Notes on Parallelogram | Grade 9 > Compulsory Maths > Geometry | KULLABS.COM

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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: 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: 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: 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: 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: 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: 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: 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: 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: 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:

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: Step 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:

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

.

### 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
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]
$$\therefore$$ 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 = 180or , 90 + ABC = 180$$\therefore$$ ABC = 180 - 90 =90 2. Being AD || BC and sum of cointerior angles. 3. ABC + DCB = 180or , 90 + DCB = 180DCB = 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 = 180or , DAB + DAB = 180or , 2 DAB = 180or , 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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70 in2
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24 ft2
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55 ft2

10 m2
36 m2
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12 m2

7 in
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2 yd
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## ASK ANY QUESTION ON Parallelogram

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##### Alisha rai

Can you give some important questions