
1.Number System
14
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11

Lecture1.12

Lecture1.13

Lecture1.14


2.Polynomials
10
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10


3.Coordinate Geometry
8
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8


4.Linear Equations
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


5.Euclid's Geometry
7
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7


6.Lines and Angles
10
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8

Lecture6.9

Lecture6.10


7.Triangles
11
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11


8.Quadrilaterals
13
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13


9.Area of Parallelogram
11
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11


10.Constructions
7
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7


11.Circles
11
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8

Lecture11.9

Lecture11.10

Lecture11.11


12.Heron's Formula
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Surface Area and Volume
16
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12

Lecture13.13

Lecture13.14

Lecture13.15

Lecture13.16


14.Statistics
15
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12

Lecture14.13

Lecture14.14

Lecture14.15


15.Probability
8
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

Chapter Notes – Quadrilaterals
(1) Prove that sum of the angles of a quadrilateral is 360∘.
Given: Quadrilateral ABCD
To Prove: ∠A+∠B+∠C+∠D=360∘
Construction: Join ACProof: In ΔABC, We have
∠1+∠4+∠6=180∘……….(i)
In ΔACD, we have
∠2+∠3+∠5=180∘……………(ii)
Adding (i) and (ii), we get
(∠1+∠2)+(∠3+∠4)+(∠5+∠6)=180∘+180∘
∠A+∠C+∠D+∠B=360∘
∠A+∠B+∠C+∠D=360∘
(2) Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Given: A parallelogram ABCD
To Prove: A diagonal, say, AC, of parallelogram ABCD divides it into congruent triangles ABC and CDA i.e.
ΔABC≅ΔCDA
Construction: Join ACProof: Since ABCD is a parallelogram. Therefore,
AB∥DC and AD∥BC
Now, AD∥BC and transversal AC intersects them at A and C respectively.
∠DAC=∠BCA …….(i) [Alternate interior angles]
Again, AB∥DC and transversal AC intersects them at A and C respectively. Therefore,
∠BAC=∠DCA ……(ii) [Alternate interior angles]
Now, in Δs ABC and CDA, we have
∠BCA=∠DAC [From (i)]
AC=AC
∠BAC=∠DCA
So, by ASA congruence criterion, we have
ΔABC≅ΔCDA
(3) Prove that two opposite angles of a parallelogram are equal.
Given: A parallelogram ABCD
To prove: ∠A=∠C and ∠B=∠DProof: Since ABCD is a parallelogram. Therefore,
AB∥DC and AD∥BC
Now, AB∥DC and transversal AD intersects them at A and D respectively.
∠A+∠D=180∘ …….(i) [Sum of Consecutive interior anglesis 180∘ ]
Again, AD∥BC and DC intersects them at D and C respectively.
∠D+∠C=180∘ ….. (ii) [Sum of Consecutive interior angles is 180∘ ]
From (i) and (ii), we get
∠A+∠D=∠D+∠C
∠A=∠C.
Similarly, ∠B=∠D.
Hence, ∠A=∠C and ∠B=∠D
(4) Prove that the diagonals of a parallelogram bisect each other.
Given: A parallelogram ABCD such that its diagonals AC and BD intersect at O.
To prove: OA=OC and OB=ODProof: Since ABCD is a parallelogram. Therefore,
AB∥DC and AD∥BC
Now, AB∥DC and transversal AC intersects them at A and C respectively.
∠BAC=∠DCA
∠BAO=∠DCO ……..(i)
Again, AB∥DC and BD intersects them at B and D respectively.
∠ABD=∠CDB
∠ABO=∠CDO ……..(ii)
Now, in Δs AOB and COD, we have
