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1. 1. T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
2. 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. 3. SIX TYPES OF QUADRILATERALS • RECTANGLE • SQUARE • PARALLELOGRAM • RHOMBUS • KITE • TRAPEZOID
4. 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. 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. 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. 7. A SQUARE IS • A rectangle having all sides of equal length
8. 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. 9. PARALLELOGRAM • A quadrilateral having two pairs of equal sides that are parallel • Opposite angles are equal • Diagonals are not equal in length
10. 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. 11. A RHOMBUS IS • A parallelogram having all sides of equal length
12. 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. 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. 14. TRAPEZOID • A quadrilateral with one pair of parallel sides