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

T R A P E Z I U M

CHARACTERISTICS

• IT HAS 4 SIDES

• IT HAS 1 PAIR OF PARALLEL LINES

THE DIFFERENCE BETWEENTRAPEZIUM AND TRAPEZOID

THE PERIMETER OF A TRAPEZIUM

THE AREA OF A TRAPEZIUM

S Q U A R E

CHARACTERISTICS

has 4 congruent sides and

4 congruent (right) angles

• opposite sides parallel

• opposite angles congruent (all right)

diagonals are congruent

AC=BD

diagonals bisect each other

• diagonals bisect opposite

angles

• all bisected angles equal 45º

• diagonals are perpendicular

SQUARE FORMULA

AREA OF SQUARE FORMULA

R H O M B U S

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 .

.

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)

BASE TIMES HEIGHT METHOD

A=bh

where , b is the base lengthh is the height

A= 5cm x 4cm = 20cm2

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

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

R E C T A N G L E

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

A= LW

THE AREA OF A RECTANGLE

P = 2L+2W

= 2(L+W)

THE PERIMETER OF A RECTANGLE

• 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

is

EVERYWHERE !

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

DEFINITION

CYCLIC QUADRILATERAL IS QUADRILATERAL

WHICH INSCRIBED IN A CIRCLE .

PROPERTIES

Since

∠ABC+∠ADC=180°,

so ∠ABC= ∠ADE

PROPERTIES

AREA OF CYCLIC QUADRILATERAL

where s is the semi-perimeter of quadrilateral

The Brahmagupta’s Formula :

I R R E G U L A RQ U A D R I L A T E R A L

• 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

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

A QUADRILATERAL WITH

OPPOSITE SIDES PARALLEL

(AND THEREFORE OPPOSITE

ANGLES EQUAL)

• 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)

THE ANGLES OF A PARALLELOGRAM SATISFY THE IDENTITIES

A=C

B=D

AND

A+B=180 DEGREES.

A PARALLELOGRAM OF BASE, B AND HEIGHT H HAS AREA

AREA= BXH

THE AREA OF A PARALLELOGRAM

K I T E

• 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.

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.

THE PERIMETER IS 2 TIMES (SIDE LENGTH A + SIDE LENGTH B):

PERIMETER = 2(A + B)

THE PERIMETER OF A KITE

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.

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.

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

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.

QUESTION :

Given that h=8 , determine

the perimeter and the area of

the trapezium.

QUESTION :

Given area of the square is

324.

Find the perimeter and the

diagonal length of the

square.

QUESTION :

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

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 :

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.

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

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

QUESTION :

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

A PARALLELOGRAM HAS AN AREA

OF 28 SQUARE CENTIMETRES. IF

ITS BASE IS 4 CENTIMETRES,

CALCULATE THE HEIGHT OF THE

PARALLELOGRAM.

QUESTION :