∠BAO=∠DCO
AB=CD
and, ∠ABO=∠CDO
So, by ASA congruence criterion
ΔAOB≅ΔCOD
OA=OC and OB=OD
Hence, OA=OC and OB=OD
(5) Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
Given: A parallelogram ABCD such that the bisectors of consecutive angles A and B intersect at P.
To prove: ∠APB=90∘Proof: Since ABCD is a parallelogram. Therefore,
AD∥BC
Now, AD∥BC and transversal AB intersects them.
∠A+∠B=180∘
12∠A+12∠B=90∘
∠1+∠2=90∘ ….(i)
AP is the bisector of ∠A and BP is the bisector of ∠B then ∠1=12∠A and ∠2=12∠B
In ΔAPB, we have
∠1+∠APB+∠2=180∘
90∘+∠APB=180∘ [From (i)]
∠APB=90∘
(6) Prove that if a diagonal of a parallelogram bisects one of the angles of the parallelogram it also bisects the second angle.
Given: A parallelogram ABCD in which diagonal AC bisects ∠A.
To prove: AC bisects ∠CProof: Since ABCD is a parallelogram. Therefore,
AB∥DC
Now, AB∥DC and AC intersects them.
∠1=∠3 ……(i) [Alternate interior angles]
Again, AD∥BC and AC intersects them.
∠2=∠4 ……(ii) [Alternate interior angles]
But, it is given that AC is the bisector of ∠A. Therefore,
∠1=∠2 ……..(iii)
From (i), (ii) and (iii), we get
∠3=∠4 ………(iv)
Hence, AC bisects ∠C.
From (ii) and (iii), we have
∠1=∠4
BC=AB [Angles opposite to equal sides are equal]
But, AB=DC and BC=AD [ ABCD is a parallelogram]
AB=BC=CD=DA
Hence, ABCD is a rhombus.
(7) Prove that the angles bisectors of a parallelogram form a rectangle.
Proof: Since ABCD is a parallelogram. Therefore,
AD∥BCNow, AD∥BC and transversal AB intersects them at A and B respectively. Therefore,
∠A+∠B=180∘ [Sum of consecutive interior angles is 180∘]
12∠A+12∠B=90∘
∠BAS+∠ABS=90∘ ….(i) [AS and BS are bisectors of ∠A and ∠B respectively]
But, in ΔABS, we have
∠BAS+∠ABS+∠ASB=180∘ [Sum of the angle of a triangle is 180∘]
90∘+∠ASB=180∘
∠ASB=90∘
∠RSP=90∘ [∠ASB and ∠RSP are vertically opposite angles ∠RSP=∠ASB]
Similarly, we can prove that
∠SRQ=90∘, ∠RQP=90∘ and ∠SPQ=90∘
Hence, PQRS is a rectangle.
(8) Prove that a quadrilateral is a parallelogram if its opposite sides are equal.
Given: A quadrilateral ABCD in which AB=CD and BC=DA
To prove: ABCD is a parallelogram.
Construction: Join AC.Proof: In Δs ACB and CAD, we have
AC=CA [Common Side]
CB=AD
AB=CD
So, by SAS criterion of congruence, we have
Δs ACB and CAD
∠CAB=∠ACD ….(i)
And, ∠ACB=∠CAD
Now, line AC intersects AB and DC at A and C, such that
∠CAB=∠ACD …..(ii)
i.e., alternate interior angles are equal.
AB∥DC …..(iii)
Similarly, line AC intersects BC and AD at C and A such that
∠ACB=∠CAD
i.e., alternate interior angles are equal.
BC∥AD …..(iv)
From (iii) and (iv), we have
AB∥DC and BC∥AD
Hence, ABCD is a parallelogram.
(9) Prove that a quadrilateral is a parallelogram if its opposite angles are equal.
Given: A quadrilateral ABCD in which ∠A=∠C and ∠B=∠D.
To prove: ABCD is a parallelogram.Proof: In quadrilateral ABCD, we have
∠A=∠C ……(i)
∠B=∠D ……(ii)
∠A+∠B=∠C+∠D …..(iii)
Since sum of the angles of a quadrilateral is 360∘
∠A+∠B+∠C+∠D=360∘ ……(iv)
(∠A+∠B)+(∠A+∠B)= 360∘
2(∠A+∠B)=360∘
(∠A+∠B)=180∘
∠A+∠B=∠C+∠D=180∘ …..(v) [∠A+∠B=∠C+∠D]
Now, line AB intersects AD and BC at A and B respectively such that
∠A+∠B=180∘
i.e. the sum of consecutive interior angles is 180∘
AD∥BC ……(vi)
Again, ∠A+∠B=180∘
∠C+∠B=180∘
Now, line BC intersects AB and DC at A and C respectively such that
∠B+∠C=180∘
i.e., the sum of consecutive interior angles is 180∘.
