Subject: Compulsory Maths
A closed figure formed by four straight lines on the same plane is called quadrilateral. Here, the properties are explained briefly with different theorem and converse of theorem. Parallelogram is a quadrilateral having opposite sides parallel.
A quadrilateral having opposite sides parallel is called the parallelogram.
In a parallelogram,
1. Opposite sides are equal.
2. Opposite angles are equal.
3. Diagonals bisect each other.
4. Area = base \(\times\) height = b \(\times\) h
Theoretical proof:
Given: MNOP is a parallelogram in which MN/ / PO and MP / / NO. ( Figure)
To prove: i) MN = PO ii) MP = NO
Construction: Join M and O
Proof:
S.N.  Statements  Reasons 
1.

In \(\triangle\)MNO and \(\triangle\)MPO i) ∠NMO = ∠POM (A) ii) MO = MO (S) iii) ∠PMO = ∠MON (A) 
i) Alternate angles, MN // PO ii) Common side of both triangles. iii) Alternate angles, MP // NO. 
2. 
\(\triangle\)MNO ≅ \(\triangle\)MPO 
2. By A.S.A. test of congruency. 
3.  MN = PO and MP = NO  3.Corresponding sides of congruent triangles. 
Proved
Theoretical proof:
Given: PORS is a parallelogram in which PS = QR and PQ = SR. ( Figure)
To prove: PQRS is a parallelogram i.e PQ // SR and PS // QR.
Construction: Join P and R.
Proof:
S.N.  Statements  Reasons 
1. 
In \(\triangle\)PQR and \(\triangle\)PSR i) PQ = SR (S) ii) QR = PS (S) iii) PR = PR (S) 
i) Given ii) Given iii) Common side of both triangles. 
2.  \(\triangle\)PQR ≅ \(\triangle\)PSR  By S.S.S. test of congruency. 
3.  ∠SPR = ∠QRP  Corresponding angles of congruent triangles. 
4.  PS // QR  The transversal PR makes equal alternate angles while cutting two lines PS and QR. 
5.  ∠QPR = ∠PRS  Corresponding angles of congruent triangles. 
6.  PQ // SR  The transversal PR makes equal alternate angles while cutting two lines PQ and SR. 
7.  ∴PQRS is a parallelogram  From statement 4 and 6, opposite sides are parallels. 
Proved
Theoretical proof:
Given: DEFG is a parallelogram in which DE // GF and DG // EF.
To prove: ∠D = ∠C and ∠E = ∠G
Proof:
S.N.  Statements  Reasons 
1.  ∠D + ∠E = 180^{o}  Sum of Cointerior angles, DG // EF. 
2.  ∠E + ∠C = 180^{o}  Sum of Cointerior angles, DE // GF. 
3.  ∠D + ∠E = ∠E + ∠G  From statements 1 and 2. 
4.  ∴ ∠D = ∠F  Eliminating the common angle E from statement 3. 
Similarly,  
5.  ∠D +∠G = 180^{o}  Sum of Cointerior angles, DE // GF. 
6.  ∠D + ∠E = ∠D + ∠G  From statements 1 and 5. 
7.  ∠E = ∠G  Eliminating the common angle D from statement 6. 
Proved
Theoretical proof:
Given: ABCD is a quadrilateral where ∠A = ∠C and ∠B = ∠D.
To prove: ABCD is a parallelogram i.e. AD  BC and AB  DC.
Proof:
S.N.  Statements  Reasons 
1.  ∠A + ∠B + ∠C + ∠D = 360^{o}  Sum of four interior angles of a quadrilateral is 360^{0}. 
2.  ∠A + ∠B + ∠A + ∠B = 360^{o}  From given 
3. 
or, 2∠A + 2∠B = 360^{o} or, 2(∠A +∠B) = 360^{o} ∴ ∠A +∠B = 180^{o} 
On simplifying statement 2.

4.  ADBC  From statement 3, the sum of cointerior angles is 180^{0}. 
5.  ∠A + ∠D + ∠A + ∠D = 360^{o}  Sustituting the value in 1 from the given. 
6. 
or, 2(∠A +∠B) = 360^{o} ∴ ∠A + ∠B = 180^{o} 
On simplifying the statement 5. 
7.  AB  DC  From statement 6, the sum of cointerior angles is 180^{0}. 
8.  ∴ ABCD is a parallelogram.  From statement 4 and 7, opposite sides parallel. 
proved
Theoretical proof:
Given: MN and OP are two line segments which are equal and parallel (i.e. MN = OP, MN // OP).
Construction: Join MO and NP and join M and P.
To prove: i) MO = NP ii) MO // NP
Proof:
S.N.  Statements  Reasons 
1. 
