MATHS IN-CLASS ACTIVITY SLIDES

1. Quadrilateral

2. Quadrilateral
● It is a four-sided polygon with four
angles
● The sum of interior angles is 360

3. Types of Quadrilateral
Square Triangle Parallelogram
Rhombus Kite Trapezium

4. Irregular
Quadrilateral
Cyclic Quadrilateral

5. SQUARE

6. 3 cm
3 cm
1 2 3
4 5 6
7 8 9
Formula
Area = Side2
= 32
= 9cm2

7. Rectangular
The formula is:
Area = L x W
L = Length
W = Width

8. 3cm
4cm
we know
L= 4cm , W= 3cm
Area= Length x Width
= L x W
= 4 x 3
= 12cm2
12 cm2

9. Kite
● Two pairs of equal length - a & a, b & b,
are adjacent to each other
● Diagonals are perpendicular to each other
● Perimeter = 2a + 2b
● Area = ½ x d1 x d2
= ½ x DB x AC

10. Area of Kite
● Area = ½ x d1 x d2
= ½ x width x length
● The remaining parts of the rectangle can
form another kite
● So, the total area needs to be divided into
half

11. Area = ½ x d1 x d2
= ½ x 4.8 x 10
= 24cm2
Area = ½ x d1 x d2
= ½ x (4+9) x (3+3)
= 39m2

12. Find the length of the diagonal of a kite whose area is 168 cm2
and
one diagonal is 14 cm.
Solution:
Given: Area of the kite (A) = 168 cm2
and one diagonal (d1) = 14 cm.
Area of Kite = ½ x d1 x d2
168 = ½ x 14 x d2
d2 = 168/7
d2 = 24cm

13. Parallelogram
● Both pairs of opposite sides are parallel
● Same opposite interior angle
● Opposite sides are equal in length and bisect each other
● The diagonals of a parallelogram bisect each other
● Each diagonal of a parallelogram separates it into two congruent
triangles

14. Parallelogram
Perimeter = 2a + 2b
Area = base x height
= b x h
Not using
value a to
calculate area!

15. Area of Parallelogram
When adjacent lengths and included angle is given,
From Theorem Hypotenuse ,
sin a = opposite / hypotenuse
sin a = h/b
Rearrange the equation, so h = b sin a
Area of Parallelogram ABCD = base x height
= a x b sin a
= ab sin a

16. Example : Area of parallelogram
Find the area of a parallelogram, two adjacent sides of which are 17cm
and 20cm and their included angle is 60 degree.
Solution :
Area of parallelogram = ab sin θ
= (17)(20) sin 60
= 340cm2
x 0.866
= 294.44cm2

17. Rhombus
● Four equal length , A=B=C=D
● Diagonals are unequal , bisect and
perpendicular to each other
● Perimeter = A+B+C+D
● Area - Altitude x Base
- a2
sin θ
- (½) ( d1
x d2
)

18. How area formula of Rhombus developed?
● Area = Altitude x Base
● Same as the area formula of a square

19. How area formula of Rhombus developed?
Diagonal AC divides the rhombus into two equal triangle ,
therefore the formula of the rhombus is given as :
Area of the rhombus = 2 x (½) ( a x a sin θ )
= a2
sin θ

20. Example : Area of rhombus
The side of a rhombus is 120m and two opposite angles are
60 degree each. Find the area.
Solution :
Area of rhombus = a2
sin θ
= (120)(120) sin 60
= 14400m2
x 0.866
= 12470.4m2

21. How area formula of Rhombus developed?
Diagonal AC & BD divide the rhombus into four equal triangles,
therefore area of rhombus given as :
Area of rhombus = 4 x ( ½ ) x ( AC/2 ) x ( BD/2 )
= ½ ( AC x BD )
= ½ ( d1
x d2
)

22. Example : Area of rhombus
The diagonal of a rhombus are 20m and 10m. Find its area.
Solution :
Area of rhombus = ½ ( d1
x d2
)
= ½ ( 20m x 10m )
= ½ ( 200m2
)
= 100m2

23. Trapezium
● 2-dimensional geometric figure with four sides
● at least one set of sides are parallel
● parallel sides are called the bases,
● other sides are called the legs

24. Properties of Trapezium
1)Perimeter of trapezium
Perimeter= a+b+c+d

25. 2) Area of Trapezium
Area=½ x h x (a+b)

