)....(1)

Similarly,
side BC || side DA and seg AC is a
transversal.
∴ ∠BCA

∠DAC ......(alternate

angle) ...(2)
In

ABC and

∠BAC
seg AC
∠BCA
∴

CDA

∠DCA

......(from 1)

seg CA .......(common side)
∠ DAC

ABC

CDA

11

........(from 2)
.......(ASA test)
9011041155 / 9011031155

∴ side AB

side CD

and side BC

......(c.s.c.t)

side DA ......(c.s.c.t.)

Hence, the opposite sides of a
parallelogram are congruent.

12
9011041155 / 9011031155

Theorem 3
If opposite sides of a quadrilateral are congruent then
the quadrilateral is parallelogram.
Given

:

In

ABCD

side AB

side DC

and side AD

To prove

:

Construction :

side BC

ABCD is a parallelogram
Draw a diagonal BD

13
9011041155 / 9011031155

Proof

:

In

ABD and

CDB

side AB

side DC

side AD

side BC

side BD

side DB

......(given)

....(common side)
∴

ABD

CDB

∴ ∠ABD

∠CDB

.....(SSS test)
......(c.a.c.t.)

∴ side AB || side CD .......(1)
...(alternate angles test)

14
9011041155 / 9011031155

Similarly, we can prove that
side AD || side BC

......(2)

....(alternate angels test)
∴ from (1) and (2) we have
ABCD is a parallelogram
Hence, if opposite sides of a
quadrilateral are congruent then it is a
parallelogram.

15
9011041155 / 9011031155

Theorem 4
The opposite angles of a parallelogram are congruent
Given

PQRS is a parallelogram

:

To prove :

∠SPQ

∠QRS

and ∠PSR

Construction :
Proof

:

∠RQP

Draw a diagonal SQ
PQRS is a parallelogram
∴ side PS || side QR and seg SQ
is a transversal
∴ ∠PSQ

∠RQS
..(Alternate angles) ...(1)

16
9011041155 / 9011031155

Also, side PQ || side SR and
seg SQ is a transversal.
∴ ∠PQS

∠RSQ
...(Alternate angles)..(2)

In

PQS and

∠PSQ

RSQ

∠RQS ......from (1)

side SQ

side QS
....(common side)

∠PQS
∴

∠RSQ

PQS

......from (2)

RSQ
......(ASA test)

∴ ∠SPQ
17

∠QRS

......(c.a.c.t.)
9011041155 / 9011031155

Similarly, we can prove by drawing
diagonal PR.
∠PSR

∠ RQP

Hence, the opposite angles of a
parallelogram are congruent.

18
9011041155 / 9011031155

Theorem 5
A quadrilateral is a parallelogram if its opposite angles
are congruent.
Given

:

PQRS is a parallelogram in which
∠SPQ
∠PQR

To prove :
Proof

:

∠QRS and
∠RSP

PQRS is a parallelogram
Let ∠SPQ = ∠QRS = xº
And ∠PQR = ∠RSP = yº ......Opposite
angle of a quadrilateral.
19
9011041155 / 9011031155

∠SPQ + ∠PQR + ∠QRS + ∠RSP= 360º.
…(Angle sum property of a
quadrilateral)
∴ x + y + x + y = 360º
∴ 2x + 2y = 360º
∴ x + y = 180º

.....(dividing by 2)

∴ ∠SPQ + ∠RSP = 180º
∴ side PQ || side SR ... (interior angles
test) ....(1)

20
9011041155 / 9011031155

Similarly, we can prove that
side PS || side QR .......(2)
∴

PQRS is a parallelogram
...... from (1) and (2)
Hence, if opposite angles of a
quadrilateral are congruent, then it
is a parallelogram.

21
9011041155 / 9011031155

Theorem 6
The diagonals of a parallelogram bisect each other.
Given

:

ABCD is a parallelogram in which
the diagonals AC and BD intersect in M.

To prove :

seg AM

seg CM

and seg BM
Proof

:

since

seg DM

ABCD is a parallelogram

side AB || side CD and AC is a
transversal.
∴ ∠BAC

∠DCA ...(Alternate angles)

i.e. ∠BAM

∠DCM ...(A – M – C)..(1)
22
9011041155 / 9011031155

Also, side AB || side DC and seg DB is
a transversal.
∴ ∠ABD

∠CDB ...(Alternate angles)

i.e. ∠ABM

∠CDM ...(B – M – D) ..(2)

Now, In

ABM and

∠BAM

∠DCM ......(from 1)

side AB
∠ABM
∴

CDM

side DC .....(opposite sides)
∠CDM ......(from 2)

ABM

CDM ......(ASA test)

23
9011041155 / 9011031155

∴ seg AM

seg CM

and seg BM

.....(c.s.c.t.)

seg DM

Hence, diagonals of a parallelogram
bisect each other.

24
9011041155 / 9011031155

Theorem 7
If the diagonals of a quadrilateral bisect each other,
then the quadrilateral is a parallelogram.
Given

:

PQRS is a quadrilateral in which
diagonals PR and QS intersect in M.
seg PM
seg QM

To prove :

Proof

:

seg RM and
seg SM

PQRS is a parallelogram

In

PMQ and

seg PM

RMS

seg RM

25

......(given)
9011041155 / 9011031155

∠PMQ

∠RMS
.....(verticallyopposite angles)

seg QM
∴

seg SM .....(given)

PMQ

RMS

∴ ∠PQM

∠RSM

i.e. ∠PQS

......(SAS test)
......(c.a.c.t)

∠RSQ ......(S – M – Q)

∴ side PQ || side SR ....(alternate
angles test) ...(1)
Similarly, we can prove that
side PS || side QR

26

.....(2)
9011041155 / 9011031155

PQRS is a parallelogram
.........from (1) and (2)
Hence, if the diagonals of a
quadrilateral bisect each other then it
is a parallelogram.

27
9011041155 / 9011031155

Theorem 8
A quadrilateral is a parallelogram if a pair of opposite
sides is parallel and congruent.
Given

LMNK is a given quadrilateral in

:

which
side LM || side NK and
side LM

To prove

:

Construction :

side NK

LMNK is a parallelogram
Draw diagonal MK

28
9011041155 / 9011031155

Proof

:

since

LMNK is a quadrilateral in

which side LM || side NK
and seg MK is a transversal.
∴∠LMK

∠NKM
..(Alternate angles) ...(1)

Now, In

KLM and

MNK

seg LM

seg NK

.....(given)

∠LMK

∠NKM

seg KM

.......(from 1)

seg MK
.......(common side)

∴

KLM

MNK
......(SAS test)

29
9011041155 / 9011031155

∴ ∠LKM

∠NMK

.......(c.a.c.t)

∴ side LK || side MN
.....(alternate
angles test) ....(2)
and side LM || side NK
...(given) .....(3)
∴ from (2) and (3) we have

LKMN is

a parallelogram
Hence, if a pair of opposite sides of a
quadrilateral is parallel and congruent
then the quadrilateral is a parallelogram.

30
9011041155 / 9011031155

Sums

1. The angles of a quadrilateral are in the ratio
2 : 3 : 5 : 8. Find all the angles of the quadrilateral.
Sol.
The angles of the quadrilateral are in ratio
2 : 3 : 5 : 8.
Let the measures of the four angles are 2xº, 3xº,
5xº and 8xº.
According to the angle sum property of a
quadrilateral, we have
2xº + 3xº + 5xº + 8xº = 360º
∴ 18xº = 360º

3600
∴x =
18
0

∴ x 0 = 200

2xº = 2 × 20º = 40º, 3xº = 3 × 20º = 60º,
5xº = 5 × 20º = 100º and 8xº = 8 × 20º = 160º
∴

The measures of the angles of the
quadrilateral are 40º, 60º, 100º and 160º.
31
9011041155 / 9011031155

2. In the figure, PQRS is a quadrilateral. The
bisectors of ∠P and ∠Q meet at the point A.
∠S = 50º, ∠R = 110º. Find the measure of∠PAQ.

Sol.
Refer above fig.
According to the angle sum property of a
quadrilateral,
∠SPQ + ∠PQR + ∠R + ∠S = 360º
∴ ∠SPQ + ∠PQR + 110º + 50º = 360º
… (substituting the given values)
32
9011041155 / 9011031155

∴ ∠SPQ + ∠PQR + 160º = 360º
∴ ∠SPQ + ∠PQR = 360º - 160º
∴ ∠SPQ + ∠PQR = 200º

… (1)

Now, it is given that Ray PA and ray QA are the
bisectors of ∠SPQ and ∠RQP respectively.
∴ ∠APQ = ½ ∠SPQ and ∠AQP = ½ ∠RQP
∴ ∠APQ + ∠AQP = ½ (∠SPQ + ∠RQP) … (2)
From (2) and (1), ∠APQ + ∠AQP = ½ (200º)
∴ ∠APQ + ∠AQP = 100º
33

… (3)
9011041155 / 9011031155

Also, In APQ, ∠PAQ + ∠APQ + ∠AQP = 180º
… (Angles of a triangle)
∴ ∠PAQ + 100º = 180º

… [From (3)]

∴ ∠PAQ = 180º - 100º

∴ ∠PAQ = 80º

∴The measure of ∠PAQ is 80º.

34
9011041155 / 9011031155

3. The sides BA and DC of quadrilateral ABCD are
extended as shown in the figure. Prove that
m + n = p + q.

Proof :

Consider ∠DAB b = a and ∠BCD = c.
According to the angle sum property of
a quadrilateral,
In quadrilateral ABCD,
a + q + c + p = 360º

35

… (1)
9011041155 / 9011031155

From (2), m + c + n + a = 360º … (3)
From (1) and (3),
a+q+c+p=m+c+n+a
∴q+p=m+n
i.e. m + n = p + q.

36
9011041155 / 9011031155

4. In a parallelogram ABCD, ∠A = xº,
∠B = (3x + 20)º. Find x, ∠C and ∠D.

Sol.
We know that the adjacent angles of a
parallelogram are supplementary.
∴ ∠A + ∠B = 180º
Put the given values
∴ xº + 3xº + 20º = 180º
∴ 4xº = 180º - 20º
∴ 4xº = 160º
37
9011041155 / 9011031155

∴ xº = 40º
3xº + 20º = 3 × 40º + 20º = 120º + 20º = 140º
∴ ∠A = 40º and ∠B = 140º
Also, the opposite angles of a parallelogram are
congruent.
∴ ∠C = ∠ A = 40º and ∠ D = ∠ B = 140º
∴ x = 40. ∠ C = 40º, ∠ D = 140º.

38
9011041155 / 9011031155

5. The perimeter of a parallelogram is 150 cm. One
of its sides is greater than the other by 25 cm. Find
the lengths of all the sides of the parallelogram.
Sol:
Consider that the smaller of the two adjacent sides
be x cm.
Then, it is given that the other side is (x + 25) cm
Now, the perimeter of a parallelogram is equal to
the sum of the lengths of four sides.
∴

x + (x + 25) + x + (x + 25) = 150

∴

4x + 50 = 150

∴

x = 25 and x + 25 = 25 + 25 = 50

∴

The lengths of the sides of the parallelogram

∴ 4x = 150 – 50 ∴ 4x = 100

are 25 cm, 50 cm, 25 cm and 50 cm.

39
9011041155 / 9011031155

6. In the figure,

WXYZ is a parallelogram. From the

information given in the figure, find the values of x
and y.

Sol.
As XY & ZW are opposite sides of a parallelogram
side XY || side ZW
seg XZ is a transversal.
∴ ∠ ZXY = ∠ XZW
∴ 4y = 28º

… (alternate angles)
… (given)

∴ y = 7º
40
9011041155 / 9011031155

Similarly, side XW || side YZ and seg XZ is the
transversal.
∴ ∠ ZXW = ∠ XZY.
∴ 10x = 60º

… (given)

∴ x = 6º.
∴The value of x is 6º and that of y is 7º.

41
9011041155 / 9011031155

7. Prove that two opposite vertices of a parallelogram
are equidistant from the diagonal not containing
these vertices.
Given :
1.

ABCD is a parallelogram.

2. Seg AM ⊥ diagonal BD and
Seg CN ⊥ diagonal BD.

We have to prove seg AM ≅ seg CN.
First, draw diagonal AC intersecting diagonal BD
in the point P.

