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WHAT ARE
QUADRILATERALS?
• A quadrilateral is a four-sided
polygon.
• If sides are parallel, the distance
between them is constant, and
they will never cross, even if
extended.
SIX TYPES OF
QUADRILATERALS
• RECTANGLE
• SQUARE
• PARALLELOGRAM
• RHOMBUS
• KITE
• TRAPEZOID
A RECTANGLE IS
• A quadrilateral having two pairs of equal
sides that are parallel
• Sides meet at right angles (90 )
• Diagonals are equal in length
RECTANGLE
 In a rectangle the diagonals, besides being equal in length
bisect each other.
Given: ABCD is a Rectangle the diagonals are AC & BD
Bisect each other at a Point O.
To Prove: (i) AC = BD
(ii) OA = OC or OB = OD
Proof: In ∆ ABC and ∆ ABD
AB = AB --------- (Common)
BC = AD ---------------------(Apposite sides ABCD)
∠ A = ∠ B = 90 (angles of the rectangles)
∆ ABC ≅ ∆ ABD S. A. S. Rule
So, AC = BD Proved
A B
CD
O
RECTANGLE
(ii) OA = OC or OB = OD
Proof: In ∆ AOB and ∆ COD
AB = CD (Sides of the ABCD)
AB || CD and transversal AC & BD
Then, ODC = OBA
And OCD = OAB
Alternate interior angles
∆ AOB ≅ ∆ COD --- A. S. A Rule
So that all congruent part are equal .
OA = OC
and OB = OD
Proved
A B
CD
O
A SQUARE IS
• A rectangle having all sides of equal
length
SQUARE
 The diagonals of a square are perpendicular bisectors of
each other
Given: ABCD is a Square, Where AC and BD is a diagonal
bisect each other at a Pont ‘O’
To Prove: ∠ AOD = ∠COD = 90
Proof: In ∆ AOD and ∆ COD
OD Common & OA =OC
AD = DC ------ Two sides of Squire
So, (By SSS) ∆ AOB ∆ COD,
∠ AOD = ∠COD
Since ∠AOD and ∠COD are a linear pair,
∠ AOD = ∠ COD = 90
A B
CD
O
PARALLELOGRAM
• A quadrilateral having two pairs of equal
sides that are parallel
• Opposite angles are equal
• Diagonals are not equal in length
PARALLELOGRAM
 The diagonals of a parallelogram bisect each other
Given: ABCD is a parallelogram where AC and BD is a
diagonal bisect each other at Point O
To Prove: AO = OC or BO= OD
Proof: In ∆ AOB and ∆ COD
AB = CD
AB || CD
CDO = OBA
And DCO = OAB ------------ Alternate interior angles
So, ∆ AOB ∆ COD, then
AO = OC or BO =OD Proved
A B
CD
O
A RHOMBUS IS
• A parallelogram having all sides of equal
length
RHOMBUS
 The diagonals of a rhombus are perpendicular
bisectors of one another.
Given: ABCD is a rhombus. Therefore it is a parallelogram
too. Since diagonals bisect each other,
OA = OC and OB = OD.
To Prove: ∠ AOD = ∠COD = 90
Proof: In ∆ AOD and ∆ COD
OD Common & OA =OC --Given
AD = DC ------ Two sides of this rhombus
So, (By SSS) ∆ AOB ∆ COD, then ∠ AOD = ∠COD,
Since ∠AOD and ∠COD are a linear pair,
∠ AOD = ∠ COD = 90
A B
CD
O
KITE
• A quadrilateral having two pairs of equal sides
• One pair of opposite equal angles
• One diagonal bisects the other.
• Diagonals intersect at right angles.
TRAPEZOID
• A quadrilateral with one pair of parallel
sides
Call us for more
information:
www.iTutor.com
1-855-694-8886
Visit

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Types of Quadrilatrals

  • 2. WHAT ARE QUADRILATERALS? • A quadrilateral is a four-sided polygon. • If sides are parallel, the distance between them is constant, and they will never cross, even if extended.
  • 3. SIX TYPES OF QUADRILATERALS • RECTANGLE • SQUARE • PARALLELOGRAM • RHOMBUS • KITE • TRAPEZOID
  • 4. A RECTANGLE IS • A quadrilateral having two pairs of equal sides that are parallel • Sides meet at right angles (90 ) • Diagonals are equal in length
  • 5. RECTANGLE  In a rectangle the diagonals, besides being equal in length bisect each other. Given: ABCD is a Rectangle the diagonals are AC & BD Bisect each other at a Point O. To Prove: (i) AC = BD (ii) OA = OC or OB = OD Proof: In ∆ ABC and ∆ ABD AB = AB --------- (Common) BC = AD ---------------------(Apposite sides ABCD) ∠ A = ∠ B = 90 (angles of the rectangles) ∆ ABC ≅ ∆ ABD S. A. S. Rule So, AC = BD Proved A B CD O
  • 6. RECTANGLE (ii) OA = OC or OB = OD Proof: In ∆ AOB and ∆ COD AB = CD (Sides of the ABCD) AB || CD and transversal AC & BD Then, ODC = OBA And OCD = OAB Alternate interior angles ∆ AOB ≅ ∆ COD --- A. S. A Rule So that all congruent part are equal . OA = OC and OB = OD Proved A B CD O
  • 7. A SQUARE IS • A rectangle having all sides of equal length
  • 8. SQUARE  The diagonals of a square are perpendicular bisectors of each other Given: ABCD is a Square, Where AC and BD is a diagonal bisect each other at a Pont ‘O’ To Prove: ∠ AOD = ∠COD = 90 Proof: In ∆ AOD and ∆ COD OD Common & OA =OC AD = DC ------ Two sides of Squire So, (By SSS) ∆ AOB ∆ COD, ∠ AOD = ∠COD Since ∠AOD and ∠COD are a linear pair, ∠ AOD = ∠ COD = 90 A B CD O
  • 9. PARALLELOGRAM • A quadrilateral having two pairs of equal sides that are parallel • Opposite angles are equal • Diagonals are not equal in length
  • 10. PARALLELOGRAM  The diagonals of a parallelogram bisect each other Given: ABCD is a parallelogram where AC and BD is a diagonal bisect each other at Point O To Prove: AO = OC or BO= OD Proof: In ∆ AOB and ∆ COD AB = CD AB || CD CDO = OBA And DCO = OAB ------------ Alternate interior angles So, ∆ AOB ∆ COD, then AO = OC or BO =OD Proved A B CD O
  • 11. A RHOMBUS IS • A parallelogram having all sides of equal length
  • 12. RHOMBUS  The diagonals of a rhombus are perpendicular bisectors of one another. Given: ABCD is a rhombus. Therefore it is a parallelogram too. Since diagonals bisect each other, OA = OC and OB = OD. To Prove: ∠ AOD = ∠COD = 90 Proof: In ∆ AOD and ∆ COD OD Common & OA =OC --Given AD = DC ------ Two sides of this rhombus So, (By SSS) ∆ AOB ∆ COD, then ∠ AOD = ∠COD, Since ∠AOD and ∠COD are a linear pair, ∠ AOD = ∠ COD = 90 A B CD O
  • 13. KITE • A quadrilateral having two pairs of equal sides • One pair of opposite equal angles • One diagonal bisects the other. • Diagonals intersect at right angles.
  • 14. TRAPEZOID • A quadrilateral with one pair of parallel sides
  • 15. Call us for more information: www.iTutor.com 1-855-694-8886 Visit