26. Derivation of the formula of trapezium
Area of Parallelogram= Base x Height
=(b1+b2) x h
Since this is the area of two trapezium we have to divide this by
two, giving
Area of Trapezium=½ (b1+ b2) x h

27. Example:
Find the area of a trapezoid with bases of 9 centimeters and 7
centimeters, and a height of 3 centimeters.
Solution:
Area = ½ x base x height
= ½ x (9 cm + 7 cm) x 3 cm
= ½ x (16 cm) x (3 cm)
= ½ x 48 cm2
= 24 cm2

28. Cyclic Quadrilateral
● Quadrilateral which
inscribed in a circle

29. Properties of Cyclic
Quadrilateral
● Corresponding angle ● External angle
Since ∠ABC+∠ADC=180°,
∠ABC= ∠ADE

30. Area of
Cyclic Quadrilateral
The Brahmagupta’s
Formula:
where s = half perimeter of
quadrilateral

31. How Does the Brahmagupta’s
Formula Derived?
By extending line AB and line DC,
intersection point P is formed.
From the properties of cyclic
quadrilateral, ∠ABC= ∠ADP and
∠BCD= ∠PAD.
Therefore, ΔPBC is similar to ΔPAD.
The ratio of the two triangles:

32. How Does the Brahmagupta’s
Formula Derived?

33. How Does the Brahmagupta’s
Formula Derived?
Proof of Heron’s Formula:
http://jwilson.coe.uga.
edu/emt725/Heron/HeronProofAlg.html

34. How Does the Brahmagupta’s
Formula Derived?

35. How Does the Brahmagupta’s
Formula Derived?

36. Then, find s by substitute e
and f.

37. Then evaluate [s-(e+a)], [s-
(f+c)] and (s-b) in terms of
a,b, c and d
You will find that
[s-(e+a)]
[s-(f+c)] (s-b)

38. Finally, evaluate the Area of Cyclic Quadrilateral
Therefore,

39. Example:
Problem 1: Find the area of a cyclic quadrilateral whose sides are 7 cm, 5 cm, 4
cm and 10 cm.
Solution: Given that a = 7 cm, b = 5 cm, c = 4 cm and d = 10 cm
s = (7+5+4+10)/2
s = 13
Using Brahmagupta's formula:
Area of cyclic quadrilateral = √(s−a)(s−b)(s−c)(s−d)
= √(13−7)(13−5)(13−4)(13−10)
= √(6)(8)(9)(3)
= √1296

40. Problem 2: Find the area of a cyclic quadrilateral with sides 1 m, 300 cm, 2 m
and 1.2 m.
Example:
Solution: Given that a =100 cm, b =300 cm, c =200 cm and d = 120 cm
s = (100+300+200+120)/2
s = 360 cm
Using Brahmagupta's formula:
Area of cyclic quadrilateral = √(s−a)(s−b)(s−c)(s−d)
=√(360−100)(360−300)(360−200)(360−120)
= √(260)(60)(160)(240)
= √599040000
= 224475.3 sq cm
= 22.448 sq m

41. Irregular Quadrilateral
• Quadrilateral that does not fit into any of
the above is considered as irregular
quadrilateral.

42. Steps to find the area of an irregular
quadrilateral
Examples
Find the area of a quadrilateral ABCD where AB = 30cm, BC = 140cm,
CD = 20cm and DA = 150cm.
B C
A
D
140 cm
150 cm
30 cm 20 cm

43. Step 1 : Divide the figure into two triangles by drawing a diagonal.
Step 2 : Calculate the area of the triangle that has the given angle.
Area of triangle BCD = (140cm) (20cm) (sin 80o
)
= 1378.73 cm2

44. Step 3 : Calculate the length of the diagonal BD using the Law of Cosines.
BD2
= 1402
+ 202
- 2(140)(20) cos 80o
BD = 137.94 cm

45. Step 5 : Calculate the area of the second triangle using Heron's Formula.
Perimeter of triangle ABD = 30 cm + 150 cm + 137.94 cm
= 317.94 cm
Half of the perimeter, s = 158.97 cm
Area of triangle ABD =
= 1966.61 cm2

46. 1966.61 cm2
Step 6 : Add the two areas of the triangles to determine the area of the
quadrilateral.
Area of Quadrilateral ABCD
= Area of triangle ABD + Area of triangle BCD
= 1966.61 cm2
+ 1378.73 cm2
= 3345.34 cm2