42
9011041155 / 9011031155

Proof :
P is the point of intersection of the diagonals of
parallelogram ABCD.
∴

seg AP ≅ seg CP

… (1)

In APM and CPN,
seg AP ≅ seg CP

… [From (1)]

∠ APM ≅ ∠ CPN
… (vertically opposite angles)
∠ AMP ≅ ∠CNP

… (each a right angle)

43
9011041155 / 9011031155

∴ APM ≅ CPN

… (SAA test)

∴ seg AM ≅ seg CN

… (c.s.c.t.)

Therefore, the opposite vertices of a parallelogram
are at equal distance from the diagonal not
containing these vertices.

44
9011041155 / 9011031155

8. The ratio of two sides of a parallelogram is 3 : 4. If
its perimeter is 112 cm, find the lengths of the
sides of the parallelogram.
Sol:
It is given that the ratio of the lengths of two sides
of the parallelogram is 3:4.
Let the lengths of those two sides be 3x cm and 4x
cm respectively.
We know that the opposite sides of a
parallelogram are congruent.
∴ The sides of the parallelogram are 3x cm, 4x cm,
3x cm and 4x cm.
Now, the perimeter of a parallelogram = the sum
of the lengths of four sides
∴

112 = 3x + 4x + 3x + 4x
… (given perimeter 112 cm)
45
9011041155 / 9011031155

∴

14x = 112

∴

x=8
3x = 3 × 8 = 24 and 4x = 4 × 8 = 32

∴

The lengths of the sides of the given
parallelogram are 24 cm, 32 cm, 24 cm and 32
cm.

46
9011041155 / 9011031155

Definitions
Parallelogram
A quadrilateral is a parallelogram if its opposite sides
are parallel.
Rectangle
A parallelogram (or a quadrilateral) in which each
angle is a right angle, is called a rectangle.
Rhombus
A quadrilateral having all sides congruent is called a
rhombus.
Square
A quadrilateral is called a square if all its sides and
angles are congruent.

47
9011041155 / 9011031155

Trapezium
If only one pair of opposite sides is parallel, the
quadrilateral is said to be a trapezium.
Isosceles trapezium
A trapezium in which nonparallel sides are congruent,
it is called as isosceles trapezium.
Kite
If in

ABCD, AB

AD and CB

CD and diagonal

AC is perpendicular bisector of diagonal BD, then
ABCD is called as a kite.

48
9011041155 / 9011031155

Theorem 9
Diagonals of a rectangle are congruent.

Theorem 10
If diagonals of a parallelogram are congruent then it is
a rectangle.

49
9011041155 / 9011031155

Theorem 11
Diagonals of a rhombus are perpendicular bisectors of
each other.

Theorem 12
If the diagonals of a quadrilateral bisect each other at
right angle then the quadrilateral is a rhombus.

50
9011041155 / 9011031155

Theorem 13
Diagonals of a square are congruent and
perpendicular bisectors of each other.

Theorem 14
If diagonals of quadrilateral are congruent and
perpendicular bisectors of each other then it is square.

51
9011041155 / 9011031155

Sums
1. Adjacent sides of a rectangle are of lengths 7 cm
and 24 cm. Find the lengths of its diagonals.
Sol.
Consider

ABCD as a rectangle in which

AB = 24 cm and BC = 7 cm.

In right-angled ABC,
By Pythagoras’ theorem,
AC² = AB² + BC²
= (24)² + (7)² = 576 + 49 = 625
∴ AC = 25

52
9011041155 / 9011031155

But, the diagonals of a rectangle are congruent.
∴ BD = AC = 25
∴ The length of diagonal AC = 25 cm and that of
BD = 25 cm.

53
9011041155 / 9011031155

2.

ABCD is a trapezium in which AB || DC. M and
N are the midpoints of side AD and side BC
respectively. If AB = 12 cm and MN = 14 cm, find
CD.

Sol.
The length of the segment joining the midpoints of
non - parallel sides of a trapezium is half the sum
of the lengths of its parallel sides.

∴ MN = ½ (AB + CD)
∴ 14 = ½ (12 + CD)
… (substituting the given values)
54
9011041155 / 9011031155

∴

28 = 12 + CD
… (Multiplying both the sides by 2)

∴ CD = 28 – 12
∴ CD = 16.

55
9011041155 / 9011031155

3. Prove that the base angles of an isosceles
trapezium are congruent.
Given

:

□ ABCD is an isosceles
trapezium in which
side AB || side CD and
side AD

To prove

:

Construction :

∠ADC

side BC.
∠BCD.

Draw seg BE || side AD and
D-E-C.

56
9011041155 / 9011031155

Proof

:

IN □ABED, side AB || side DE.
… (given)
seg BE || side AD
… (construction )
∴ □ABED is a parallelogram.
The opposite sides and the
opposite angles of a
parallelogram are congruent.
∴ side AD
and ∠D

57

side BE … (1)
∠ABE

… (2)
9011041155 / 9011031155

side AD ≅ side BC … (given)
… (3)
From (1) and (3),
side BE ≅ side BC
∴ ∠ BEC ≅ ∠C …(Isosceles
triangle theorem) … (4)
side AB || side DC

… (given)

side BE is the transversal.
∴ ∠ABE ≅ ∠BEC
… (alternate angles) … (5)
From (2), (4) and (5),
∠D ≅ ∠C.
i.e. ∠ADC ≅ ∠BCD.
[Similarly, we can prove that
∠DAB
58

∠CBA.]
9011041155 / 9011031155

4. The lengths of the diagonals PR and QS of a
rhombus are 20 cm and 48 cm respectively. Find
the length of each side of the rhombus.
Sol.
We know that the diagonals of a rhombus bisect
each other at right angles.

∴ MQ = ½ QS = ½ × 48 cm
∴ MQ = 24 cm
MP = ½ PR = ½ × 20 cm
∴ MP = 10 cm.

∠ PMQ = 90º

59
9011041155 / 9011031155

In right-angled PMQ,
By Pythagoras’ theorem
PQ² = PM² + QM²
= (10)² + (24)²
= 100 + 576
= 676
∴ PQ = 26.
But all the sides of a rhombus are congruent.
∴ The length of each side of the given rhombus
is 26 cm.

60
9011041155 / 9011031155

5.

ABCD is a kite in which diagonal AC and
diagonal BD intersect at point O.
∠ OBC = 20º and ∠OCD = 40º.
Find : i. ∠ABC

ii. ∠ADC

iii. ∠BAD.

Sol.
Refer above fig,
The diagonal joining the unequal sides of a kite
bisects the other diagonal at right angles.
∴ Each angle at O is a right angle.
Also, side BA

side BC.
61
9011041155 / 9011031155

Side DA

side DC and seg AO

seg CO

… (perpendicular bisector theorem)
… (1)
In BOC, ∠BOC = 90º, ∠OBC = 20º
… (given)
∴ ∠OCB = 180º - (90º + 20º)
= 180º - 110º
∴ ∠OCB = 70º

… (2)

62
9011041155 / 9011031155

In BAC, BA = BC

… [From (1)]

∴ ∠BCA = ∠BAC

… (angles opposite to
equal sides)

i.e. ∠OCB = ∠OAB
∴ ∠OAB = 70º

… [From (2)]

In DAC, DA = DC
∴ ∠DCA = ∠DAC

… [From (1)]

… (angles opposite to
equal sides)

i.e. ∠DCO = ∠DAO

63

… (3)
9011041155 / 9011031155

∴ ∠DAO = 40º
… [Given : ∠DCO = 40º]

… (4)

From (3) and (4),
∠OAB + ∠DAO = 70º + 40º = 110º … (5)
∠BAO + ∠DAO = ∠BAD
… (Angle addition postulate)

… (6)

From (5) and (6),
∠BAD = 110º.
In DAC, ∠ADC + ∠DAC + ∠DCA = 180º
… (angles of a triangle)
∴ ∠ ADC + 40º + 40º = 180º
… [From (4) and given]
64
9011041155 / 9011031155

∴ ∠ADC = 180º - 80º
∴ ∠ADC = 100º

… (7)

In BAC, ∠ABC + ∠BAC + ∠BCA = 180º
… (angles of a triangle)
∴ ∠ABC + 70º + 70º = 180º
… [From (2) and 3)]
∴ ∠ABC = 180º - 140º
∴ ∠ABC = 40º
Ans. : i. ∠ABC = 40º

ii. ∠ADC = 100º

iii. ∠BAD = 110º.
65
9011041155 / 9011031155

Theorem 15
Diagonals of an isosceles trapezium are congruent.
Given

ABCD is an isosceles trapezium in

:

which seg AC and seg BD are
diagonals side AD

side BC and side

AB || side DC

To Prove

:

Construction :

seg AC

seg BD

Draw seg AM ⊥ seg DC,
seg BN ⊥ seg DC

Proof

:

In ∆AMD and ∆BNC
side AD

side BC
66

….(given)
9011041155 / 9011031155

∠AMD
seg AM

∠BNC

…..(90º each)

seg BN …..(perpendicular
distance between
two parallel lines)

∴ ∆AMD

∆BNC ……(hypotenuse
side test)

∴ ∠ADM

∠BCN …..(c. a. c. t.)

i.e. ∠ADC

∠BCD .(D - M - C)…(1)

Now, In ∆ADC and ∆BCD
side AD
∠ADC

side BC
∠BCD

67

….(given)
…..(from (1))
9011041155 / 9011031155

side DC
∴ ∆ADC

side CD …..(common side)
∆BCD

∴ seg AC

…..(SAS test)

seg BD ……(c.s.c.t.)

Hence, diagonals of an isosceles
trapezium are congruent.

68
9011041155 / 9011031155

Theorem 16
(Intercepts made, by three parallel lines)
If three parallel lines make congruent intercepts on a
transversal then they make congruent intercepts on
any other transversal.
Given

:

line

|| line m || line n

and lines t1 and t2 are transversals.
seg AB

Prove

:

Construction :

seg BC

seg DE

seg EF

Draw a line GI parallel to the line t1
through E.
69
9011041155 / 9011031155

Proof

:

seg AG || seg BE ……(given)
seg AB || seg GE ……. (construction)
∴

ABEG is a parallelogram.

∴ seg AB

seg GE…(opposite sides
of a parallelogram …(1)

Similarly,
∴ seg BC

BCIE is a parallelogram.
seg EI …..(opposite sides
of a parallelogram….(2)

But seg AB

seg BC ….. (given)...(3)

Hence seg GE

seg EI ….(4) from
(1), (2) and (3)

70
9011041155 / 9011031155

Now, In ∆GED and ∆IEF
∠DGE

∠FIE …(alternate angles)

seg GE
∠GED

seg EI …..(from (4))
∠IEF

……(vertically
opposite angles)

∴ ∆GED
∴ seg DE

∆IEF ……(ASA test)
seg EF ……(c.s.c.t.)

Hence, if three parallel lines make
congruent intercepts on a transversal
then they make congruent intercepts
on any other transversal.
71
9011041155 / 9011031155

Theorem 17
Mid - Point theorem
The line segment joining the midpoints of any two
sides of a triangle is parallel to the third side and is
half of it.
Given

:

In ∆ABC, P and Q are midpoints of
sides AB and AC respectively.

To prove :

i.

seg PQ || side BC

ii. seg PQ =

1
side BC
2

Construction : Take the point R on ray PQ such that
seg PQ

seg QR. Join C and R
72
9011041155 / 9011031155

Proof

:

In ∆AQP and ∆CQR
seg AQ
∠AQP

seg CQ ……(given)
∠CQR ….(vertically opposite
angles)

seg PQ

seg RQ ……(construction)

∴ ∆AQP

∆CQR

…… (SAS test)

seg AP

seg CR ……(c.s.c.t.)

seg BP

seg CR ...(

and ∠PAQ

∠RCQ

∴ seg AP || seg CR …

AP = PB)…(1)
……(c.a.c.t)
(alternate
angles test)

∴ seg BP || seg CR …(AP = PB)....(2)
73
9011041155 / 9011031155

PBCR is a parallelogram (from (1)
and (2))
∴ side PR || side BC …..(opposite
sides of a parallelogram)
seg PQ || side BC

……(P - Q - P)

1
Also, seg PQ = side PR
2
…(construction)

∴ seg PQ =

1
side BC …(
2

PR = BC)

Hence, In a triangle, line segment
joining mid - points of any two sides is
parallel to the third side and is half of
it.
74
9011041155 / 9011031155

Theorem 18
Converse of mid - point theorem
If a line drawn through the mid - point of one side of a
triangle is parallel to second side then it bisects the
third side.