AB∥DC ……(vii)
From (vi) and (vii), we get
AD∥BC and AB∥DC.
Hence, ABCD is a parallelogram.
(10) Prove that the diagonals of a quadrilateral bisect each other, if it is a parallelogram.
Proof: In Δs AOD and COB, we have
AO=OC
OD=OB
∠AOD=∠COBSo, by SAS criterion of congruence, we have
ΔAOD≅ΔCOB
∠OAD=∠OCB
Now, line AC intersects AD and BC at A and C respectively such that
∠OAD=∠OCB
i.e., alternate interior angles are equal.
AD∥BC
Similarly, AB∥CD
Hence, ABCD is a parallelogram.
(11) Prove that a quadrilateral is a parallelogram if its one Pair of opposite sides are equal and parallel.
Given: A quadrilateral ABCD in which AB=CD and AB∥CD.
To prove: ΔABCD is a parallelogram.
Construction: Join AC.Proof: In Δs ABC and CDA, we have
AB=DC
AC=AC
And, ∠BAC=∠DCA
So, by SAS criterion of congruence, we have
ΔABC≅ΔCDA
∠BCA=∠DAC
Thus, line AC intersects AB and DC at A and C respectively such that
∠DAC=∠BCA
i.e., alternate interior angles are equal.
AD∥BC .
Thus, AB∥CD and AD∥BC.
Hence, quadrilateral ABCD is a parallelogram.
(12) Prove that each of the four angles of a rectangle is a right angle.
Given: A rectangle ABCD such that ∠A=90∘
To prove: ∠A=∠B=∠C=∠D=90∘Proof: Since ABCD is a rectangle.
ABCD is a parallelogram
AD∥BC
Now, AD∥BC and line AB intersects them at A and B.
∠A+∠B=180∘
90∘+∠B=180∘
∠B=90∘
Similarly, we can show that ∠C=90∘ and ∠D=90∘
Hence, ∠A=∠B=∠C=∠D=90∘
(13) Prove that each of the four sides of a rhombus of the same length.
Given: A rhombus ABCD such that AB=BC.
To prove: AB=BC=CD=DA.Proof: Since ABCD is rhombus
ABCD is a parallelogram
AB=CD and BC=AD
But, AB=BC
AB=BC=CD=DA
Hence, all the four sides of a rhombus are equal.
(14) Prove that the diagonals of a rectangle are of equal length.
Given: A rectangle ABCD with AC and BD as its diagonals.
To prove: AC=BD.Proof: Since ABCD is a rectangle
ABCD is a parallelogram such that one of its angles, say , ∠A is a right angle.
AD=BC and ∠A=90∘
Now, AD∥BC and AB intersects them at A and B respectively.
∠A+∠B=180∘
90∘+∠B=180∘
∠B=90∘]
In Δs ABD and BAC, we have
AB=BA
∠A=∠B
And, AD=BC
So, by SAS criterion of congruence, we have
ΔABD≅ΔBAC}
BD=AC
Hence, AC=BD
(15) Prove that diagonals of a parallelogram are equal if and only if it is a rectangle.
Proof: In Δs ABC and DCB, we have
AB=DC
BC=CB
And, AC=DBSo, by SAS criterion of congruence, we have
ΔABC≅ΔDCB
∠ABC=∠DCB
But, AB∥DC and BC cuts them.
∠ABC+∠DCB=180∘
2∠ABC=180∘
∠ABC=90∘
Thus, ∠ABC=∠DCB=90∘.
Hence, ABCD is a rectangle.
(16) Prove that the diagonals of a rhombus are perpendicular to each other.
Given: A rhombus ABCD whose diagonals AC and BD intersect at O.
To prove: ∠BOC=∠DOC=∠AOD=∠AOB=90∘4Proof: We know that a parallelogram is a rhombus, if ll of its sides are equal. So, ABCD is a rhombus. This implies that ABCD is a parallelogram such that
AB=BC=CD=DA …..(i)
Since the diagonals of a parallelogram bisect each other.
OB=OD and OA=OC …..(ii)
Now, in Δs BOC and DOC, we have
BO=OD
BC=DC
OC=OC
So, by SSS criterion of congruence, we have
ΔBOC≅ΔDOC
∠BOC=∠DOC
But, ∠BOC+∠DOC=180∘
∠BOC=∠DOC=90∘
Similarly, ∠AOB=∠AOD=90∘
Hence, ∠AOB=∠BOC=∠COD=∠DOA=90∘
(17) Prove that diagonals of a parallelogram are perpendicular if and only if it is a rhombus.
Given: A parallelogram ABCD in which AC⊥BD.
To prove: Parallelogram ABCD is a rhombus.Proof: Suppose AC and BD intersect at O. Since the diagonals of a parallelogram bisect each other. So, we have
