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 = 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
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
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 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
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
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 90o.
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 90o.
Its properties are :
1. All sides are equal
2. Each angle is 90o,
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 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
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 co-interior 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 co-iterior 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 ≅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 |
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 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 |
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 ≅Δs |
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 co-interior 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 co-interior 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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