In ∆ABC, A - D - B if AD = DB and line DE || side BC
then, line DE bisects side AC. i.e. AE = CE
Try to write the proof.

75
9011041155 / 9011031155

Sums
14. In the figure, □PQRS and □LMNR are rectangles,
where M is the midpoint of PR. Prove that:
i. SL = LR
ii. LN = ½ SQ.

Proof

:

In above fig.
□LMNR is a rectangle.
∴ LM || RN

… (Opposite

sides of a rectangle)
i.e. LM || RQ

… (R-N-Q)
… (1)

76
9011041155 / 9011031155

RQ || SP

… (opposite sides
of a rectangle)

…(2)

From (1) and (2), we have
LM || RQ || SP

… (3)

In PSR, it is given that M is the
midpoint of PR and LM || RQ
… [From (1)]
∴ By the converse of midpoint
theorem, point L is the midpoint of
side SR.
∴ SL = LR

77

… (4)
…i
9011041155 / 9011031155

Again, LR || MN and LR || PQ
… (opposite sides of
rectangle LMNR and
rectangle PQRS
respectively)
∴ MN || PQ

… (property for
parallel sides)

∴ In RPQ,
M is the midpoint of PR.

78
9011041155 / 9011031155

∴ By the converse of midpoint
theorem,
point N is the midpoint of RQ.
… (5)
From (4) and (5), L and N are the
midpoints of SR and RQ
respectively.,
∴ In RSQ, by midpoint theorem,
LN = ½ SQ
Hence proved.

79

… ii
9011041155 / 9011031155

15. In the figure, seg PD is the median of PQR. Point

PM 1
G is the midpoint of seg PD. Show that
= .
PR 3

Proof

:

Draw seg DN || seg QM.
We are given that, In PDN, G is
the midpoint of seg PD.
Seg QM (i.e. seg GM) || seg DN
… (by construction)

80
9011041155 / 9011031155

∴ By the converse of midpoint
theorem, M is the midpoint of seg
PN.
∴ PM = MN

… (1)

In QRM, As PD is median, D is the
midpoint of side QR
And seg QM || seg DN
… (construction)
∴ By the converse of midpoint
theorem, N is the midpoint of
seg MR.
∴ MN = NR
81

… (2)
9011041155 / 9011031155

From (1) and (2), PM = MN = NR
… (3)
Now, PR = PM + MN + NR
… (4)
∴ PR = 3PM
… [From (3) and (4)]

i.e. 3PM = PR

82

PM 1
∴
= .
PR 3
9011041155 / 9011031155

16. Prove that a diagonal of a rhombus bisects two
opposite angles.

Proof :

Refer above fig.
ABCD is a rhombus & AC is its
diagonal.
In ABC and ADC,
side AB
side BC

side AD and
side DC
… (sides of a rhombus)

side AC

side AC
… (common side)

∴ ABC

ADC
83

… (SSS test)
9011041155 / 9011031155

∴ ∠BAC
and ∠BCA

∠DAC

… (1)

∠DCA

… (2)
… (c.a.c.t.)

seg AC bisects ∠BAD.
…[from(1)]
seg AC bisects ∠BCD.
…[From(2)]
Therefore diagonal of a rhombus
bisects two opposite angels.

84
9011041155 / 9011031155

17. In the figure,

BHIR is a kite. If ∠HIR = 50º, find a.

∠HRI and b. ∠BHI.

Sol.
Consider abive fig. □BHIR is a kite.
a. In IRH, side IR = side IH
∴ ∠HRI = ∠RHI

… (given)

… (angles opposite to
equal sides)

… (1)

∠I + ∠HRI + ∠RHI = 180º
… (angles of a triangle)
85
9011041155 / 9011031155

∴ 50º + ∠HRI + ∠HRI = 180º
… [Given and from (1)]
∴ 2∠HRI = 180º - 50º
∴ 2∠HRI = 130º
∴ ∠HRI = 65º

… (2)

∴ ∠RHI = 65º … [From (1) and (2)] … (3)

86
9011041155 / 9011031155

b.

BHR is an equilateral triangle.
… (given)
∴ ∠BHR = 60º

… (angle of an

equilateral triangle)

… (4)

∠BHI = ∠BHR + ∠RHI
… (angle addition postulate)
= 60º + 65º … [From (4) and (3)]
∴ ∠BHI = 125º

87
9011041155 / 9011031155

18. In CBS, seg BC

seg SC. Ray CE bisects

exterior ∠DCS. Ray SE || ray BC. Prove that
CBSE is a parallelogram.

Proof :

We are given that
seg CB
∴ ∠CBS

seg CS
∠CSB
… (angles opposite to
congruent sides) … (1)

∠DCS is the exterior angle of CBS.
88
9011041155 / 9011031155

∴ ∠DCS = ∠CBS + ∠CSB
… (Remote interior angle
theorem)

… (2)

From (1) and (2), .
∠CBS = ∠CSB = ½∠DCS
… (3)
Now, it is given that Ray CE bisects
∠DCS.
∴ ∠DCE = ∠ECS = ½∠DCS
… (4)

89
9011041155 / 9011031155

From (3) and (4), ∠CSB = ∠ECS
∴ seg BS || seg CE
… (Alternate angles test
for parallel lines)
seg SE || seg BC … (given)
∴ Both the pairs of opposite sides
of
∴

CBSE are parallel.

CBSE is a parallelogram.

90
9011041155 / 9011031155

19. In the figure,

LAXM is a parallelogram. Point I is

the midpoint of diagonal LX. PQ is a line passing
through I. P and Q are the points of intersection
with the sides LA and MX respectively.
Prove that seg PI

Proof

:

seg IQ.

In LPI and XQI,
∠LIP & ∠XIQ are vertically
opposite angles
∴∠LIP

∠XIQ

seg LI

seg XI

… (As I is the midpoint of LX)
91
9011041155 / 9011031155

∠PLI

∠QXI
… (Alternate angles)

∴ LPI

XQI

∴ seg PI

seg QI

Hence proved.

92

… (ASA test)
… (c.s.c.t.)
9011041155 / 9011031155

20.

MNOP is a rhombus. Q is a point in the interior
of the rhombus such that QM = QO. Prove that Q
lies on diagonal NP.

Proof :

In MPN and OPN,
seg PM

seg PO and

seg MN

seg ON
… (sides of a rhombus)

seg PN

sg PN … (common side)

∴By SSS test
∴ MPN

OPN
93
9011041155 / 9011031155

∴ ∠MPN

∠OPN

… (c.a.c.t.)

i.e. diagonal PN is the bisector of
∠MPO … (1)
In MPQ and OPQ,
seg MP

seg OP
… (sides of a rhombus)

seg QM

sg QO

… (given)

seg PQ

seg PQ … (common
side)

∴ By SSS test
∴ MPQ

OPQ

∴∠MPQ

∠OPQ

94

… (c.a.c.t.)
9011041155 / 9011031155

i.e. seg PQ is the bisector of ∠MPO
… (2)
Point Q is in the interior of

MNOP.

… (3)
The bisector of an angle is unique.
∴From (1), (2) and (3) bisectors PN
and PQ of ∠MPO are one and the
same.
∴ Point Q lies on the diagonal NP.
Hence proved.

95
9011041155 / 9011031155

21.

ABCD is a square. P and Q are the points such
that seg AQ

seg DP.

Prove that seg AQ ⊥ seg DP.

Proof :

In DAP and ABQ,
∠DAP

∠ABQ
… (Each is a right angle)

It is given that
hypotenuse DP

hypotenuse AQ

side DA ≅ side AB
… (side of a square)
∴By hypotenuse-side theorem,
96
9011041155 / 9011031155

∴ DAP

ABQ

∴ ∠DPA

∠AQB

… (c.a.c.t.)
and ∠ADP

… (1)

∠BAQ
… (c.a.c.t.)

… (2)

From (1), Assume
∠DPA = ∠AQB = α

… (3)

From (2), Assume
∠ADP = ∠BAQ = β

97

… (4)
9011041155 / 9011031155

Then, In DAP,
α+β+

DAP = 180º

… (Angles of a triangle)
∴ α + β + 90º = 180º
… (∠DAP = 90º)
∴ α +β = 90º

… (5)

Now, In APT,
α + β + ∠ATP = 180º
… (Angles of a triangle)
∴ 90º + ∠ATP = 180º
… [From (5)]
∴ ∠ATP = 90º
98
9011041155 / 9011031155

Point T is the point of
intersection of seg AQ and seg
PD.
∴ seg AQ ⊥ seg DP.
Hence proved!!

99
9011041155 / 9011031155

22.

ABCD is a kite. AB = AD and CB = CD. Prove
that
i.

diagonal AC ⊥ diagonal BD.

ii. diagonal AC bisects diagonal BD.

Proof :

We are given, AB = AD
∴ Point A is equidistant from points
B and D of seg BD.
Also, CB = CD

… (1)
… (given)

∴ Point C is equidistant from points B
and D of seg BD.
100

… (2)
9011041155 / 9011031155

From (1) and (2), points A and C are
equidistant from points B and D of
seg BD.
Therefore by perpendicular bisector
theorem, AC is the perpendicular
bisector of BD.
i.e.
i. diagonal AC ⊥ diagonal BD and
ii. diagonal AC bisects diagonal BD.
Hence proved!!

101
9011041155 / 9011031155

23. Let points A and B be on one side of line ℓ. Draw
seg AD ⊥ line ℓ and seg BE ⊥ line ℓ. Let point C
be the midpoint of seg AB. Prove that
seg CD

seg CE.

Construction :

Draw seg CM ⊥ lin l.

Proof

Refer above fig.

:

We can have,
Seg AD, seg BE and seg CM
are perpendiculars to line l
∴ seg AD || seg BE || seg CM.
102
9011041155 / 9011031155

Seg CA

seg CB

… (As C is the
midpoint of seg AB)
Now, Seg CA and seg CB
are the intercepts made by
three parallel lines AD, BE
and CM.
As DM & ME are intercepts
made by the same parallel
lines on other transversal,
we have
∴ intercept DM

103

intercept ME
9011041155 / 9011031155

i.e. seg DM

seg ME

… (1)

In CDM and CEM,
seg DM

seg EM
… [From (1)]

∠CMD

∠CME
… (Each a right
angle : construction)

seg CM

seg CM
… (common side)

∴ CDM

CEM
… (SAS test)

∴ seg CD
104

seg CE … (c.s.c.t.)
9011041155 / 9011031155

24. □ABCD is a parallelogram. P, Q, R and S are the
points on sides AB, BC, CD and DA respectively
such that seg AP
Prove that

Proof :

seg BQ

seg CR

seg DS.

PQRS is a parallelogram.

AB = AP + PB

… (A-P-B)

… (1)

CD = CR + RD

… (C-R-D)

… (2)

side AB

side CD

… (opposite sides of a parallelogram)
∴ AB = CD

… (3)

105
9011041155 / 9011031155

From (1), (2) and (3),
AP + PB = CR + RD

… (4)

seg AP

… (given)

seg CR

∴ AP = CR

… (5)

From (4) and (5), PB = RD
∴ seg PB

seg RD

… (6)

In PBQ and RDS,
seg PB

seg RD

… [From (6)]

Now, ∠B &∠D are opposite angles of a
parallelogram
∴∠B

∠D

106
9011041155 / 9011031155

seg BQ

seg DS

… (given)

∴By SAS test
∴ PBQ

RDS

∴ seg PQ ≅ seg RS … (c.s.c.t.)
Similarly, seg PS
proved.

seg QR can be
… (8)

From (7) and (8), opposite sides of
PQRS are congruent.
∴

PQRS is a parallelogram.

Hence proved!!

107
9011041155 / 9011031155

Ask Your Doubts
call
For inquiry and registration, call 9011041155
/ 9011031155.

