Quadrilateral

Quadrilateral● It is a four-sided polygon with four

angles● The sum of interior angles is 360

Types of Quadrilateral

Square Triangle Parallelogram

Rhombus Kite Trapezium

Irregular Quadrilateral

Cyclic Quadrilateral

SQUARE

3 cm

3 cm

1 2 3

4 5 6

7 8 9

Formula

Area = Side2

= 32

= 9cm2

Rectangular

The formula is:

Area = L x W

L = Length

W = Width

3cm

4cm

we know L= 4cm , W= 3cm

Area= Length x Width= L x W= 4 x 3= 12cm2

12 cm2

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

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

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

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

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

Parallelogram

Perimeter = 2a + 2bArea = base x height

= b x h

Not using value a to calculate area!

Area of Parallelogram

When adjacent lengths and included angle is given, From Theorem Hypotenuse ,

sin a = opposite / hypotenusesin a = h/b

Rearrange the equation, so h = b sin aArea of Parallelogram ABCD = base x height

= a x b sin a = ab sin a

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

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 )

How area formula of Rhombus developed?

● Area = Altitude x Base● Same as the area formula of a square

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

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

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 )

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

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

Properties of Trapezium1)Perimeter of trapezium

Perimeter= a+b+c+d

2) Area of Trapezium

Area=½ x h x (a+b)

Derivation of the formula of trapezium

Area of Parallelogram= Base x Height =(b1+b2) x hSince this is the area of two trapezium we have to divide this by two, giving

Area of Trapezium=½ (b1+ b2) x h

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

Cyclic Quadrilateral

● Quadrilateral which inscribed in a circle

Properties of Cyclic Quadrilateral

● Corresponding angle ● External angle

Since ∠ABC+∠ADC=180°, ∠ABC= ∠ADE

Area of Cyclic Quadrilateral

The Brahmagupta’s Formula:

where s = half perimeter of quadrilateral

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:

How Does the Brahmagupta’s Formula Derived?

How Does the Brahmagupta’s Formula Derived?

Proof of Heron’s Formula:http://jwilson.coe.uga.edu/emt725/Heron/HeronProofAlg.html

How Does the Brahmagupta’s Formula Derived?

How Does the Brahmagupta’s Formula Derived?

Then, find s by substitute e and f.

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)

Finally, evaluate the Area of Cyclic Quadrilateral

Therefore,

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)/2s = 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

= 36 sq cm

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)/2s = 360 cmUsing 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

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

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

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

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

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

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