Note on Parallelogram

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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 of properties of a parallelogram

Theorem 7: The opposite sides of a parallelogram are equal

Theoretical proof:

Theorem-7

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

Theorem 8: The quadrilateral having opposite sides equal is a parallelogram

Theoretical proof:

08

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

Theorem 9: Opposite angles of a parallelogram are equal

Theoretical proof:

Theorem-9

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 = 180o Sum of Co-interior angles, DG // EF.
2. ∠E + ∠C = 180o Sum of Co-interior 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 = 180o Sum of Co-interior 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

Theorem 10: Quadrilateral having opposite angles equal is a parallelogram

Theoretical proof:

Thorem-10

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 = 360o Sum of four interior angles of a quadrilateral is 3600.
2. ∠A + ∠B + ∠A + ∠B = 360o From given
3.

or, 2∠A + 2∠B = 360o

or, 2(∠A +∠B) = 360o

∴ ∠A +∠B = 180o

On simplifying statement 2.

 

4. AD||BC From statement 3, the sum of co-interior angles is 1800.
5. ∠A + ∠D + ∠A + ∠D = 360o Sustituting the value in 1 from the given.
6.

or, 2(∠A +∠B) = 360o

∴ ∠A + ∠B = 180o

On simplifying the statement 5.
7. AB || DC From statement 6, the sum of co-interior angles is 1800.
8. ∴ ABCD is a parallelogram. From statement 4 and 7, opposite sides parallel.

proved

Theorem 11: Two line segments joining the end points towards the same side of two equal and parallel line segments are also equal and parallel

Theoretical proof:

Theorem-11

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

Theorem 12: Two line segments joining the opposite end points of two equal and parallel line segments bisect each other

Theoretical proof:

THEOREM-12

 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

Theorem 13: The diagonals of a parallelogram bisect each other 

Theoretical proof:

Theorem-13

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

Theorem 14: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram

Theoretical proof:

Theorem-14

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

Mid- point theorem

Theorem 15: A straight line segment drawn through the mid-point of one of a triangle and parallel to another side bisects the third side

Experimental verification:

15

Step 1: Draw three triangles ABC of different positions and sizes with BC as the base in a different orientation.

Step 2: Mark the mid-point 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 mid-point of one of a triangle and parallel to another side bisects the third side.

Theoretical proof:

1

Given: In \(\triangle\)ABC, E is the mid-point 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 mid-point 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

Theorem 16: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it

Experimental verification:

theorem-16Step 1: Draw three triangles MNO of different positions and sizes in different orientations.

Step 2: Find the mid-points 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:

The-16

Given: X and Y are the mid-points 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

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Very Short Questions

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.

A rectangle is a parallelogram in which each angle is a right angle.
Its properties are :
1. Each angle is 90o.
2. Opposite sides and angles are equal.
3. Lentgh of the diagonals are equal.
4. Diagonals bisect each other.

A square is a parallelogram in which all sides are equal and each angle is 90o.
Its properties are :
1. All sides are equal
2. Each angle is 90o,
3. Diagonals are equal .

ADC = ABC [opposite angles of a parm]
n = 75
BAD + ABC = 180 [ sum of co-iterioir 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

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 co-interior angles and AB || DC]
or , ABC = 180 - BCD = 180 - 114
= 66
\(\therefore\) ADC = ABC = 66
Hence , 66 , 114 , 66 , 114

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

Here , QPS + PQR = 180 [sum of co-iterior angles and PS || QR]
or , (5x + 50 = 94x - 50 = 180
or , 9x = 180
\(\therefore\) x = \(\frac{180}{9}\) = 20

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]

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 ≅Δs
4. ΔBOC≅ =ΔDOC 4. From similar statements and reasons as above (1) AND (2)
5. BC = DC 5 correspondig sides of ≅Δs
6. ΔDOC ≅ΔAOD 6. from similar statements and reasos as above a and 2
7. DC = AD 7. corresponding sides of ≅Δs
8. AB = BC = DC = AD 8, from statements 3 , 5 and 7

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 co-interior agles.
4. ADC = 90 4. from statements and reasons as above.
5. DAB = ABC = DCB = ADc 5. from statements 1 , 2 , 3 , ad 4

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 ≅Δs

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 co-interior angles is 180
5. ABCD is a square 5. In parallelogram ABCD , A is a right angle

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  • Find the area of a parallelogram with a base of 7 inches and a height of 10 inches.

    60 in2
    30 in2
    45 in2
    70 in2
  • The area of a parallelogram is 24 square centimeters and the base is 4 centimeters. Find the height.

    6 cm
    12 cm
    9 cm
    5 cm
  • Find the area of a parallelogram with a base of 8 feet and a height of 3 feet.

    45 ft2
    24 ft2
    55 ft2
    23 ft2
  • Find the area of parallelogram with a base of 4 meters and a height of 9 meters.

    10 m2
    36 m2
    12 m2
    24 m2
  • The area of a parallelogram is 64 square inches and the height is 16 inches. Find the base.

    7 in
    4 in
    10 in
    5 in
  • A parallelogram-shaped garden has an area of 42 square yards and a height of 6 yards. Find the base.

     

    4 yd
    2 yd
    5 yd
    7 yd
  • A parallelogram has an area of 54 square centimeters and a base of 6 centimeters. Find the height.

    5 cm
    10 cm
    6 cm
    9 cm
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theorem 3


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