maths ppt on quadrilaterals from sumedha tiwari

1. By:- Sumedha Tiwari
IX-C

2.  4 vertices
• 4 sides .B
• 4 angles A.
D.
.
C

3. .B
A. o
.B 360
A.
D.
.
C
D.
The sum of ALL the angles of a
quadrilateral is 360o .
C

4. General Quadrilateral
Trapezoid
4 sides, 4 angles
Only 1 pair of parallel
sides

5. General Quadrilateral
Parallelogram Trapezoid
4 sides, 4 angles
Opposite sides are parallel Only 1 pair of parallel
and congruent sides

6. General Quadrilateral
Parallelogram Trapezoid
4 sides, 4 angles
Opposite sides are parallel Only 1 pair of parallel
and congruent sides
Rectangle
A parallelogram with 4
right angles

7. General Quadrilateral
Parallelogram Trapezoid
4 sides, 4 angles
Opposite sides are parallel Only 1 pair of parallel
and congruent sides
Rectangle
A parallelogram with 4
right angles
Square
A rectangle with 4
congruent sides

8. General Quadrilateral
Parallelogram Trapezoid
4 sides, 4 angles
Opposite sides are parallel Only 1 pair of parallel
and congruent sides
Rectangle
A parallelogram with 4
right angles
Rhombus
A parallelogram with 4
congruent sides
Square
A rectangle with 4
congruent sides

9. Diagonal of a parallelogram
divides it into two congruent
triangles . Proof

10. GIVEN : ABCD is a parallelogram. A .B
AB=CD AD=BC
TO PROVE : ABD  CBD
PROOF :In ABD & CBD
(1)AB = CD {Opposite sides of
parallelogram}
D C
(2)AD = BC
(3)BD =BD {Common side}
Thus ABD  CBD
HENCE PROVED

11. In a parallelogram each
pair of opposite sides are
equal Proof

12. A B
 GIVEN : ABCD is a parallelogram.
TO PROVE : AB =CD
AD =BC
PROOF :In ABD & CBD D C
(1)LADB=LABC (Alternative interior angles)
(2)BD = BD ( Common side )
(3)LBDC= LABD (Alternate interior angles )
ABD  BCD
Thus Proved AB = CD (CPCTE)
AD = BC (CPCTE)
(HENCE PROVED)

13. In a parallelogram each
pair of opposite angles are
equal
Proof

14. A B
 GIVEN : ABCD is a parallelogram.
A
 AB=CD AD=BC
 TO PROVE : LA = LC LB = LD
 PROOF :In ABD & CBD D C
 (1)AB = CD Opposite sides of parallelogram
 (2)AD = BC
 (3)BD =BD Common side
 Thus ABD  CBD
 LC = LA (C.P.C.T.E)
 Similarly , LB = LD (C.P.C.T.E)
 HENCE PROVED

15. Diagonal of a
parallelogram bisect each
other. Proof

16.  GIVEN : ABCD is a parallelogram
 AD=BC A B
 TO PROVE: AO=OC
 DO=OB
O
 PROOF: In AOD & BOC D
(1)LDAO=LBCO (Alt. interior angle)
C
(2)LADO=LCBO (Alt. interior angle)
(3)AD=BC (Given)
Thus AOD  BOC
AO=OC (C.P.C.T.E)
DO=OB (C.P.C.T.E)
HENCE PROVED

17.  The line segment joining the mid-points of two
sides of a triangle is parallel to the third side and is
A
half of it. 3
1 E
D F
2
4
B C
Proof

18.  Given: In ABC & EFC
 AD=DB(mid pt)
 AE=EC(mid pt)
 To prove: DE IIBC and DE=1/2 BC
 Construct: Draw a line from C to extended point F such that it is
parallel to AB.
 In ABC & EFC
 L1=L2 (V.O.A)
 AE=EC
 DE=EF
 So ABC  EFC by SAS

19.  AD=CF (C.P.C.T)
 L3 = L4 (C.P.C.T)
 But they are alternate Interior angles for lines AB and CF
 Thus, AB IICF or DB IIFC -(1)
 AD=CF (proved)
 Also AD=DB (given)
 Thus, DB=FC -(2)
 From (1) and(2)
 DBCF is a parallelogram.
 Thus, the other pair DF is parallel to BC and DF=BC (By
construction E is the mid-pt of DF)
 Thus, DE=1/2 BC

20.  Thank you