108

More Related Content

What's hot

Math
Math Math
Math
sarah321
 
Maths herons formula
Maths   herons formulaMaths   herons formula
Maths herons formula
PODAR INTERNATIONAL SCHOOL
 
Rational numbers ppt
Rational numbers pptRational numbers ppt
Rational numbers ppt
Mathukutty1
 
Magic square
Magic squareMagic square
Magic square
Meeran Banday
 
Cubes & cuboids
Cubes & cuboidsCubes & cuboids
Cubes & cuboids
Srilekha G
 
Career opportunities after 10th std (Career after Class 10)
Career opportunities after 10th std (Career after Class 10)Career opportunities after 10th std (Career after Class 10)
Career opportunities after 10th std (Career after Class 10)
Ritika Dhameja
 
The Comprehensive Guide on Branches of Mathematics
The Comprehensive Guide on Branches of MathematicsThe Comprehensive Guide on Branches of Mathematics
The Comprehensive Guide on Branches of Mathematics
Stat Analytica
 
Hcf & lcm
Hcf & lcmHcf & lcm
Hcf & lcm
weerawattk
 
Playing with numbers Class-8
Playing with numbers Class-8Playing with numbers Class-8
Playing with numbers Class-8
Javed Alam
 
Number System
Number SystemNumber System
Number System
samarthagrawal
 
Introduction 3D shapes
Introduction 3D shapesIntroduction 3D shapes
Introduction 3D shapes
Inne Milanisti Sihasale
 
The Evolution of the Number System
The Evolution of the Number System  The Evolution of the Number System
The Evolution of the Number System
immanueljohnisaac
 
Area and its boundary
Area and its boundaryArea and its boundary
Area and its boundary
Snju Lehri
 
Lesson plan multiple and factors.ppt v 3
Lesson plan  multiple and factors.ppt v 3Lesson plan  multiple and factors.ppt v 3
Lesson plan multiple and factors.ppt v 3
Kavita Grover
 
number system ppt
number system ppt number system ppt
number system ppt
Akash dixit
 
Quadrilaterals
QuadrilateralsQuadrilaterals
Quadrilaterals
Home
 
Whole brain teaching
Whole brain teachingWhole brain teaching
Whole brain teaching
lmaad Schools Lekki
 

What's hot (17)

Math
Math Math
Math
 
Maths herons formula
Maths   herons formulaMaths   herons formula
Maths herons formula
 
Rational numbers ppt
Rational numbers pptRational numbers ppt
Rational numbers ppt
 
Magic square
Magic squareMagic square
Magic square
 
Cubes & cuboids
Cubes & cuboidsCubes & cuboids
Cubes & cuboids
 
Career opportunities after 10th std (Career after Class 10)
Career opportunities after 10th std (Career after Class 10)Career opportunities after 10th std (Career after Class 10)
Career opportunities after 10th std (Career after Class 10)
 
The Comprehensive Guide on Branches of Mathematics
The Comprehensive Guide on Branches of MathematicsThe Comprehensive Guide on Branches of Mathematics
The Comprehensive Guide on Branches of Mathematics
 
Hcf & lcm
Hcf & lcmHcf & lcm
Hcf & lcm
 
Playing with numbers Class-8
Playing with numbers Class-8Playing with numbers Class-8
Playing with numbers Class-8
 
Number System
Number SystemNumber System
Number System
 
Introduction 3D shapes
Introduction 3D shapesIntroduction 3D shapes
Introduction 3D shapes
 
The Evolution of the Number System
The Evolution of the Number System  The Evolution of the Number System
The Evolution of the Number System
 
Area and its boundary
Area and its boundaryArea and its boundary
Area and its boundary
 
Lesson plan multiple and factors.ppt v 3
Lesson plan  multiple and factors.ppt v 3Lesson plan  multiple and factors.ppt v 3
Lesson plan multiple and factors.ppt v 3
 
number system ppt
number system ppt number system ppt
number system ppt
 
Quadrilaterals
QuadrilateralsQuadrilaterals
Quadrilaterals
 
Whole brain teaching
Whole brain teachingWhole brain teaching
Whole brain teaching
 

Similar to 9th Maths - Quadrilateral and Its Types

quadrilateral
quadrilateralquadrilateral
quadrilateral
Wendy Mendoza
 
Shivam goyal ix e
Shivam goyal ix eShivam goyal ix e
Shivam goyal ix e
shivamgoyal786
 
Heep205
Heep205Heep205
Circle
CircleCircle
Circle
letsenjoy
 
Slideshare
SlideshareSlideshare
Slideshare
letsenjoy
 
Angles in a circle and cyclic quadrilateral --GEOMETRY
Angles in a circle and cyclic quadrilateral  --GEOMETRYAngles in a circle and cyclic quadrilateral  --GEOMETRY
Angles in a circle and cyclic quadrilateral --GEOMETRY
indianeducation
 
Triangles
Triangles Triangles
Triangles
jatin bisht
 
Triangles For Class 10 CBSE NCERT
Triangles For Class 10 CBSE NCERTTriangles For Class 10 CBSE NCERT
Triangles For Class 10 CBSE NCERT
Let's Tute
 
Semana 1 geo y trigo
Semana 1  geo y trigo Semana 1  geo y trigo
Semana 1 geo y trigo
watson baltazar gonzales
 
Best International School Hyderabad
Best International School HyderabadBest International School Hyderabad
Best International School Hyderabad
OpenmindsHyderabad
 
PPT ON TRIANGLES FOR CLASS X
PPT ON TRIANGLES FOR CLASS XPPT ON TRIANGLES FOR CLASS X
PPT ON TRIANGLES FOR CLASS X
Miku09
 
ch6.pdf
ch6.pdfch6.pdf
ch6.pdf
irrfanalikhan
 
Circle 10 STB.pptx
Circle 10 STB.pptxCircle 10 STB.pptx
Circle 10 STB.pptx
Vinod Gupta
 
triangles class9.pptx
triangles class9.pptxtriangles class9.pptx
triangles class9.pptx
KirtiChauhan62
 
Triangle
TriangleTriangle
angles 7TH GRADE.pptx msth exercises and book
angles 7TH GRADE.pptx msth exercises and bookangles 7TH GRADE.pptx msth exercises and book
angles 7TH GRADE.pptx msth exercises and book
GuediaLozano
 
Geometry unit 6.4
Geometry unit 6.4Geometry unit 6.4
Geometry unit 6.4
Mark Ryder
 
Triangle
TriangleTriangle
mathsproject-
mathsproject-mathsproject-
mathsproject-
abhichowdary16
 
triangles ppt.pptx
triangles ppt.pptxtriangles ppt.pptx
triangles ppt.pptx
abhichowdary16
 

Similar to 9th Maths - Quadrilateral and Its Types (20)

quadrilateral
quadrilateralquadrilateral
quadrilateral
 
Shivam goyal ix e
Shivam goyal ix eShivam goyal ix e
Shivam goyal ix e
 
Heep205
Heep205Heep205
Heep205
 
Circle
CircleCircle
Circle
 
Slideshare
SlideshareSlideshare
Slideshare
 
Angles in a circle and cyclic quadrilateral --GEOMETRY
Angles in a circle and cyclic quadrilateral  --GEOMETRYAngles in a circle and cyclic quadrilateral  --GEOMETRY
Angles in a circle and cyclic quadrilateral --GEOMETRY
 
Triangles
Triangles Triangles
Triangles
 
Triangles For Class 10 CBSE NCERT
Triangles For Class 10 CBSE NCERTTriangles For Class 10 CBSE NCERT
Triangles For Class 10 CBSE NCERT
 
Semana 1 geo y trigo
Semana 1  geo y trigo Semana 1  geo y trigo
Semana 1 geo y trigo
 
Best International School Hyderabad
Best International School HyderabadBest International School Hyderabad
Best International School Hyderabad
 
PPT ON TRIANGLES FOR CLASS X
PPT ON TRIANGLES FOR CLASS XPPT ON TRIANGLES FOR CLASS X
PPT ON TRIANGLES FOR CLASS X
 
ch6.pdf
ch6.pdfch6.pdf
ch6.pdf
 
Circle 10 STB.pptx
Circle 10 STB.pptxCircle 10 STB.pptx
Circle 10 STB.pptx
 
triangles class9.pptx
triangles class9.pptxtriangles class9.pptx
triangles class9.pptx
 
Triangle
TriangleTriangle
Triangle
 
angles 7TH GRADE.pptx msth exercises and book
angles 7TH GRADE.pptx msth exercises and bookangles 7TH GRADE.pptx msth exercises and book
angles 7TH GRADE.pptx msth exercises and book
 
Geometry unit 6.4
Geometry unit 6.4Geometry unit 6.4
Geometry unit 6.4
 
Triangle
TriangleTriangle
Triangle
 
mathsproject-
mathsproject-mathsproject-
mathsproject-
 
triangles ppt.pptx
triangles ppt.pptxtriangles ppt.pptx
triangles ppt.pptx
 

More from Ednexa

Recommendation letters
Recommendation lettersRecommendation letters
Recommendation letters
Ednexa
 
Physics Measurements Notes for JEE Main 2015
Physics Measurements Notes for JEE Main 2015 Physics Measurements Notes for JEE Main 2015
Physics Measurements Notes for JEE Main 2015
Ednexa
 
Important Points on Elasticity for JEE Main 2015
Important Points on Elasticity for JEE Main 2015 Important Points on Elasticity for JEE Main 2015
Important Points on Elasticity for JEE Main 2015
Ednexa
 
Important points on Gravitation for JEE Main 2015
Important points on Gravitation for JEE Main 2015Important points on Gravitation for JEE Main 2015
Important points on Gravitation for JEE Main 2015
Ednexa
 
Current Electricity Notes for JEE Main 2015 - Part II
Current Electricity Notes for JEE Main 2015 - Part IICurrent Electricity Notes for JEE Main 2015 - Part II
Current Electricity Notes for JEE Main 2015 - Part II
Ednexa
 
Aldehydes Ketones and Carboxylic Acids - JEE Main 2015
Aldehydes Ketones and Carboxylic Acids - JEE Main 2015Aldehydes Ketones and Carboxylic Acids - JEE Main 2015
Aldehydes Ketones and Carboxylic Acids - JEE Main 2015
Ednexa
 
Physics - Current Electricity Notes for JEE Main 2015
Physics - Current Electricity Notes for JEE Main 2015Physics - Current Electricity Notes for JEE Main 2015
Physics - Current Electricity Notes for JEE Main 2015
Ednexa
 
Enhancement in Food Production Exercises - MH-CET 2015
Enhancement in Food Production Exercises - MH-CET 2015 Enhancement in Food Production Exercises - MH-CET 2015
Enhancement in Food Production Exercises - MH-CET 2015
Ednexa
 
Properties of Solids and Liquids Notes - JEE Main 2015
Properties of Solids and Liquids Notes - JEE Main 2015Properties of Solids and Liquids Notes - JEE Main 2015
Properties of Solids and Liquids Notes - JEE Main 2015
Ednexa
 
12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015 12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015
Ednexa
 
Organisms and Environment Exercise - MH-CET 2015
Organisms and Environment Exercise - MH-CET 2015Organisms and Environment Exercise - MH-CET 2015
Organisms and Environment Exercise - MH-CET 2015
Ednexa
 
Communication System Theory for JEE Main 2015
Communication System Theory for JEE Main 2015 Communication System Theory for JEE Main 2015
Communication System Theory for JEE Main 2015
Ednexa
 
Genetic Engineering and Genomics Notes - MH-CET 2015
Genetic Engineering and Genomics Notes - MH-CET 2015 Genetic Engineering and Genomics Notes - MH-CET 2015
Genetic Engineering and Genomics Notes - MH-CET 2015
Ednexa
 
Notes and Important Points on Circular Motion for JEE Main 2015
Notes and Important Points on Circular Motion for JEE Main 2015Notes and Important Points on Circular Motion for JEE Main 2015
Notes and Important Points on Circular Motion for JEE Main 2015
Ednexa
 
Notes and Important Points on Electrochemistry - JEE Main 2015
Notes and Important Points on Electrochemistry - JEE Main 2015Notes and Important Points on Electrochemistry - JEE Main 2015
Notes and Important Points on Electrochemistry - JEE Main 2015
Ednexa
 
