1. Quadrilateral 1 pair of // opp. Sides One of the diagionals is axis of symmetry 2 diagionals are  2 pairs of equal adjacent sides Sum of interior angles is 180 0 2 pairs of opposite sides are equal.(opp. sides of // gram) 2 pairs of opposite angles are equal (opp.  s of // gram) Diagonals bisect each other (diag. Of // gram) 2 pairs of opp.// sides 4 right angles Diagonals are equal Properties of trapesium Properties of // gram Diagonals bisects each interior angle Kite Trapezium Rectangle Rhombus 4 equal sides Properties of // gram and kite Angles between each diagional and each side is 45 0 45 0 Properties of rhombus/rectangle 4 right angles and 4 equal sides Parallelogram Square

2. Trapeziums Definition : 1 pair of parallel sides Properties: Sum of interior angles is 180 0

3. Parallelogram Definition : 2 pairs of opp. parallel sides Properties: 2 pairs of opposite sides are equal. (opp. sides of // gram) 2 pairs of opposite angles are equal (opp.  s of // gram) Diagonals bisect each other (diag. Of // gram)

4. Conditions for Parallelogram If 2 pairs of opposite angles are equal then the quadrilateral is parallelogram. (opp.  s of eq.) If diagonals bisect each other then the quadrilateral is parallelogram (diag. Bisect each other) If 2 pairs of opposite sides are equal then the quadrilateral is parallelogram. (opp. sides eq.) If 1 pair of opposite sides is equal and parallel then the quadrilateral is parallelogram (opp. sides eq. and //)

5. Rhombus Definition : a // gram or a kite of 4 equal sides Properties: 2 pairs of opposite sides are equal. (opp. sides of // gram) 2 pairs of opposite angles are equal (opp.  s of // gram) Diagonals bisect each other (diag. Of // gram) Diagonals bisects each interior angle Diagonals are 

6. Rectangle Definition : a parallelogram of 4 right angles Properties: 2 pairs of opposite sides are equal. (opp. sides of // gram) 2 pairs of opposite angles are equal (opp.  s of // gram) Diagonals bisect each other (diag. Of // gram) Diagonals are equal

7. Square Definition : a // gram of 4 right angles and 4 equal sides Properties: 2 pairs of opposite sides are equal. (opp. sides of // gram) 2 pairs of opposite angles are equal (opp.  s of // gram) Diagonals bisect each other (diag. Of // gram) Diagonals are equal Diagonals are  45 0 Angles between each diagonal and each side is 45 0

8. Example 1: In the figure, PQRS is a kite (a)Find x and y. (b)Find the perimeter of the kite PQRS P R S Q x+1 y+3 8 x+y PQ = PS (given) x+1 = y+3 x-y=2(1) QR=SR (given) x+y=8(2) (1)+(2),2x=10 x=5 Put x=5 into (1), 5-y=2 y=3 (a) (b) PQ = x+1=5+1=6  PQ+PS+SR+QR = 6 + 6 + 8 + 8 =28

9. Example 2: In the figure, ABCD is a kite. E is a point of intersection of diagonals AC and BD, AE=9 cm, EC=16 cm and DE=EB=12 cm (a)Find the area of ABCD. (b)Find the perimeter of ABCD (a)  ABC=  ADC (axis of symmetry AC)  AED=90 0 In  ADE, AD 2 =AE 2 +DE 2 =9 2 +12 2 =225 cm 2 (Pyth theorem) AD=15 cm In  CDE, DC 2 =DE 2 +EC 2 =12 2 +16 2 =400 cm 2 (Pyth theorem) DC=20 cm  Perimeter of ABCD=AD+AB+ DC+CB = 15 + 15 + 20 + 20 =70 cm A C D B 9 16 12 E Area of  ADC = Area of kite ABCD=Area of  ABC+Area of  ADC = 150+150 =300 cm 2 (b)