In \(\triangle\)MOP and \(\triangle\)MNP i) MN = OP (S) ii) ∠MPO = ∠NMP (A) iii) MP = MP (S) 
i) Given ii) MN // OP, alternate angles. iii) Common side of both the triangles. 
2.  \(\triangle\)MOP ≅ \(\triangle\)MNP  By S.A.S. axiom. 
3.  MO = NP  Corresponding sides of congruent triangles. 
4.  ∠OMP = ∠MPN  Corresponding angles of congruent triangles. 
5.  ∴ MO // NP  Equal Alternate formed in a pair of lines MO and NP. 
Proved
Theoretical proof:
Given: PQ and RS are two equal and parallel line segments (i.e. PQ = RS, PQ // RS).
Construction: Join PS and RQ. Then, the line segments PS and RQ intersect at the point O.
To prove: i) PO = OS ii) RO = OQ
Proof:
S.N.  Statements  Reasons 
1. 
In \(\triangle\)POQ and \(\triangle\)ROS i) ∠OPQ = ∠OSR (A) ii) PQ = RS (S) iii) ∠PQO = ∠SRO (A) 
i) PQ // RS being alternate angles. ii) Given iii) Alternate angles, PQ // RS. 
2.  \(\triangle\)POQ ≅ \(\triangle\)ROS  By A.S.A. axiom. 
3.  PO = OS and RO = OQ  Corresponding sides of congruent triangles. 
Proved
Theoretical proof:
Given: MNOP is a parallelogram (i.e. MN // PO and MP // NO) in which diagonals MO and NP intersect at X.
To prove: MO = XO and NX = XP.
Proof:
S.N.  Statements  Reasons 
1. 
In \(\triangle\)MXN and \(\triangle\)PXN i) ∠XNM = ∠XPO (A) ii) MN = OP (S) iii) ∠XMN = ∠XOP (A) 
i) Alternate angles, MN // PO. ii) Opposite sides of a parallelogram. iii) Alternate angles, MN // PO. 
2.  \(\triangle\)MXN ≅ \(\triangle\)PXO  By A.S.A. axiom. 
3.  MX = XO and NX = XP  Corresponding sides of congruent triangles are equal. 
Proved
Theoretical proof:
Given: DEFG is a quadrilateral in which two diagonals DF and BD bisect each other at O. (i.e. DO = OF and EO = OG).
To prove: DEFG is a parallelogram i.e. DE // GF and DG // EF.
Proof:
S.N.  Statements  Reasons 
1. 
In \(\triangle\)DOE and \(\triangle\)GOF i) DO = OF (S) ii) ∠DOE = ∠FOG (A) iii) EO = OG (S) 
i) Given ii) Vertically opposite angles. iii) Given 
2.  \(\triangle\)DOE ≅ \(\triangle\)GOF 
By S.A.S axiom. 
3.  DE = GF  Corresponding sides of congruent triangles. 
4.  ∠DEO = ∠OGF  Corresponding angles of congruent triangles. 
5.  DE // GF  From statement 4 alternate angles are equal. 
6.  DG = EF and DG // EF 
Two line segments joining the end points on the same side of two equal and parallel line segments are also equal and parallel. 
7.  ∴ DEFG is a parallelogram.  Form statement 5 and 6. 
Proved
Experimental verification:
Step 1: Draw three triangles ABC of different positions and sizes with BC as the base in a different orientation.
Step 2: Mark the midpoint of side AB in each triangle as P and draw a line parallel to BC such that it cuts the side AC at Q.
Step 3: Measure the sizes of AQ and QC in each figure and complete the table below:
Figure  AQ  QC  Result 
i)  
ii)  
iii0 
Conclusion: A straight line segment drawn through the midpoint of one of a triangle and parallel to another side bisects the third side.
Theoretical proof:
Given: In \(\triangle\)ABC, E is the midpoint of the side AB and EF//BC.
To prove: AF = FC
Construction: Produce EF to O such that CD // BE.
Proof:
S.N.  Statements  Reasons 
1.  BCOE is a paralleogram.  BE//CO and BC//EO 
2.  BE = CO  Opposite sides of the parallelogram. 
3.  BE = EA  E is the midpoint of AB. 
4.  ∴ EA = CO  
5. 
In \(\triangle\)AEF and \(\triangle\)COF i) ∠AFE = ∠CFO (A) ii) ∠AEF = ∠COF (A) iii) EA = CO (S) iv)∴ \(\triangle\)AEF ≅ \(\triangle\)COF 
i) Vertically opposite angles. ii) BA//CO and being alternate angles. iii) From statement 4. iv) By S.A.A. axiom. 
6.  AF = FC  Corresponding sides of congruent triangles are equal. 
7.  EF bisects the side AC at F.  From statement 6. 
Proved
Experimental verification:
Step 1: Draw three triangles MNO of different positions and sizes in different orientations.