Physics Sound and Waves for JEE Main 2015 - Part I
Physics Sound and Waves for JEE Main 2015 - Part IPhysics Sound and Waves for JEE Main 2015 - Part I
Physics Sound and Waves for JEE Main 2015 - Part I
Ednexa
 
Notes and Important Points on Solid State - JEE Main 2015
Notes and Important Points on Solid State - JEE Main 2015 Notes and Important Points on Solid State - JEE Main 2015
Notes and Important Points on Solid State - JEE Main 2015
Ednexa
 
Study material 12th Physics - Wave Theory of Light Part II
 Study material 12th Physics - Wave Theory of Light Part II  Study material 12th Physics - Wave Theory of Light Part II
Study material 12th Physics - Wave Theory of Light Part II
Ednexa
 
Physics in day to day life Notes for JEE Main 2015
Physics in day to day life Notes for JEE Main 2015 Physics in day to day life Notes for JEE Main 2015
Physics in day to day life Notes for JEE Main 2015
Ednexa
 
Summarized notes on Interference and Diffraction for JEE Main
Summarized notes on Interference and Diffraction for JEE MainSummarized notes on Interference and Diffraction for JEE Main
Summarized notes on Interference and Diffraction for JEE Main
Ednexa
 

More from Ednexa (20)

Recommendation letters
Recommendation lettersRecommendation letters
Recommendation letters
 
Physics Measurements Notes for JEE Main 2015
Physics Measurements Notes for JEE Main 2015 Physics Measurements Notes for JEE Main 2015
Physics Measurements Notes for JEE Main 2015
 
Important Points on Elasticity for JEE Main 2015
Important Points on Elasticity for JEE Main 2015 Important Points on Elasticity for JEE Main 2015
Important Points on Elasticity for JEE Main 2015
 
Important points on Gravitation for JEE Main 2015
Important points on Gravitation for JEE Main 2015Important points on Gravitation for JEE Main 2015
Important points on Gravitation for JEE Main 2015
 
Current Electricity Notes for JEE Main 2015 - Part II
Current Electricity Notes for JEE Main 2015 - Part IICurrent Electricity Notes for JEE Main 2015 - Part II
Current Electricity Notes for JEE Main 2015 - Part II
 
Aldehydes Ketones and Carboxylic Acids - JEE Main 2015
Aldehydes Ketones and Carboxylic Acids - JEE Main 2015Aldehydes Ketones and Carboxylic Acids - JEE Main 2015
Aldehydes Ketones and Carboxylic Acids - JEE Main 2015
 
Physics - Current Electricity Notes for JEE Main 2015
Physics - Current Electricity Notes for JEE Main 2015Physics - Current Electricity Notes for JEE Main 2015
Physics - Current Electricity Notes for JEE Main 2015
 
Enhancement in Food Production Exercises - MH-CET 2015
Enhancement in Food Production Exercises - MH-CET 2015 Enhancement in Food Production Exercises - MH-CET 2015
Enhancement in Food Production Exercises - MH-CET 2015
 
Properties of Solids and Liquids Notes - JEE Main 2015
Properties of Solids and Liquids Notes - JEE Main 2015Properties of Solids and Liquids Notes - JEE Main 2015
Properties of Solids and Liquids Notes - JEE Main 2015
 
12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015 12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015
 
Organisms and Environment Exercise - MH-CET 2015
Organisms and Environment Exercise - MH-CET 2015Organisms and Environment Exercise - MH-CET 2015
Organisms and Environment Exercise - MH-CET 2015
 
Communication System Theory for JEE Main 2015
Communication System Theory for JEE Main 2015 Communication System Theory for JEE Main 2015
Communication System Theory for JEE Main 2015
 
Genetic Engineering and Genomics Notes - MH-CET 2015
Genetic Engineering and Genomics Notes - MH-CET 2015 Genetic Engineering and Genomics Notes - MH-CET 2015
Genetic Engineering and Genomics Notes - MH-CET 2015
 
Notes and Important Points on Circular Motion for JEE Main 2015
Notes and Important Points on Circular Motion for JEE Main 2015Notes and Important Points on Circular Motion for JEE Main 2015
Notes and Important Points on Circular Motion for JEE Main 2015
 
Notes and Important Points on Electrochemistry - JEE Main 2015
Notes and Important Points on Electrochemistry - JEE Main 2015Notes and Important Points on Electrochemistry - JEE Main 2015
Notes and Important Points on Electrochemistry - JEE Main 2015
 
Physics Sound and Waves for JEE Main 2015 - Part I
Physics Sound and Waves for JEE Main 2015 - Part IPhysics Sound and Waves for JEE Main 2015 - Part I
Physics Sound and Waves for JEE Main 2015 - Part I
 
Notes and Important Points on Solid State - JEE Main 2015
Notes and Important Points on Solid State - JEE Main 2015 Notes and Important Points on Solid State - JEE Main 2015
Notes and Important Points on Solid State - JEE Main 2015
 
Study material 12th Physics - Wave Theory of Light Part II
 Study material 12th Physics - Wave Theory of Light Part II  Study material 12th Physics - Wave Theory of Light Part II
Study material 12th Physics - Wave Theory of Light Part II
 
Physics in day to day life Notes for JEE Main 2015
Physics in day to day life Notes for JEE Main 2015 Physics in day to day life Notes for JEE Main 2015
Physics in day to day life Notes for JEE Main 2015
 
Summarized notes on Interference and Diffraction for JEE Main
Summarized notes on Interference and Diffraction for JEE MainSummarized notes on Interference and Diffraction for JEE Main
Summarized notes on Interference and Diffraction for JEE Main
 

Recently uploaded

The basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptxThe basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptx
heathfieldcps1
 
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem studentsRHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
Himanshu Rai
 
clinical examination of hip joint (1).pdf
clinical examination of hip joint (1).pdfclinical examination of hip joint (1).pdf
clinical examination of hip joint (1).pdf
Priyankaranawat4
 
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
PECB
 
Walmart Business+ and Spark Good for Nonprofits.pdf
Walmart Business+ and Spark Good for Nonprofits.pdfWalmart Business+ and Spark Good for Nonprofits.pdf
Walmart Business+ and Spark Good for Nonprofits.pdf
TechSoup
 
PIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf IslamabadPIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf Islamabad
AyyanKhan40
 
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
National Information Standards Organization (NISO)
 
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
Nguyen Thanh Tu Collection
 
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptxChapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Denish Jangid
 
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
GeorgeMilliken2
 
Cognitive Development Adolescence Psychology
Cognitive Development Adolescence PsychologyCognitive Development Adolescence Psychology
Cognitive Development Adolescence Psychology
paigestewart1632
 
Wound healing PPT
Wound healing PPTWound healing PPT
Wound healing PPT
Jyoti Chand
 
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptxPengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Fajar Baskoro
 
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
Nguyen Thanh Tu Collection
 
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdfবাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
eBook.com.bd (প্রয়োজনীয় বাংলা বই)
 
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptxNEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
iammrhaywood
 
คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1
คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1
คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1
สมใจ จันสุกสี
 
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptxChapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Mohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
 
UGC NET Exam Paper 1- Unit 1:Teaching Aptitude
UGC NET Exam Paper 1- Unit 1:Teaching AptitudeUGC NET Exam Paper 1- Unit 1:Teaching Aptitude
UGC NET Exam Paper 1- Unit 1:Teaching Aptitude
S. Raj Kumar
 
How to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 InventoryHow to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 Inventory
Celine George
 

Recently uploaded (20)

The basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptxThe basics of sentences session 6pptx.pptx
The basics of sentences session 6pptx.pptx
 
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem studentsRHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
 
clinical examination of hip joint (1).pdf
clinical examination of hip joint (1).pdfclinical examination of hip joint (1).pdf
clinical examination of hip joint (1).pdf
 
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
 
Walmart Business+ and Spark Good for Nonprofits.pdf
Walmart Business+ and Spark Good for Nonprofits.pdfWalmart Business+ and Spark Good for Nonprofits.pdf
Walmart Business+ and Spark Good for Nonprofits.pdf
 
PIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf IslamabadPIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf Islamabad
 
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
 
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
BÀI TẬP DẠY THÊM TIẾNG ANH LỚP 7 CẢ NĂM FRIENDS PLUS SÁCH CHÂN TRỜI SÁNG TẠO ...
 
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptxChapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering.pptx
 
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
 
Cognitive Development Adolescence Psychology
Cognitive Development Adolescence PsychologyCognitive Development Adolescence Psychology
Cognitive Development Adolescence Psychology
 
Wound healing PPT
Wound healing PPTWound healing PPT
Wound healing PPT
 
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptxPengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
 
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
BÀI TẬP BỔ TRỢ TIẾNG ANH LỚP 9 CẢ NĂM - GLOBAL SUCCESS - NĂM HỌC 2024-2025 - ...
 
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdfবাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
 
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptxNEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
 
คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1
คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1
คำศัพท์ คำพื้นฐานการอ่าน ภาษาอังกฤษ ระดับชั้น ม.1
 
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptxChapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
 
UGC NET Exam Paper 1- Unit 1:Teaching Aptitude
UGC NET Exam Paper 1- Unit 1:Teaching AptitudeUGC NET Exam Paper 1- Unit 1:Teaching Aptitude
UGC NET Exam Paper 1- Unit 1:Teaching Aptitude
 
How to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 InventoryHow to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 Inventory
 