OA=OC ……(i)
Now, in ΔsAOD and COD, we have
OA=OC
∠AOD≅∠COD
OD=OD
So, by SAS criterion of congruence, we have
ΔAOD≅ΔCOD
AD=CD ……(ii)
Since ABCD is a parallelogram.
AB=CD and AD=CD
AB=CD=AD=BC
Hence, parallelogram ABCD is a rhombus.
(18) Prove that the diagonals of a square are equal and perpendicular to each other.
Given: A square ABCD.
To prove: AC=BD and AC⊥BD.Proof: In Δs ADB and BCA, we have
AD=BC
∠BAD=∠ABC
And, AB=BA
So, by SAS criterion of congruence, we have
ΔADB≅ΔBCA
AC=BD
Now, in Δs AOB and AOD, we have
OB=OD
AB=AD
And, AO=AO
So, by SSS criterion of congruence, we have
ΔAOB≅ΔAOD
∠AOB=∠AOD
But, ∠AOB+∠AOD=180∘
∠AOB=∠AOD=90∘
AO⊥BD
AC⊥BD
Hence, AC=BD and AC⊥BD
(19) Prove that if the diagonals of a parallelogram are equal and intersect at right angles, then it is square
Given: A parallelogram ABCD in which AC=BD and AC⊥BD
To prove: ABCD is a square.Proof: In Δs AOB and AOD, we have
AO=AO
∠AOB=∠AOD
And OB=OD
So, by SAS criterion of congruence, we have
ΔAOB≅ΔAOD
AB=AD
But, AB=CD and AD=BC
AB=BC=CD=DA …….(i)
Now, in Δs ABD and BAC, we have
AB=BA
AD=BC
And, BD=AC
So, by SSS criterion of congruence, we have
ΔABD≅ΔBAC
∠DAB=∠CBA
But, ∠DAB+∠CBA=180∘
∠DAB=∠CBA=90∘ ……(ii)
From (i) and (ii), we obtain that ABCD is a parallelogram whose all side are equal and all angles are right angles.
Hence, ABCD is a square.
(20) Prove that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
Given: A ΔABC in which D and E are the mid points of sides AB and AC respectively. DE is joined
To prove: DE∥BC and DE=12BC
Construction: Produce the line segment DE to F, such that DE=EF. Join FCProof: In Δs AED and CEF, we have
AE=CE
∠AED=∠CEF
And, DE=EF
So, by SAS criterion of congruence, we have
ΔAED≅ΔCFE
AD∥CF …(i)
And, ∠ADE=∠CFE ……(ii)
Now, D is the midpoint of AB
AD=DB
DB=CF …..(iii)
Now, DF intersects AD and FC at D and F respectively such that
∠ADE=∠CFE
i.e. alternate interior angles are equal.
AD∥FC
DB∥CF …..(iv)
From (iii) and (iv), we find that DBCF is a quadrilateral such that one pair of sides are equal and parallel.
DBCF is a parallelogram.
DF∥BC and DF=BC
But, D,E,F are collinear andDE=EF.
DE∥BC and DE=12BC
(21) Prove that a line through the midpoint of a side of a triangle parallel to another side bisects the third side.
Proof: We have to prove that E is the midpoint of AC. If possible, let E be not the midpoint of AC. Let E prime be the midpoint AC. Join DE prime.Now, in ΔABC, D is the midpoint of AB and E prime is the midpoint of AC. We have,
DE′∥BC …..(i)
Also, DE∥BC ….(ii)
From (i) and (ii), we find that two intersecting lines DE and DE’ are both parallel to Line BC.
This is contradiction to the parallel line axiom.
So, our supposition is wrong. Hence, E is the midpoint of AC.
(22) Prove that the quadrilateral formed by joining the midpoints of the sides of a quadrilateral, in order, is a parallelogram.
Given: ABCD is a quadrilateral in which P,Q,R,S are the midpoints of sides AB, BC, CD and DA respectively.
To prove: PQRS is a parallelogram.Proof: In ΔABC, P and Q are the midpoints of sides AB and BC respectively.
PQ∥AC and PQ=12AC ….(i)
In ΔADC, R and S are the midpoints of CD and AD respectively.
RS∥AC and RS=12AC …….(ii)
From (i) and (ii), we have
PQ=RS and PQ∥RS
Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.
Hence, PQRS is a parallelogram.