maths class activity

1. Q.U.A.D.R.I.L.A.T.E.R.A.L

4. T R A P E Z I U M

5. CHARACTERISTICS
• IT HAS 4 SIDES
• IT HAS 1 PAIR OF PARALLEL LINES

6. THE DIFFERENCE BETWEEN
TRAPEZIUM AND TRAPEZOID

7. THE PERIMETER OF A TRAPEZIUM

8. THE AREA OF A TRAPEZIUM

9. S Q U A R E

10. CHARACTERISTICS
has 4 congruent sides and
4 congruent (right) angles
• opposite sides parallel
• opposite angles congruent (all right)

11. diagonals are congruent
AC=BD
diagonals bisect each other
• diagonals bisect opposite
angles
• all bisected angles equal 45º
• diagonals are perpendicular

12. SQUARE FORMULA

13. AREA OF SQUARE FORMULA

14. R H O M B U S

15. DEFINITION
A FOUR –SIDED FLAT SHAPE WHOSE
SIDES ARE ALL THE SAME LENGTH AND
WHOSE OPPOSITE SIDES ARE PARALLEL.
 ALL SIDES HAVE EQUAL LENGTH
 DIAGONALS ARE UNEQUAL , BISECT AND PERPENDICULAR TO
EACH OTHER .
.

16. Area
• Altitude x Base ( the ‘base times height’
method)
• s2 sin A ( the trigonometry method )
• (½) ( d1 x d2 ) / (½) ( p x q ) ( the
diagonals method )
PERIMETER
4S (S+S+S+S)

17. BASE TIMES HEIGHT METHOD
A=bh
where ,
b is the base length
h is the height
A= 5cm x 4cm
= 20cm2

18. TRIGONOMETRY METHOD
A= a² sin A
where
a is the length of a
A is the interior angle
 Rhombus is formed by two equal triangles
Example :
1. The side of a rhombus is 140m and two
opposite angles are 60 degree each. Find the
area.
A = 140² sin 60
= 19600m² x 0.866
= 16973.60m²

19. THE DIAGONALS METHOD
A= (1/2) x (d1 x d2)
Where
d1 is the length of diagonal
d2 is the length of another diagonal
Example :
1. The diagonals of a rhombus are 40m and
20m. Find its area .
A = (1/2) (d1 x d2)
= (1/2) (40m x 20m)
= 400m²

20. R E C T A N G L E

21. CHARACTERISTICS
• ANGLE SUM OF QUADRILATERAL OF 360 DEGREES
• 2 SETS OF PARALLEL LINES
• 2 SETS OF 2 SETS EQUAL
• ALL ANGLES ARE RIGHT ANGLES
• 4 CORNERS

22. A= LW
THE AREA OF A RECTANGLE

23. P = 2L+2W
= 2(L+W)
THE PERIMETER OF A RECTANGLE

24. • DIAGONAL HALF A RECTANGLE
• D= SQUARE ROOT(LENGTH SQUARE+ WIDTH SQUARE)
• PYTHAGORAS THEOREM CAN ALSO BE APPLY TO LOOK FOR THE LENGTH OF THE DIAGONAL
THE DIAGONAL OF A RECTANGLE

25. is
EVERYWHERE !

26. C Y C L I C
Q U A D R I L A T E R A L

27. DEFINITION
CYCLIC QUADRILATERAL IS QUADRILATERAL
WHICH INSCRIBED IN A CIRCLE .

28. PROPERTIES

29. Since
∠ABC+∠ADC=180°,
so ∠ABC= ∠ADE
PROPERTIES

30. AREA OF CYCLIC QUADRILATERAL
where s is the semi-perimeter of quadrilateral
The Brahmagupta’s Formula :

31. I R R E G U L A R
Q U A D R I L A T E R A L

32. • IRREGULAR QUADRILATERAL DOES
NOT HAVE ANY SPECIAL PROPERTIES
• IRREGULAR QUADRILATERAL IS ONE
WHERE THE SIDES ARE UNEQUAL OR
THE ANGLES ARE UNEQUAL OR
BOTH
CHARACTERISTICS

33. P A R A L L E L O G R A M

34. A QUADRILATERAL WITH
OPPOSITE SIDES PARALLEL
(AND THEREFORE OPPOSITE
ANGLES EQUAL)

35. • OPPOSITE SIDES ARE CONGRUENT (AB = DC).
• OPPOSITE ANGELS ARE CONGRUENT (B = D).
• CONSECUTIVE ANGLES ARE SUPPLEMENTARY (A +B = 180°).
• IF ONE ANGLE IS RIGHT, THEN ALL ANGLES ARE RIGHT.
• THE DIAGONALS OF A PARALLELOGRAM BISECT EACH OTHER.
• EACH DIAGONAL OF A PARALLELOGRAM SEPARATES IT INTO TWO CONGRUENT
TRIANGLES. (ABC AND ACD)