Step 2: Find the midpoints of the sides MN and MO and mark them as X and Y.
Step 3: Join the points X and Y.
Step 4: Measure the corresponding angles∠MXY and∠XNO. And also measure the lengths of XY and NO and fill up the table.
Figure  ∠MXY  ∠XNO  Result  XY (cm)  NO (cm)  Result 
i)  
ii)  
iii) 
Proved
Theoretical proof:
Given: X and Y are the midpoints of the side MN and MO respectively i.e. MX = NX and MY = YO.
To prove: XY//NO and XY = 1/2NO
Construction: Produce XY to Z such that OZ//NX.
Proof:
S.N.  Statements  Reasons 
1. 
In \(\triangle\)MXY and \(\triangle\)OYZ i) ∠XYM = ∠OYZ (A) ii) MY = YO (S) iii) ∠XMY = ∠YOZ (A) iv) \(\triangle\)MXY ≅ \(\triangle\)OYZ 
i) Vertically opposite angles are equal. ii) Given iii) OZ // NM and being alternate angles. iv) A.S.A axiom. 
2.  MX = OZ  Corresponding sides of congruent triangles. 
3.  MX = NX  Given 
4.  ∴ NX = OZ  From statements 2 and 3. 
5.  NX // OZ  By construction. 
6.  ∴ XZ // NO i.e. XY // NO and XZ = NO  Being NX = OZ and NX//OZ 
7.  XY = YZ  Corresponding sides of congruent triangles. 
8.  XZ = XY + YZ  Whole part axiom. 
9. 
XY = 1/2 XZ or, XY = 1/2 MN 
From statement 8 and 6. 
Proved
A quadrilateral having opposite sides parallel is called the parallelogram.
In a parallelogram ,
1. Opposite sides are equal.
2. Opposite angles are equal.
3. Diagonals bisect each other.
4. Area = base \(\times\) height = b \(\times\) h
What is a parallelogram ? Write its 3 Properties.
A parallelogram is a quadrilateral in which opposite sides are parallel.
Its properties are:
1. Opposite sides are equal.
2. Consecutive angles are supplementary.
3. Diagonals Bisect each other.
What is a rectangle ? Write its 4 properties.
A rectangle is a parallelogram in which each angle is a right angle.
Its properties are :
1. Each angle is 90^{o. }2. Opposite sides and angles are equal.
3. Lentgh of the diagonals are equal.
4. Diagonals bisect each other.
What is a square ? What are its properties ?
A square is a parallelogram in which all sides are equal and each angle is 90^{o. }
Its properties are :
1. All sides are equal
2. Each angle is 90^{o, }3. Diagonals are equal .
Find the angles a , b , c , d e , m , n and p in each of the following figures.
ADC = ABC [opposite angles of a parm]
n = 75
BAD + ABC = 180 [ sum of coiterioir angles AD  BC]
or , m = 75 = 180
m = 180  75 = 105
DCE = ADC [alternate angles and AD  BE]
or . P = n
\(\therefore\) P = 75
Hence , m = 105 , n = 75 and p = 75
Fid the measurement of all the four angles of the given parallelograms.
ACB = ACD [in a rhombus , diagonals bisects the vertex]
= 57
\(\therefore\) BCD = ACB + ACD
= 57 + 57
= 114
\(\therefore\) BAD = BCD = 114
ABC + BCD = 180 [sum of cointerior angles and AB  DC]
or , ABC = 180  BCD = 180  114
= 66
\(\therefore\) ADC = ABC = 66
Hence , 66 , 114 , 66 , 114
In the figure , ABCD is a parallelogram and ΔEBC is an equilateral triangle. Find the value of DCE.
CBE = 60 [being BC = BE = CE]
DAB = CBE [corresponding angles ad AD  BC]
\(\therefore\) DAB = 60 [opposite agles of a parallelogram]
BCD = DAB = 60
Now , DCE = BCE + BCD = 60 + 60 = 120
In the give paralleogram PQRS , QPS = (5x + 5) and PQR = (4x  5) , find te value of x.
Here , QPS + PQR = 180 [sum of coiterior angles and PS  QR]
or , (5x + 50 = 94x  50 = 180
or , 9x = 180
\(\therefore\) x = \(\frac{180}{9}\) = 20
T is a point o side QR of the given parallelogram PQRS. if PQ = PT , QPT = x
and PQT = 4x , then find the vallue of PSR
PTQ = PQT \(\therefore\) PQ = PT
= 4x
PTQ + PQT + QPT = 180 [sum of all three interior angles of a triangle]
or , 4x + 4x + x = 180
or , 9x = 180
or , x = \(\frac{180}{9}\) = 20
\(\therefore\) PSR = PQT = 4 \(\times\) 20 = 80 [opposite angles of a parallelogram]
A figure formed by joining ends of two line segmet which bisect each other perpendicularly is a rhombus.
