Dictionary Rank of a Word | Permutations & Combinations

 PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

Theorems on Quadrilaterals Ch-8 Class-IX

 Theorems on Parallelograms Ch-8 Class-IX

Explanation of all theorems on Parallelograms chapter 8 class IX, Theorem 8.1, 8.2, 8.3, 8.4, 8.5, 8.6, 8.7, 8.8, 8.9, 8.10, Mid point theorem and its converse. All theorems of chapter 8 class IX.


Theorem 8.1: Prove that a diagonal of a parallelogram divides it into two congruent triangles.

Given: In Parallelogram ABCD, AC is the diagonal

To Prove:  △ACD ≌ △ABC


Proof: In △ACD and △ABC, 
∠1 = ∠2  .........   (Alternate angles
∠3 = ∠4  ..........  (Alternate interior angles
AC = AC ........   (Common Sides

⇒ By ASA ≌ rule  

△ACD ≌ △ABC


 Theorem 8.2: In a parallelogram, opposite sides are equal.

Given: ABCD is a parallelogram

To Prove : AB = CD and BC = AD

Proof: In ACD and ABC, 

1 = 2  .........   (Alternate angles

3 = 4  ..........  (Alternate interior angles

AC = AC ........   (Common Sides

By ASA  rule  

ACD ABC

AB = CD and BC = AD  …..  By CPCT


Theorem 8.3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Given: In quadrilateral ABCD, AB = CD and BC = AD

To prove: ABCD is a parallelogram.

Proof:

In ACD and ABC, 

AD = BC  ..........  (Given

CD = AB .........   (Given

AC = AC ........   (Common Sides

By SSS  rule, 

ACD ABC

1 = 2  and 3 = 4  …..  By CPCT

Now 1 = ⇒ alternate angles are equal

⇒ AD BC  …….. (i)

Now 3 = ⇒ alternate angles are equal

⇒ AB CD …….. (ii)

From (i) and (ii) we  have

In quadrilateral ABCD both pair of opposite sides are parallel.

Hence ABCD is a Parallelogram.


 Theorem 8.4:  In a parallelogram opposite angles are equal.

Given: ABCD is a parallelogram
To prove: AD = BC and CD = AB

Proof: ABCD is a parallelogram

 AD BC

A + B = 180o ………. (i)

    ABCD is a parallelogram

  AB CD

B + C = 180o ………. (ii)

From (i) and (ii)

A + B = B + C

A = C

Similarly : B = D


Theorem 8.5: If in a quadrilateral each pair of opposite angles is equal, then it is a parallelogram.

Given: In quadrilateral ABCD , A = C and B = D

To prove: ABCD is a parallelogram

Proof: Since sum of all angles of a quadrilateral is equal to 360o

A + B + C + D = 360o

A + B +A + B = 360o

2(A + B) = 360o

A + B = 180o

⇒ Adjacent angles are supplementary

   AD BC

Similarly : AB CD

Since both pair of opposite sides are parallel

Hence ABCD is a Parallelogram


Theorem 8.6: The diagonals of a parallelogram bisects each other.

Given: ABCD is a Parallelogram.

To Prove: OA = OC and OB = OD

Proof: In AOD and BOC, 

1 = 2  .........   (Alternate angles

3 = 4  ..........  (Alternate angles

AD = BC ........   (Common Sides

By ASA  rule 

AOD ≌ BOC

OA = OC and OB = OD  …..  By CPCT


Theorem 8.7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Given: In quadrilateral ABCD, OA = OC and OB = OD

To Prove : ABCD is a parallelogram.


Proof: In AOD and BOC, 

1 = 2  .........   (Vertically Opposite angles

OA = OC  ..........  (Given

OB = OD ........   (Given

By ASA  rule  

AOD  BOC

3 = 4  and AD = BC  …..  By CPCT

3 = 4 ⇒ Alternate angles are equal

AD  BC

Now AD  BC and AD = BC 

⇒ One pair of opposite sides are equal and parallel

⇒ ABCD is a Parallelogram.


Theorem: 8.8  A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel.

Given: In quadrilateral ABCD, AD  BC and AD = BC 

To Prove: ABCD is a parallelogram

Proof: In ACD and △ABC, 

AD = BC ........ (Given

1 = 2  .........   (Vertically Opposite angles

AC = AC .... (Common Side

 By SAS  rule  

ACD  △ABC

3 = 4  …….. By CPCT

3 = 4 ⇒ Alternate angles are equal

AB  CD

Now AD  BC and AB  CD

⇒ Both pair of opposite sides are parallel

Hence ABCD is a parallelogram

Mid Point Theorem

 Theorem 8.9: If a line is drawn through the mid points of two sides of a triangle then the line is parallel to the third side and is half of it.

Given: In ABC, D and E are the mid points of sides AB and AC

To Prove: DE ॥ BC and DE =  BC


Construction: Extend DE to F such that DE = EF and join CF
Proof: In ADE and CEF
DE = EF  .........  (By Construction
AE = CE  ........  (E is the mid point of  AC
∠1 = ∠2  ........  (Vertically opposite angles

 By SAS  rule  

ADE  △CEF

∠3 = ∠4  ........  By CPCT

∠3 = ∠4 ⇒ Alternate angles are equal

⇒ AB ॥ CF   or  BD ॥ CF ......  (i)

AD = CF ....... By CPCT

But AD = BD ......  (Given

⇒ BD = CF   ...............  (ii)

From (i) and (ii) we have

BD = CF  and  BD ॥ CF

⇒ One pair of opposite sides are equal and parallel

∴ BCFD is a parallelogram

⇒ DF = BC and DF ॥ BC 



⇒  DF = BC and  
Hence prove the required theorem

Converse of Mid Point Theorem

Theorem 8.10:  In a triangle if a line is drawn through the mid point of one side and is parallel to the second side, then it bisects the third side.

Given: In ABC,  point D is the mid point of side AB and DE ॥ BC
To Prove: E is the mid point of AC
Construction: If possible let E is not the mid point of AC, so let us take another point F the mid point of AC.


Proof: In ABC, 
            D is the mid point of AB ........ (Given)
            F is the mid point of AC ......... (By Construction)
Therefore by mid point theorem  DF ॥ BC

But DE ॥ BC ........(Given)
This is possible only if DE and DF are coincident lines
⇒ E and F are coincide with each other.
⇒ E and F represents the same point.
Hence DF ॥ BC      DE ॥ BC
⇒ Point E is the mid - point of AC or AE = EC

Hence prove the required theorem



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