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
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
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
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
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
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
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.
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 :
