Famous geometry theorems.

              

  Famous geometry theorems.

 

  • Today in this article I will discuss on some geometry theorems of quadrilaterals  for class 9 and class 10 which are very famous in mathematics education.


    •   Prove that if one  pair of opposite side of a quadrilateral is equal and parallel then it is a parallelogram.



    Famous geometry theorems.



    Answer:-
    In the picture, ABCD is a quadrilateral, whose AB ∥ CD ,also AB = CD, To prove that,ABCD is a parallelogram.


    Proof:- ∵ AB∥ CD and AC is the transversal
      
       ∴ ∠BAC = <ACD [∵ alternate angles are 
                                   equal].....(1)
    Now,   in between Δs ABC and ACD we get,
      ∵ ∠BAC = <ACD [ from….(1)]
     ∵  AB = CD [ ∵ given]
    and,  AC = AC [∵ common side of the Δs]
      ∴ Δ ABC ≌ Δ ACD [∵by SAS congruence 
                                   rule]
    ∴  BC = AD              [∵ by CPCT]
    and <ACB = <CAD


    but , <ACB and < CAD are alternate angles and AC is the transversal between them,
       ∴  BC ∥ AD 
    Now, in quadrilateral ABCD, we find,
        ∵ BC ∥ AD, BC= AD [∵ Proved]
    and AB∥ CA, AB= CD [∵ given]
    So,
       Opposite sides of quadrilateral ABCD are equal and parallel.
        ∴ ABCD is a parallelogram ,  Proved.

    Construction:- Diagonal AC is joined.


    •  Prove  that if opposite sides of a quadrilateral are equal, then it is a parallelogram        

    Famous geometry theorems.,



    Answer:- 
     In the picture, ABCD is a quadrilateral,its AB= CD and BC = AD To prove, ABCD is a parallelogram. 

     
    Construction:- Diagonal AC is joined. 
     
    Proof:-
    ∵ In between Δs ABC and ADC
    we get, 
     
    ∵ AB = CD [ ∵ given] 
     ∵ BC = AD [∵ given] 
     and AC = AC [ ∵Common side of the Δs] 
     ∴ Δ ABC ≌ Δ ADC [ ∵by SSS congruence                            rule]  
    ∴ ∠BAC = < ADC [by CPCT] ,
    but they are alternate angles and AC is the transversal between them, ∴ AB∥CD, similarly we can prove BC∥AD 
     
    Now in quadrilateral, ABCD 
    we find AB = CD [∵given] and BC = AD and AB∥ CD [ ∵ proved] also BC∥ AD 
     So, opposite sides of the quadrilateral ABCD are equal and parallel 
     ∴ ABCD is a parallelogram, proved. 


    •  Prove that diagonals of a parallelogram bisect eachother.


     
    Famous geometry theorems.



    Answer:- 
    In the picture, ABCD is a parallelogram, AC and BD are its diagonals intersect each other at point " O" 
     
    To prove, AO = OC and BO = OD 
     
    Proof:- 
     ∵ In between Δs AOB and COD 
     we get, 
     ∵ AB = CD [∵ Opposite sides of the parallelogram] 
     ∵ ∠AOB =<COD[∵vertically opposite  angles]
     and,
      <OAB = <OCD [∵alternate angles]
                                rule]
    ∴ AO = OC
    and             [ ∵ by CPCT]
      BO = OD 
                     Proved.

    ∴ Δ AOB ≌ ΔCOD [∵by ASA congruence 



    • Prove that diagonals of a rectangle are equal.


    Famous geometry theorems.




    Answer:- 
    In the picture, PQRS is a rectangle, PR and QS are its two diagonals. 
     To prove:- PR = QS 
     Proof:- ∵ In between Δs PQR and PQS
     
    we get, 
     ∵ QR = PS [∵ Opposite sides of the rectangle] 
     ∵ PQ = PQ [∵ common side of the Δs] 
     
    and
     
    <PQR = <QPS [∵ABCD is rectangle,so both angles are 90°]

    ∴ Δ PQR ≌ Δ PQS

    ∴ PR = QS [by CPCT]
                    Proved.





    • Prove that if one angle of a rectangle is 90°, then all angles of the rectangle will be right angle.


    Famous geometry theorems.




    Answer:-

    Let, XYZP is a rectangle,its <X=90° 

    To prove;-<X=<Y=<Z=<P=90°= right angle

    Proof:- ∵ ABCD is a rectangle,


         ∴∠X +∠Y=180°[∵sum of adjacent angles of a rectangle is 180°]
         => 90° + <Y = 180°[ <X=90°given]
        => <Y = 180°- 90°
       => <Y= 90°
    Similarly,
       we can prove, <Z =<P=90°
    Now, in rectangle XYZP
       <X=<Y=<Z=<P=90°= right angle ,proved

    If you love my article and my blog then please support me and share it.

                

    FREQUENTLY ASKED QUESTIONS:-
    Write two properties of parallelogram?

    Answer :- Two properties of parallelogram are :- (i) Opposite sides of the parallelogram are equal and parallel. (ii) Diagonals of a parallelogram bisect each other at their point of contact.

    Write three properties of a rectangle.

    Answer:-Three properties of a rectangle are :- (i) Diagonals of a rectangle are equal. (ii) Each angle of a rectangle is 90° (iii) Opposite sides are equal and parallel.

    Write two similarities between rhombus and square.

    Answer :- Two similarities between rhombus and square are :- (i) All sides of both are equal. (ii) Both's diagonals bisect each-other perpendicularly.

    What is the relation of areas between a triangle and a rectangle if both are standing on the same base and in between same parallels?

    Answer:- The relation is, area of triangle = ½ area of rectangle

    Name the quadrilateral whose one pair of sides are parallel.

    Answer:- Trapezium








    Share this post:-







    Please give your valueable feedback:--