9th Maths - Quadrilateral and Its Types

  • 1. 9011041155 / 9011031155 • Live Webinars (online lectures) with recordings. • Online Query Solving • Online MCQ tests with detailed solutions • Online Notes and Solved Exercises • Career Counseling 1
  • 2. 9011041155 / 9011031155 Quadrilaterals We know that quadrilateral is a closed figure with four sides. But by definition suppose P, Q, R and S are four points in a plane such that i. 3 of these points are not collinear ii. If sides PQ, QR, RS, SP are such that any of the two such segments have a common point, it is an endpoint only. iii. If we draw a line containing any one of these segments the remaining two points lie on the same side of this line and iv. If seg PR and seg QS intersect in the points other than P, Q, R, S then the union of 4 segments PQ, QR, RS, SP is called a quadrilateral. is a symbol of ‘quadrilateral’. 2
  • 3. 9011041155 / 9011031155 (And Interior of the quadrilateral is a convex set but quadrilateral is not a convex set) Each quadrilateral has 4 sides, 4 angles, 4 vertices, 2 diagonals, 4 pairs of adjacent angles, 2 pairs of opposite sides, 4 pairs of adjacent or consecutive sides and 2 pairs of opposite angles. There are main 8 types of quadrilaterals (as shown in the figures they are i. Parallelogram ii. Rectangle iii. Rhombus iv. Square v. Trapezium (non parallel sides are not congruent) vi. Isosceles trapezium vii. Kite viii. Kite (with all side, congruent) 3
  • 4. 9011041155 / 9011031155 Table S. No. Quadrilateral Type 1. Parallelogram 2. Rectangle 3. Rhombus 4. Square 4
  • 5. 9011041155 / 9011031155 5. Trapezium 6. Isosceles Trapezium 7. Kite I 8. Kite II 5
  • 6. 9011041155 / 9011031155 Property of a quadrilateral Theorem 1 [Angle sum property of a quadrilateral] The sum of the measures of all angles of a quadrilateral is 360º Given : To prove : Construction : ABCD is a given quadrilateral ∠A + ∠B + ∠C + ∠D = 360º Join the points A and C 6
  • 7. 9011041155 / 9011031155 Proof : In ABC, ∠BAC + ∠ ABC + ∠BCA = 180º .....sum of angles of ABC .... (1) and, In ADC, ∠DAC + ∠ADC + ∠DCA = 180º .....sum of angles of ADC .... (2) 7
  • 8. 9011041155 / 9011031155 Adding (1) and (2) we get ∠BAC + ∠ABC + ∠BCA + ∠DAC + ∠ADC + ∠DCA = 180º + 180º ∴ ∠BAC + ∠DAC + ∠ABC + ∠BCA + ∠DCA + ∠ADC = 360º ......(3) But ∠BAC + ∠DAC = ∠BAD ....Angle addition property....(4) and ∠BCA + ∠DCA = ∠BCD ....Angle addition property....(5) ∴ from (3), (4) and (5) we get 8
  • 9. 9011041155 / 9011031155 ∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º i.e. ∠A + ∠B + ∠C + ∠D = 360º Hence, sum of the measures of all angles of a quadrilateral is 360º 9
  • 10. 9011041155 / 9011031155 Theorem 2 The opposite sides of a parallelogram are congruent. Given ABCD is a parallelogram in which : side AB || side CD and side BC || side DA : Construction : Proof : side AB side CD and side BC To prove side DA Draw diagonal AC. ABCD is a parallelogram in which side AB || side CD and seg AC is a transversals. 10
  • 11. 9011041155 / 9011031155 ∴∠BAC ∠DCA ...(alternate angle)....(1) Similarly, side BC || side DA and seg AC is a transversal. ∴ ∠BCA ∠DAC ......(alternate angle) ...(2) In ABC and ∠BAC seg AC ∠BCA ∴ CDA ∠DCA ......(from 1) seg CA .......(common side) ∠ DAC ABC CDA 11 ........(from 2) .......(ASA test)
  • 12. 9011041155 / 9011031155 ∴ side AB side CD and side BC ......(c.s.c.t) side DA ......(c.s.c.t.) Hence, the opposite sides of a parallelogram are congruent. 12
  • 13. 9011041155 / 9011031155 Theorem 3 If opposite sides of a quadrilateral are congruent then the quadrilateral is parallelogram. Given : In ABCD side AB side DC and side AD To prove : Construction : side BC ABCD is a parallelogram Draw a diagonal BD 13
  • 14. 9011041155 / 9011031155 Proof : In ABD and CDB side AB side DC side AD side BC side BD side DB ......(given) ....(common side) ∴ ABD CDB ∴ ∠ABD ∠CDB .....(SSS test) ......(c.a.c.t.) ∴ side AB || side CD .......(1) ...(alternate angles test) 14
  • 15. 9011041155 / 9011031155 Similarly, we can prove that side AD || side BC ......(2) ....(alternate angels test) ∴ from (1) and (2) we have ABCD is a parallelogram Hence, if opposite sides of a quadrilateral are congruent then it is a parallelogram. 15
  • 16. 9011041155 / 9011031155 Theorem 4 The opposite angles of a parallelogram are congruent Given PQRS is a parallelogram : To prove : ∠SPQ ∠QRS and ∠PSR Construction : Proof : ∠RQP Draw a diagonal SQ PQRS is a parallelogram ∴ side PS || side QR and seg SQ is a transversal ∴ ∠PSQ ∠RQS ..(Alternate angles) ...(1) 16
  • 17. 9011041155 / 9011031155 Also, side PQ || side SR and seg SQ is a transversal. ∴ ∠PQS ∠RSQ ...(Alternate angles)..(2) In PQS and ∠PSQ RSQ ∠RQS ......from (1) side SQ side QS ....(common side) ∠PQS ∴ ∠RSQ PQS ......from (2) RSQ ......(ASA test) ∴ ∠SPQ 17 ∠QRS ......(c.a.c.t.)
  • 18. 9011041155 / 9011031155 Similarly, we can prove by drawing diagonal PR. ∠PSR ∠ RQP Hence, the opposite angles of a parallelogram are congruent. 18
  • 19. 9011041155 / 9011031155 Theorem 5 A quadrilateral is a parallelogram if its opposite angles are congruent. Given : PQRS is a parallelogram in which ∠SPQ ∠PQR To prove : Proof : ∠QRS and ∠RSP PQRS is a parallelogram Let ∠SPQ = ∠QRS = xº And ∠PQR = ∠RSP = yº ......Opposite angle of a quadrilateral. 19
  • 20. 9011041155 / 9011031155 ∠SPQ + ∠PQR + ∠QRS + ∠RSP= 360º. …(Angle sum property of a quadrilateral) ∴ x + y + x + y = 360º ∴ 2x + 2y = 360º ∴ x + y = 180º .....(dividing by 2) ∴ ∠SPQ + ∠RSP = 180º ∴ side PQ || side SR ... (interior angles test) ....(1) 20
  • 21. 9011041155 / 9011031155 Similarly, we can prove that side PS || side QR .......(2) ∴ PQRS is a parallelogram ...... from (1) and (2) Hence, if opposite angles of a quadrilateral are congruent, then it is a parallelogram. 21
  • 22. 9011041155 / 9011031155 Theorem 6 The diagonals of a parallelogram bisect each other. Given : ABCD is a parallelogram in which the diagonals AC and BD intersect in M. To prove : seg AM seg CM and seg BM Proof : since seg DM ABCD is a parallelogram side AB || side CD and AC is a transversal. ∴ ∠BAC ∠DCA ...(Alternate angles) i.e. ∠BAM ∠DCM ...(A – M – C)..(1) 22
  • 23. 9011041155 / 9011031155 Also, side AB || side DC and seg DB is a transversal. ∴ ∠ABD ∠CDB ...(Alternate angles) i.e. ∠ABM ∠CDM ...(B – M – D) ..(2) Now, In ABM and ∠BAM ∠DCM ......(from 1) side AB ∠ABM ∴ CDM side DC .....(opposite sides) ∠CDM ......(from 2) ABM CDM ......(ASA test) 23
  • 24. 9011041155 / 9011031155 ∴ seg AM seg CM and seg BM .....(c.s.c.t.) seg DM Hence, diagonals of a parallelogram bisect each other. 24
  • 25. 9011041155 / 9011031155 Theorem 7 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Given : PQRS is a quadrilateral in which diagonals PR and QS intersect in M. seg PM seg QM To prove : Proof : seg RM and seg SM PQRS is a parallelogram In PMQ and seg PM RMS seg RM 25 ......(given)
  • 26. 9011041155 / 9011031155 ∠PMQ ∠RMS .....(verticallyopposite angles) seg QM ∴ seg SM .....(given) PMQ RMS ∴ ∠PQM ∠RSM i.e. ∠PQS ......(SAS test) ......(c.a.c.t) ∠RSQ ......(S – M – Q) ∴ side PQ || side SR ....(alternate angles test) ...(1) Similarly, we can prove that side PS || side QR 26 .....(2)
  • 27. 9011041155 / 9011031155 PQRS is a parallelogram .........from (1) and (2) Hence, if the diagonals of a quadrilateral bisect each other then it is a parallelogram. 27
  • 28. 9011041155 / 9011031155 Theorem 8 A quadrilateral is a parallelogram if a pair of opposite sides is parallel and congruent. Given LMNK is a given quadrilateral in : which side LM || side NK and side LM To prove : Construction : side NK LMNK is a parallelogram Draw diagonal MK 28
  • 29. 9011041155 / 9011031155 Proof : since LMNK is a quadrilateral in which side LM || side NK and seg MK is a transversal. ∴∠LMK ∠NKM ..(Alternate angles) ...(1) Now, In KLM and MNK seg LM seg NK .....(given) ∠LMK ∠NKM seg KM .......(from 1) seg MK .......(common side) ∴ KLM MNK ......(SAS test) 29
  • 30. 9011041155 / 9011031155 ∴ ∠LKM ∠NMK .......(c.a.c.t) ∴ side LK || side MN .....(alternate angles test) ....(2) and side LM || side NK ...(given) .....(3) ∴ from (2) and (3) we have LKMN is a parallelogram Hence, if a pair of opposite sides of a quadrilateral is parallel and congruent then the quadrilateral is a parallelogram. 30
  • 31. 9011041155 / 9011031155 Sums 1. The angles of a quadrilateral are in the ratio 2 : 3 : 5 : 8. Find all the angles of the quadrilateral. Sol. The angles of the quadrilateral are in ratio 2 : 3 : 5 : 8. Let the measures of the four angles are 2xº, 3xº, 5xº and 8xº. According to the angle sum property of a quadrilateral, we have 2xº + 3xº + 5xº + 8xº = 360º ∴ 18xº = 360º 3600 ∴x = 18 0 ∴ x 0 = 200 2xº = 2 × 20º = 40º, 3xº = 3 × 20º = 60º, 5xº = 5 × 20º = 100º and 8xº = 8 × 20º = 160º ∴ The measures of the angles of the quadrilateral are 40º, 60º, 100º and 160º. 31
  • 32. 9011041155 / 9011031155 2. In the figure, PQRS is a quadrilateral. The bisectors of ∠P and ∠Q meet at the point A. ∠S = 50º, ∠R = 110º. Find the measure of∠PAQ. Sol. Refer above fig. According to the angle sum property of a quadrilateral, ∠SPQ + ∠PQR + ∠R + ∠S = 360º ∴ ∠SPQ + ∠PQR + 110º + 50º = 360º … (substituting the given values) 32
  • 33. 9011041155 / 9011031155 ∴ ∠SPQ + ∠PQR + 160º = 360º ∴ ∠SPQ + ∠PQR = 360º - 160º ∴ ∠SPQ + ∠PQR = 200º … (1) Now, it is given that Ray PA and ray QA are the bisectors of ∠SPQ and ∠RQP respectively. ∴ ∠APQ = ½ ∠SPQ and ∠AQP = ½ ∠RQP ∴ ∠APQ + ∠AQP = ½ (∠SPQ + ∠RQP) … (2) From (2) and (1), ∠APQ + ∠AQP = ½ (200º) ∴ ∠APQ + ∠AQP = 100º 33 … (3)
  • 34. 9011041155 / 9011031155 Also, In APQ, ∠PAQ + ∠APQ + ∠AQP = 180º … (Angles of a triangle) ∴ ∠PAQ + 100º = 180º … [From (3)] ∴ ∠PAQ = 180º - 100º ∴ ∠PAQ = 80º ∴The measure of ∠PAQ is 80º. 34
  • 35. 9011041155 / 9011031155 3. The sides BA and DC of quadrilateral ABCD are extended as shown in the figure. Prove that m + n = p + q. Proof : Consider ∠DAB b = a and ∠BCD = c. According to the angle sum property of a quadrilateral, In quadrilateral ABCD, a + q + c + p = 360º 35 … (1)