10. Example 3: In the figure, ABCD is a parallelogram. Find x and y. AD//BC (Given) x+68 0 =180 0 (prop. Of trapezium)  x=112 0 (150 0 -y)+2y=180 0 (prop. Of trapezium) 150 0 +y=180 0  y=180 0 -150 0 =30 0 A B D C 150 0 -y 68 0 x 2y

11. Example 4: In the figure, ABCD is a parallelogram. Find x and y.  DAB=  DCB (opp.  s of // gram) x+20 0 =3x-10 0 2x=30 0 x=15 0  DAB+  CBA=180 0 (int.  s, AD//BC) x+20 0 +y=180 0 15 0 +20 0 +y=180 0 y=145 0 x+20 0 y 3x+10 0 A B C D

12. Example 5: In the figure, ABCD is a isosceles trapezium with AB=DC. Find x, y and z 126 0 x y z A BC D AD//BC (Given) x+126 0 =180 0 (prop. Of trapezium)  x=54 0 Construct AE // DC E a  AD//EC and AE//DC ADCE is a parallelogram (Definition of // gram)   ADCE is a parallelogram (proof) AE=DC (opp.sides of // gram)  In  ABE, AE=DC (proof) AB=AC (given)  AB=AE  y=a (base  s. isos  ) a= x (corr.  s. AE//DC)  y=x =54 0 y+z=180 0 (prop. Of trapesium) z= 180 0 -54 0 = 126 0

13. A N M B C MID-POINT THEOREM IF AM = MB and AN =NC then (a) MN // BC (b) MN = (Abbreviation: Mid-point theorem)

14. Example 13: In the figure, ABC is a triangle, find x and y. DE//AC (mid-point theorem) x =  EDB =42 0 (corr.  s, DE//AC) (mid-point theorem) C E B D A y 6 42 0 x CE=BE (given) AD=DB (given)

15. Example 14: Prove that BPQR is a parallelgram AR=RB (given) (mid-point theorem) AQ=QC (given) (opp-sides eq. And //) A Q C P B R

16. Ex 11D 1(b) D A M B CN y cm x cm 5 cm AM=AC (given) BN=NC (given) (mid-point theorem) BM=MD (given) BN=NC (given) (mid-point theorem)

17. Ex 11D 2(b) AP=BP (given) AQ=CQ (given) (mid-point theorem) P Q B C A a 110 0 46 0 (corr.  s. PQ//BC) In  APQ,  APQ+  PAQ+ a = 180 0 46 0 +110 0 +a=180 0 a=24 0 (adj  s. on a st line)

18. A F E B D C 10 9 8 3(a) 3(b) B D C 9 A F E 60 70 50

19. 4. A Q P C R B 6 8 (mid-point theorem) AQ=QB (given) AP=PC (given) (mid-point theorem) BP=PA (given) CR=RB (given) Area of ABC 

20. INTERCEPT THEOREM A B C D P Q X Y transversal intercept

21. INTERCEPT THEOREM A B C D E F If AB//CD//EF then (intercept theorem)

22. INTERCEPT THEOREM A C D E F (intercept theorem) Construct GB through A such that BG//CD//EF B G GB//CD//EF (given) Proved:

23. Example 15. AP//BQ//CR, AB=BC, AP=11 and CR=5. Find BQ. A B C PQR 5 11 S Join AR to cut BQ at S AP//BQ//CR(given) (intercept theorem) (given) (proved) (mid-pt theorem) (proved) (mid-pt theorem) BQ=BS+SQ = 2.5+5.5=8

24. Example 16. AB and DC are straight lined. Find x and y. Join DE through A and // BC DE//PQ//BC(given) (intercept theorem) A P Q B C (a) Proved: E D (b) AB=6, PB=2 and AQ=9. Find QC (proved)

25. Example 16. Find QR and CD. AP//BQ//CR(given) (intercept theorem) BQ//CR//DS(given) A P R 8 D S B Q 32 6 C