36. THE ANGLES OF A PARALLELOGRAM SATISFY THE IDENTITIES
A=C
B=D
AND
A+B=180 DEGREES.

37. A PARALLELOGRAM OF BASE, B AND HEIGHT H HAS AREA
AREA= BXH
THE AREA OF A PARALLELOGRAM

38. K I T E

39. • IT LOOKS LIKE A KITE. IT HAS TWO PAIRS
OF SIDES.
• EACH PAIR IS MADE UP OF ADJACENT
SIDES (THE SIDES THEY MEET) THAT ARE
ALSO EQUAL IN LENGTH.
• THE ANGLES ARE EQUAL WHERE THE
PAIRS MEET.
• DIAGONALS (DASHED LINES) MEET AT A
RIGHT ANGLE, AND ONE OF THE
DIAGONAL BISECTS (CUTS EQUALLY IN
HALF) THE OTHER.​

40. KITES HAVE A COUPLE OF
PROPERTIES THAT WILL HELP US
IDENTIFY THEM FROM OTHER
QUADRILATERALS:
(1) THE DIAGONALS OF A KITE
MEET AT A RIGHT ANGLE.
(2) KITES HAVE EXACTLY ONE PAIR
OF OPPOSITE ANGLES THAT ARE
CONGRUENT.

41. THE PERIMETER IS 2 TIMES (SIDE LENGTH A + SIDE LENGTH B):
PERIMETER = 2(A + B)
THE PERIMETER OF A KITE

42. THE AREA OF A KITE
1ST METHOD: USING THE "DIAGONALS" METHOD.
The Area is found by multiplying the lengths of the
diagonals and then dividing by 2:
x and y refers to the
length of the diagonals.

43. 2ND METHOD: USING TRIGONOMETRY.
When you have the lengths of all sides and a measurement of the angle
between a pair of two unequal sides, the area of a standard kite is
written as: Area = a b sin C
a and b refer to length of two
unequal sides.
C refers to the angle between
two different sides.
sin refers to the sine function in
trigonometry.

44. FOR A KITE THAT IS NOT A SQUARE OR A RHOMBUS,
WHAT IS THE MAXIMUM NUMBER OF RIGHT ANGLES IT
COULD HAVE?
QUESTION :
A. 1
B. 2
C. 4

45. SOLUTION :
A kite has either zero right angles, one right angle or
two right angles:
If there were four right angles, then it would be a square.
So the maximum number is 2.

46. QUESTION :
Given that h=8 , determine
the perimeter and the area of
the trapezium.

47. QUESTION :
Given area of the square is
324.
Find the perimeter and the
diagonal length of the
square.

48. QUESTION :
Find the area of the rhombus
having each side equal to 17 cm
and one of its diagonals equal to
16 cm.

49. ABCD is a rhombus in which AB = BC = CD
= DA = 17 cm
AC = 16 cm
Therefore, AO = 8 cm
In ∆ AOD,
AD2 = AO2 + OD2
⇒ 172 = 82 + OD2
⇒ 289 = 64 + OD2
⇒ 225 = OD2
⇒ OD = 15
Therefore, BD = 2 OD
= 2 × 15
= 30 cm
Now, area of rhombus
= 1/2 × d1 × d2
= 1/2 × 16 × 30
= 240 cm2
SOLUTION :

50. QUESTION
THE DIAGONAL D OF A RECTANGLE HAS A
LENGTH OF 100 FEET AND ITS LENGTH Y IS
TWICE ITS WIDTH X (SEE FIGURE BELOW).
FIND ITS AREA.

51. EXAMPLE :
Find the area of a cyclic quadrilateral whose sides
are 36m , 77m , 75m , 40m.
Solution : Given a=36m, b=77m , c=75m , d=40m
s = (36+77+75+40)/2
= ( 228)/2
=114m
Using Brahmagupta’s Formula :
Area of cyclic quadrilateral =
√(s−a)(s−b)(s−c)(s−d)
A= √(114-36)(114−77)(114−75)(114-40)
= √ (78)(37)(39)(74)
= √ 8328996
= 2886 m2

52. The diagram shows a quadrilateral ABCD. The area of
triangle BCD is 12 cm2 and
BCD is acute. Calculate
(a) BCD,
(b) the length, in cm, of BD,
(c) ABD,
(d) the area, in cm2, quadrilateral ABCD.
QUESTION :

53. SOLUTION :
(b) Using cosine rule,
BD2 = BC2 + CD2 – 2 (7)(4) cos 59o
BD2 = 72 + 42 – 2 (7)(4) cos 59o
BD2 = 65 – 28.84
BD2 = 36.16
BD= √36.16
BD = 6.013 cm
(c) Using sine rule,
(d) Area of quadrilateral ABCD
= Area of triangle ABD + Area of
triangle BCD
= ½ (AB)(BD) sin B + 12 cm
= ½ (10) (6.013) sin 124.82 + 12
= 24.68 + 12
= 36.68 cm²
(a) Given area of triangle BCD = 12 cm2
½ (BC)(CD) sin C = 12
½ (7) (4) sin C = 12
14 sin C = 12
sin C = 12/14 = 0.8571
C = 59o
BCD = 59o

54. A PARALLELOGRAM HAS AN AREA
OF 28 SQUARE CENTIMETRES. IF
ITS BASE IS 4 CENTIMETRES,
CALCULATE THE HEIGHT OF THE
PARALLELOGRAM.
QUESTION :