  • 36. 9011041155 / 9011031155 From (2), m + c + n + a = 360º … (3) From (1) and (3), a+q+c+p=m+c+n+a ∴q+p=m+n i.e. m + n = p + q. 36
  • 37. 9011041155 / 9011031155 4. In a parallelogram ABCD, ∠A = xº, ∠B = (3x + 20)º. Find x, ∠C and ∠D. Sol. We know that the adjacent angles of a parallelogram are supplementary. ∴ ∠A + ∠B = 180º Put the given values ∴ xº + 3xº + 20º = 180º ∴ 4xº = 180º - 20º ∴ 4xº = 160º 37
  • 38. 9011041155 / 9011031155 ∴ xº = 40º 3xº + 20º = 3 × 40º + 20º = 120º + 20º = 140º ∴ ∠A = 40º and ∠B = 140º Also, the opposite angles of a parallelogram are congruent. ∴ ∠C = ∠ A = 40º and ∠ D = ∠ B = 140º ∴ x = 40. ∠ C = 40º, ∠ D = 140º. 38
  • 39. 9011041155 / 9011031155 5. The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the lengths of all the sides of the parallelogram. Sol: Consider that the smaller of the two adjacent sides be x cm. Then, it is given that the other side is (x + 25) cm Now, the perimeter of a parallelogram is equal to the sum of the lengths of four sides. ∴ x + (x + 25) + x + (x + 25) = 150 ∴ 4x + 50 = 150 ∴ x = 25 and x + 25 = 25 + 25 = 50 ∴ The lengths of the sides of the parallelogram ∴ 4x = 150 – 50 ∴ 4x = 100 are 25 cm, 50 cm, 25 cm and 50 cm. 39
  • 40. 9011041155 / 9011031155 6. In the figure, WXYZ is a parallelogram. From the information given in the figure, find the values of x and y. Sol. As XY & ZW are opposite sides of a parallelogram side XY || side ZW seg XZ is a transversal. ∴ ∠ ZXY = ∠ XZW ∴ 4y = 28º … (alternate angles) … (given) ∴ y = 7º 40
  • 41. 9011041155 / 9011031155 Similarly, side XW || side YZ and seg XZ is the transversal. ∴ ∠ ZXW = ∠ XZY. ∴ 10x = 60º … (given) ∴ x = 6º. ∴The value of x is 6º and that of y is 7º. 41
  • 42. 9011041155 / 9011031155 7. Prove that two opposite vertices of a parallelogram are equidistant from the diagonal not containing these vertices. Given : 1. ABCD is a parallelogram. 2. Seg AM ⊥ diagonal BD and Seg CN ⊥ diagonal BD. We have to prove seg AM ≅ seg CN. First, draw diagonal AC intersecting diagonal BD in the point P. 42
  • 43. 9011041155 / 9011031155 Proof : P is the point of intersection of the diagonals of parallelogram ABCD. ∴ seg AP ≅ seg CP … (1) In APM and CPN, seg AP ≅ seg CP … [From (1)] ∠ APM ≅ ∠ CPN … (vertically opposite angles) ∠ AMP ≅ ∠CNP … (each a right angle) 43
  • 44. 9011041155 / 9011031155 ∴ APM ≅ CPN … (SAA test) ∴ seg AM ≅ seg CN … (c.s.c.t.) Therefore, the opposite vertices of a parallelogram are at equal distance from the diagonal not containing these vertices. 44
  • 45. 9011041155 / 9011031155 8. The ratio of two sides of a parallelogram is 3 : 4. If its perimeter is 112 cm, find the lengths of the sides of the parallelogram. Sol: It is given that the ratio of the lengths of two sides of the parallelogram is 3:4. Let the lengths of those two sides be 3x cm and 4x cm respectively. We know that the opposite sides of a parallelogram are congruent. ∴ The sides of the parallelogram are 3x cm, 4x cm, 3x cm and 4x cm. Now, the perimeter of a parallelogram = the sum of the lengths of four sides ∴ 112 = 3x + 4x + 3x + 4x … (given perimeter 112 cm) 45
  • 46. 9011041155 / 9011031155 ∴ 14x = 112 ∴ x=8 3x = 3 × 8 = 24 and 4x = 4 × 8 = 32 ∴ The lengths of the sides of the given parallelogram are 24 cm, 32 cm, 24 cm and 32 cm. 46
  • 47. 9011041155 / 9011031155 Definitions Parallelogram A quadrilateral is a parallelogram if its opposite sides are parallel. Rectangle A parallelogram (or a quadrilateral) in which each angle is a right angle, is called a rectangle. Rhombus A quadrilateral having all sides congruent is called a rhombus. Square A quadrilateral is called a square if all its sides and angles are congruent. 47
  • 48. 9011041155 / 9011031155 Trapezium If only one pair of opposite sides is parallel, the quadrilateral is said to be a trapezium. Isosceles trapezium A trapezium in which nonparallel sides are congruent, it is called as isosceles trapezium. Kite If in ABCD, AB AD and CB CD and diagonal AC is perpendicular bisector of diagonal BD, then ABCD is called as a kite. 48
  • 49. 9011041155 / 9011031155 Theorem 9 Diagonals of a rectangle are congruent. Theorem 10 If diagonals of a parallelogram are congruent then it is a rectangle. 49
  • 50. 9011041155 / 9011031155 Theorem 11 Diagonals of a rhombus are perpendicular bisectors of each other. Theorem 12 If the diagonals of a quadrilateral bisect each other at right angle then the quadrilateral is a rhombus. 50
  • 51. 9011041155 / 9011031155 Theorem 13 Diagonals of a square are congruent and perpendicular bisectors of each other. Theorem 14 If diagonals of quadrilateral are congruent and perpendicular bisectors of each other then it is square. 51
  • 52. 9011041155 / 9011031155 Sums 1. Adjacent sides of a rectangle are of lengths 7 cm and 24 cm. Find the lengths of its diagonals. Sol. Consider ABCD as a rectangle in which AB = 24 cm and BC = 7 cm. In right-angled ABC, By Pythagoras’ theorem, AC² = AB² + BC² = (24)² + (7)² = 576 + 49 = 625 ∴ AC = 25 52
  • 53. 9011041155 / 9011031155 But, the diagonals of a rectangle are congruent. ∴ BD = AC = 25 ∴ The length of diagonal AC = 25 cm and that of BD = 25 cm. 53
  • 54. 9011041155 / 9011031155 2. ABCD is a trapezium in which AB || DC. M and N are the midpoints of side AD and side BC respectively. If AB = 12 cm and MN = 14 cm, find CD. Sol. The length of the segment joining the midpoints of non - parallel sides of a trapezium is half the sum of the lengths of its parallel sides. ∴ MN = ½ (AB + CD) ∴ 14 = ½ (12 + CD) … (substituting the given values) 54
  • 55. 9011041155 / 9011031155 ∴ 28 = 12 + CD … (Multiplying both the sides by 2) ∴ CD = 28 – 12 ∴ CD = 16. 55
  • 56. 9011041155 / 9011031155 3. Prove that the base angles of an isosceles trapezium are congruent. Given : □ ABCD is an isosceles trapezium in which side AB || side CD and side AD To prove : Construction : ∠ADC side BC. ∠BCD. Draw seg BE || side AD and D-E-C. 56
  • 57. 9011041155 / 9011031155 Proof : IN □ABED, side AB || side DE. … (given) seg BE || side AD … (construction ) ∴ □ABED is a parallelogram. The opposite sides and the opposite angles of a parallelogram are congruent. ∴ side AD and ∠D 57 side BE … (1) ∠ABE … (2)
  • 58. 9011041155 / 9011031155 side AD ≅ side BC … (given) … (3) From (1) and (3), side BE ≅ side BC ∴ ∠ BEC ≅ ∠C …(Isosceles triangle theorem) … (4) side AB || side DC … (given) side BE is the transversal. ∴ ∠ABE ≅ ∠BEC … (alternate angles) … (5) From (2), (4) and (5), ∠D ≅ ∠C. i.e. ∠ADC ≅ ∠BCD. [Similarly, we can prove that ∠DAB 58 ∠CBA.]
  • 59. 9011041155 / 9011031155 4. The lengths of the diagonals PR and QS of a rhombus are 20 cm and 48 cm respectively. Find the length of each side of the rhombus. Sol. We know that the diagonals of a rhombus bisect each other at right angles. ∴ MQ = ½ QS = ½ × 48 cm ∴ MQ = 24 cm MP = ½ PR = ½ × 20 cm ∴ MP = 10 cm. ∠ PMQ = 90º 59
  • 60. 9011041155 / 9011031155 In right-angled PMQ, By Pythagoras’ theorem PQ² = PM² + QM² = (10)² + (24)² = 100 + 576 = 676 ∴ PQ = 26. But all the sides of a rhombus are congruent. ∴ The length of each side of the given rhombus is 26 cm. 60
  • 61. 9011041155 / 9011031155 5. ABCD is a kite in which diagonal AC and diagonal BD intersect at point O. ∠ OBC = 20º and ∠OCD = 40º. Find : i. ∠ABC ii. ∠ADC iii. ∠BAD. Sol. Refer above fig, The diagonal joining the unequal sides of a kite bisects the other diagonal at right angles. ∴ Each angle at O is a right angle. Also, side BA side BC. 61
  • 62. 9011041155 / 9011031155 Side DA side DC and seg AO seg CO … (perpendicular bisector theorem) … (1) In BOC, ∠BOC = 90º, ∠OBC = 20º … (given) ∴ ∠OCB = 180º - (90º + 20º) = 180º - 110º ∴ ∠OCB = 70º … (2) 62
  • 63. 9011041155 / 9011031155 In BAC, BA = BC … [From (1)] ∴ ∠BCA = ∠BAC … (angles opposite to equal sides) i.e. ∠OCB = ∠OAB ∴ ∠OAB = 70º … [From (2)] In DAC, DA = DC ∴ ∠DCA = ∠DAC … [From (1)] … (angles opposite to equal sides) i.e. ∠DCO = ∠DAO 63 … (3)
  • 64. 9011041155 / 9011031155 ∴ ∠DAO = 40º … [Given : ∠DCO = 40º] … (4) From (3) and (4), ∠OAB + ∠DAO = 70º + 40º = 110º … (5) ∠BAO + ∠DAO = ∠BAD … (Angle addition postulate) … (6) From (5) and (6), ∠BAD = 110º. In DAC, ∠ADC + ∠DAC + ∠DCA = 180º … (angles of a triangle) ∴ ∠ ADC + 40º + 40º = 180º … [From (4) and given] 64
  • 65. 9011041155 / 9011031155 ∴ ∠ADC = 180º - 80º ∴ ∠ADC = 100º … (7) In BAC, ∠ABC + ∠BAC + ∠BCA = 180º … (angles of a triangle) ∴ ∠ABC + 70º + 70º = 180º … [From (2) and 3)] ∴ ∠ABC = 180º - 140º ∴ ∠ABC = 40º Ans. : i. ∠ABC = 40º ii. ∠ADC = 100º iii. ∠BAD = 110º. 65
  • 66. 9011041155 / 9011031155 Theorem 15 Diagonals of an isosceles trapezium are congruent. Given ABCD is an isosceles trapezium in : which seg AC and seg BD are diagonals side AD side BC and side AB || side DC To Prove : Construction : seg AC seg BD Draw seg AM ⊥ seg DC, seg BN ⊥ seg DC Proof : In ∆AMD and ∆BNC side AD side BC 66 ….(given)
  • 67. 9011041155 / 9011031155 ∠AMD seg AM ∠BNC …..(90º each) seg BN …..(perpendicular distance between two parallel lines) ∴ ∆AMD ∆BNC ……(hypotenuse side test) ∴ ∠ADM ∠BCN …..(c. a. c. t.) i.e. ∠ADC ∠BCD .(D - M - C)…(1) Now, In ∆ADC and ∆BCD side AD ∠ADC side BC ∠BCD 67 ….(given) …..(from (1))
  • 68. 9011041155 / 9011031155 side DC ∴ ∆ADC side CD …..(common side) ∆BCD ∴ seg AC …..(SAS test) seg BD ……(c.s.c.t.) Hence, diagonals of an isosceles trapezium are congruent. 68
  • 69. 9011041155 / 9011031155 Theorem 16 (Intercepts made, by three parallel lines) If three parallel lines make congruent intercepts on a transversal then they make congruent intercepts on any other transversal. Given : line || line m || line n and lines t1 and t2 are transversals. seg AB Prove : Construction : seg BC seg DE seg EF Draw a line GI parallel to the line t1 through E. 69
  • 70. 9011041155 / 9011031155 Proof : seg AG || seg BE ……(given) seg AB || seg GE ……. (construction) ∴ ABEG is a parallelogram. ∴ seg AB seg GE…(opposite sides of a parallelogram …(1) Similarly, ∴ seg BC BCIE is a parallelogram. seg EI …..(opposite sides of a parallelogram….(2) But seg AB seg BC ….. (given)...(3) Hence seg GE seg EI ….(4) from (1), (2) and (3) 70