Give , AC and BD are bisected at a point O perpendicularly
Here , AO = OC , OB = OD , AC⊥BD
To prove AB = BC = CD = AD
statements  reasons 
1 In ΔAOB and ΔBOC , (i) AO = OC (ii) AOB = BOC (iii) BO = AO  1 (i) From given (ii) both are a right angle (iii) Common sides 
2. ΔAOB≅ΔBOC  2. by S.A.S fact 
3. AB = BC  3. correspoding sides of ≅Δ^{s} 
4. ΔBOC≅ =ΔDOC  4. From similar statements and reasons as above (1) AND (2) 
5. BC = DC  5 correspondig sides of ≅Δ^{s} 
6. ΔDOC ≅ΔAOD  6. from similar statements and reasos as above a and 2 
7. DC = AD  7. corresponding sides of ≅Δ^{s} 
8. AB = BC = DC = AD  8, from statements 3 , 5 and 7 
All angles of a rectangle are right angle.
Given , ABCD is a rectangle
i.e. AD  BC ,AB  DC and DAB = 90
To prove ABC = BCD = ADC = DAB = 90
Statements  Reasons 
1 DAB = 90  1. from given 
2. DAB + ABC = 180 or , 90 + ABC = 180 \(\therefore\) ABC = 180  90 =90  2. Being AD  BC and sum of cointerior angles. 
3. ABC + DCB = 180 or , 90 + DCB = 180 DCB = 90  3. being AB  DC and sum of cointerior agles. 
4. ADC = 90  4. from statements and reasons as above. 
5. DAB = ABC = DCB = ADc  5. from statements 1 , 2 , 3 , ad 4 
Diagonals of a rectangle are equal .
Given : ABCD is a rectangle and AC and BD are diagnolas.
To prove AC = BD
Statements  Reasons 
1. In ABC and BCD (i) AB= DC (ii) ABC = DCB (iii) BC = BC  1. (i) Opposite sides of rectangle are equal (ii) both are right angle (iii) Common sides 
2. ABC≅BCD  2. By SAS fact 
3. AC = BD  3. Corresponding sides of ≅Δ^{s } 
If the diagonals of a parallelogram (in D which adjacent sides are not equal) are equal , then it is a rectangle.
Given : In a parallelogram ABCD , diagonals AC = BD
To prove ; ABCD is a rectangle.
Statements  Reasons 
1. In ABD and ABC (i) AB = AB (ii) AD = BC (iii) BD = AC  1. (i) Common sides (ii) Opposite sides of a parallelogram (iii) From Given 
2. ABD ≅ ABC  2. By S.S.S fact. 
3. DAB = ABC  3. Corresponding angles of ≅ triangles. 
4. DAB + ABC = 180 or , DAB + DAB = 180 or , 2 DAB = 180 or , DAB = 180 / 2 = 90  4. Being AD  BC , sum of cointerior angles is 180 
5. ABCD is a square  5. In parallelogram ABCD , A is a right angle 
If the diagonals of a rhombus are equal , then it is a square.
given , In retcangle ABCD , AB = BC = CD = DA and diagonals BD and AC intersect at point O.
To prove: ABCD is a square.
Statements  Reasons 
1. ABC and ABD (i). BC = AD (ii) AB = AB (iii) AC =BD  1. (i) From given (ii) Common sides (iii) From given 
2. ABD≅ABC  2. By S.S.S fact 
3. ABC = BAD  3. Corresponding angles of ≅triangles. 
4. ABC + BAD = 180 or , BAD + BAD =180 or , 2BAD = 180 or , BAD = 90  4. Rhombus is also parallelogram. Here , the sum of cointerior angles is 180 and AD  BC. 
5. ABCD is a square  5. In , ABCD , BAD is 90 and all sides are equal. 
The lines joining the mid points of the opposite sides of a quadrilateral bisect each other.
Given ,
In quaqdrilateral ABCD , L , M , P and Q are the midpoints of AB , BC , CD and DA respectively in which PL and QM are joined and intersect at point O.
To prove : PL and Qm bisects each other.
Construction: Join A and C.
Statements  Reasons 
1. PQ  AC  1. In , ADC , the line PQ joining the mid point Q of AD and P of DC is parallel to the third side AC. 
2. LM  AC  2. In ABC , the line LM joining the mid point L of AB and M of BC is parallel to tht third side AC. 
3. PQ  ML  3. From above statements 1 and 2. 
4. QL  PM  4. By joining D and B and similarly from statements and reasons as above 1 and 2. 
5. QLMP is a parallelogarm.  5. Opposite sides are parallel. 
6. PL and Qm are bisected at Q.  6. Diagonals of a parm , bisects each other. 
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