  • 71. 9011041155 / 9011031155 Now, In ∆GED and ∆IEF ∠DGE ∠FIE …(alternate angles) seg GE ∠GED seg EI …..(from (4)) ∠IEF ……(vertically opposite angles) ∴ ∆GED ∴ seg DE ∆IEF ……(ASA test) seg EF ……(c.s.c.t.) Hence, if three parallel lines make congruent intercepts on a transversal then they make congruent intercepts on any other transversal. 71
  • 72. 9011041155 / 9011031155 Theorem 17 Mid - Point theorem The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it. Given : In ∆ABC, P and Q are midpoints of sides AB and AC respectively. To prove : i. seg PQ || side BC ii. seg PQ = 1 side BC 2 Construction : Take the point R on ray PQ such that seg PQ seg QR. Join C and R 72
  • 73. 9011041155 / 9011031155 Proof : In ∆AQP and ∆CQR seg AQ ∠AQP seg CQ ……(given) ∠CQR ….(vertically opposite angles) seg PQ seg RQ ……(construction) ∴ ∆AQP ∆CQR …… (SAS test) seg AP seg CR ……(c.s.c.t.) seg BP seg CR ...( and ∠PAQ ∠RCQ ∴ seg AP || seg CR … AP = PB)…(1) ……(c.a.c.t) (alternate angles test) ∴ seg BP || seg CR …(AP = PB)....(2) 73
  • 74. 9011041155 / 9011031155 PBCR is a parallelogram (from (1) and (2)) ∴ side PR || side BC …..(opposite sides of a parallelogram) seg PQ || side BC ……(P - Q - P) 1 Also, seg PQ = side PR 2 …(construction) ∴ seg PQ = 1 side BC …( 2 PR = BC) Hence, In a triangle, line segment joining mid - points of any two sides is parallel to the third side and is half of it. 74
  • 75. 9011041155 / 9011031155 Theorem 18 Converse of mid - point theorem If a line drawn through the mid - point of one side of a triangle is parallel to second side then it bisects the third side. In ∆ABC, A - D - B if AD = DB and line DE || side BC then, line DE bisects side AC. i.e. AE = CE Try to write the proof. 75
  • 76. 9011041155 / 9011031155 Sums 14. In the figure, □PQRS and □LMNR are rectangles, where M is the midpoint of PR. Prove that: i. SL = LR ii. LN = ½ SQ. Proof : In above fig. □LMNR is a rectangle. ∴ LM || RN … (Opposite sides of a rectangle) i.e. LM || RQ … (R-N-Q) … (1) 76
  • 77. 9011041155 / 9011031155 RQ || SP … (opposite sides of a rectangle) …(2) From (1) and (2), we have LM || RQ || SP … (3) In PSR, it is given that M is the midpoint of PR and LM || RQ … [From (1)] ∴ By the converse of midpoint theorem, point L is the midpoint of side SR. ∴ SL = LR 77 … (4) …i
  • 78. 9011041155 / 9011031155 Again, LR || MN and LR || PQ … (opposite sides of rectangle LMNR and rectangle PQRS respectively) ∴ MN || PQ … (property for parallel sides) ∴ In RPQ, M is the midpoint of PR. 78
  • 79. 9011041155 / 9011031155 ∴ By the converse of midpoint theorem, point N is the midpoint of RQ. … (5) From (4) and (5), L and N are the midpoints of SR and RQ respectively., ∴ In RSQ, by midpoint theorem, LN = ½ SQ Hence proved. 79 … ii
  • 80. 9011041155 / 9011031155 15. In the figure, seg PD is the median of PQR. Point PM 1 G is the midpoint of seg PD. Show that = . PR 3 Proof : Draw seg DN || seg QM. We are given that, In PDN, G is the midpoint of seg PD. Seg QM (i.e. seg GM) || seg DN … (by construction) 80
  • 81. 9011041155 / 9011031155 ∴ By the converse of midpoint theorem, M is the midpoint of seg PN. ∴ PM = MN … (1) In QRM, As PD is median, D is the midpoint of side QR And seg QM || seg DN … (construction) ∴ By the converse of midpoint theorem, N is the midpoint of seg MR. ∴ MN = NR 81 … (2)
  • 82. 9011041155 / 9011031155 From (1) and (2), PM = MN = NR … (3) Now, PR = PM + MN + NR … (4) ∴ PR = 3PM … [From (3) and (4)] i.e. 3PM = PR 82 PM 1 ∴ = . PR 3
  • 83. 9011041155 / 9011031155 16. Prove that a diagonal of a rhombus bisects two opposite angles. Proof : Refer above fig. ABCD is a rhombus & AC is its diagonal. In ABC and ADC, side AB side BC side AD and side DC … (sides of a rhombus) side AC side AC … (common side) ∴ ABC ADC 83 … (SSS test)
  • 84. 9011041155 / 9011031155 ∴ ∠BAC and ∠BCA ∠DAC … (1) ∠DCA … (2) … (c.a.c.t.) seg AC bisects ∠BAD. …[from(1)] seg AC bisects ∠BCD. …[From(2)] Therefore diagonal of a rhombus bisects two opposite angels. 84
  • 85. 9011041155 / 9011031155 17. In the figure, BHIR is a kite. If ∠HIR = 50º, find a. ∠HRI and b. ∠BHI. Sol. Consider abive fig. □BHIR is a kite. a. In IRH, side IR = side IH ∴ ∠HRI = ∠RHI … (given) … (angles opposite to equal sides) … (1) ∠I + ∠HRI + ∠RHI = 180º … (angles of a triangle) 85
  • 86. 9011041155 / 9011031155 ∴ 50º + ∠HRI + ∠HRI = 180º … [Given and from (1)] ∴ 2∠HRI = 180º - 50º ∴ 2∠HRI = 130º ∴ ∠HRI = 65º … (2) ∴ ∠RHI = 65º … [From (1) and (2)] … (3) 86
  • 87. 9011041155 / 9011031155 b. BHR is an equilateral triangle. … (given) ∴ ∠BHR = 60º … (angle of an equilateral triangle) … (4) ∠BHI = ∠BHR + ∠RHI … (angle addition postulate) = 60º + 65º … [From (4) and (3)] ∴ ∠BHI = 125º 87
  • 88. 9011041155 / 9011031155 18. In CBS, seg BC seg SC. Ray CE bisects exterior ∠DCS. Ray SE || ray BC. Prove that CBSE is a parallelogram. Proof : We are given that seg CB ∴ ∠CBS seg CS ∠CSB … (angles opposite to congruent sides) … (1) ∠DCS is the exterior angle of CBS. 88
  • 89. 9011041155 / 9011031155 ∴ ∠DCS = ∠CBS + ∠CSB … (Remote interior angle theorem) … (2) From (1) and (2), . ∠CBS = ∠CSB = ½∠DCS … (3) Now, it is given that Ray CE bisects ∠DCS. ∴ ∠DCE = ∠ECS = ½∠DCS … (4) 89
  • 90. 9011041155 / 9011031155 From (3) and (4), ∠CSB = ∠ECS ∴ seg BS || seg CE … (Alternate angles test for parallel lines) seg SE || seg BC … (given) ∴ Both the pairs of opposite sides of ∴ CBSE are parallel. CBSE is a parallelogram. 90
  • 91. 9011041155 / 9011031155 19. In the figure, LAXM is a parallelogram. Point I is the midpoint of diagonal LX. PQ is a line passing through I. P and Q are the points of intersection with the sides LA and MX respectively. Prove that seg PI Proof : seg IQ. In LPI and XQI, ∠LIP & ∠XIQ are vertically opposite angles ∴∠LIP ∠XIQ seg LI seg XI … (As I is the midpoint of LX) 91
  • 92. 9011041155 / 9011031155 ∠PLI ∠QXI … (Alternate angles) ∴ LPI XQI ∴ seg PI seg QI Hence proved. 92 … (ASA test) … (c.s.c.t.)
  • 93. 9011041155 / 9011031155 20. MNOP is a rhombus. Q is a point in the interior of the rhombus such that QM = QO. Prove that Q lies on diagonal NP. Proof : In MPN and OPN, seg PM seg PO and seg MN seg ON … (sides of a rhombus) seg PN sg PN … (common side) ∴By SSS test ∴ MPN OPN 93
  • 94. 9011041155 / 9011031155 ∴ ∠MPN ∠OPN … (c.a.c.t.) i.e. diagonal PN is the bisector of ∠MPO … (1) In MPQ and OPQ, seg MP seg OP … (sides of a rhombus) seg QM sg QO … (given) seg PQ seg PQ … (common side) ∴ By SSS test ∴ MPQ OPQ ∴∠MPQ ∠OPQ 94 … (c.a.c.t.)
  • 95. 9011041155 / 9011031155 i.e. seg PQ is the bisector of ∠MPO … (2) Point Q is in the interior of MNOP. … (3) The bisector of an angle is unique. ∴From (1), (2) and (3) bisectors PN and PQ of ∠MPO are one and the same. ∴ Point Q lies on the diagonal NP. Hence proved. 95
  • 96. 9011041155 / 9011031155 21. ABCD is a square. P and Q are the points such that seg AQ seg DP. Prove that seg AQ ⊥ seg DP. Proof : In DAP and ABQ, ∠DAP ∠ABQ … (Each is a right angle) It is given that hypotenuse DP hypotenuse AQ side DA ≅ side AB … (side of a square) ∴By hypotenuse-side theorem, 96
  • 97. 9011041155 / 9011031155 ∴ DAP ABQ ∴ ∠DPA ∠AQB … (c.a.c.t.) and ∠ADP … (1) ∠BAQ … (c.a.c.t.) … (2) From (1), Assume ∠DPA = ∠AQB = α … (3) From (2), Assume ∠ADP = ∠BAQ = β 97 … (4)
  • 98. 9011041155 / 9011031155 Then, In DAP, α+β+ DAP = 180º … (Angles of a triangle) ∴ α + β + 90º = 180º … (∠DAP = 90º) ∴ α +β = 90º … (5) Now, In APT, α + β + ∠ATP = 180º … (Angles of a triangle) ∴ 90º + ∠ATP = 180º … [From (5)] ∴ ∠ATP = 90º 98
  • 99. 9011041155 / 9011031155 Point T is the point of intersection of seg AQ and seg PD. ∴ seg AQ ⊥ seg DP. Hence proved!! 99
  • 100. 9011041155 / 9011031155 22. ABCD is a kite. AB = AD and CB = CD. Prove that i. diagonal AC ⊥ diagonal BD. ii. diagonal AC bisects diagonal BD. Proof : We are given, AB = AD ∴ Point A is equidistant from points B and D of seg BD. Also, CB = CD … (1) … (given) ∴ Point C is equidistant from points B and D of seg BD. 100 … (2)
  • 101. 9011041155 / 9011031155 From (1) and (2), points A and C are equidistant from points B and D of seg BD. Therefore by perpendicular bisector theorem, AC is the perpendicular bisector of BD. i.e. i. diagonal AC ⊥ diagonal BD and ii. diagonal AC bisects diagonal BD. Hence proved!! 101
  • 102. 9011041155 / 9011031155 23. Let points A and B be on one side of line ℓ. Draw seg AD ⊥ line ℓ and seg BE ⊥ line ℓ. Let point C be the midpoint of seg AB. Prove that seg CD seg CE. Construction : Draw seg CM ⊥ lin l. Proof Refer above fig. : We can have, Seg AD, seg BE and seg CM are perpendiculars to line l ∴ seg AD || seg BE || seg CM. 102
  • 103. 9011041155 / 9011031155 Seg CA seg CB … (As C is the midpoint of seg AB) Now, Seg CA and seg CB are the intercepts made by three parallel lines AD, BE and CM. As DM & ME are intercepts made by the same parallel lines on other transversal, we have ∴ intercept DM 103 intercept ME
  • 104. 9011041155 / 9011031155 i.e. seg DM seg ME … (1) In CDM and CEM, seg DM seg EM … [From (1)] ∠CMD ∠CME … (Each a right angle : construction) seg CM seg CM … (common side) ∴ CDM CEM … (SAS test) ∴ seg CD 104 seg CE … (c.s.c.t.)
  • 105. 9011041155 / 9011031155 24. □ABCD is a parallelogram. P, Q, R and S are the points on sides AB, BC, CD and DA respectively such that seg AP Prove that Proof : seg BQ seg CR seg DS. PQRS is a parallelogram. AB = AP + PB … (A-P-B) … (1) CD = CR + RD … (C-R-D) … (2) side AB side CD … (opposite sides of a parallelogram) ∴ AB = CD … (3) 105
  • 106. 9011041155 / 9011031155 From (1), (2) and (3), AP + PB = CR + RD … (4) seg AP … (given) seg CR ∴ AP = CR … (5) From (4) and (5), PB = RD ∴ seg PB seg RD … (6) In PBQ and RDS, seg PB seg RD … [From (6)] Now, ∠B &∠D are opposite angles of a parallelogram ∴∠B ∠D 106
  • 107. 9011041155 / 9011031155 seg BQ seg DS … (given) ∴By SAS test ∴ PBQ RDS ∴ seg PQ ≅ seg RS … (c.s.c.t.) Similarly, seg PS proved. seg QR can be … (8) From (7) and (8), opposite sides of PQRS are congruent. ∴ PQRS is a parallelogram. Hence proved!! 107
  • 108. 9011041155 / 9011031155 Ask Your Doubts call For inquiry and registration, call 9011041155 / 9011031155. 108