6.7 Quadrilateral:
6.7.1
Properties of Quadrilaterals 
Definition: A ‘quadrilateral’
is a geometric figure on a plane enclosed by 4 line segments.
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Properties:
Each diagonal divides the quadrilateral in to 2
triangles (
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Adjacent Sides ( Have common vertex) |
Opposite Sides ( Do not have common vertex) |
Adjacent Angles ( Have common side) |
Opposite Angles ( Do not have common side) |
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(AB,BC) : B common vertex (BC,CD) : C common vertex (CD,DA) : D common vertex (DA,AB) : A common vertex |
(AB,CD) (AD,BC) |
( (
( ( |
( ( |
We notice
that the sum of all four interior angles in a quadrilateral is 3600.
Can we prove this mathematically?
6.7.1
Problem 1:.The
four angles of a quadrilateral are in the ratio of 4:5:6:9. Find all the angles.
Solution:
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The sum of all angles in a quadrilateral is 3600.
Since the angles are in the ratio of 4:5:6:9, let they
be 4x, 5x, 6x and 9x.
i.e. 24x = 3600 x = 360/24 = 150 4x = 4*15 = 600 5x = 5*15 =750 6x = 6*15 =900 9x= 9*15 = 1350 The angles are 600, 750, 900 and 1350 |
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6.7.1 Problem 2: If the angles of a quadrilateral are 3x, 3x+150, 3x+300 and 900.Find all the angles.
Solution:
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We know that sum
of interior angles of a quadrilateral is 360o
i.e.9x=2250 i.e. x=250 Therefore the angles are 3x = 750, 3x+150
=750+150=900, 3x+300 = 1050 Verify that the sum of these four angles = 3600 |
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We have seen earlier that a triangle totally has six
elements (3 sides and 3 angles). We have also learnt to construct triangles
with just three elements given (3 sides, 1 side
and 2 angles, 2 sides
and 1 included angle).
Note: With just three angles we cannot construct a unique triangle.
Compared to a triangle, quadrilateral has 10 elements (4
sides, 4 angles, 2 diagonals). We cannot draw a quadrilateral uniquely with 4
elements. We need at least 5 elements to draw unique quadrilaterals. However,
it will not be possible to construct a unique quadrilateral with 1 side and 4
angles.
We need any one group of the following set of elements, to
construct a unique quadrilateral.
6.7.1
Table 1:
|
No |
Given no. of Sides |
Given no. of Diagonals |
Given no. of Angles |
No. of elements required |
|
1 |
2 |
2 |
1 |
5 |
|
2 |
2 |
1 |
2 |
5 |
|
3 |
4 |
1 |
- |
5 |
|
4 |
4 |
- |
1 |
5 |
|
5 |
3 |
- |
2(Included) |
5 |
|
6 |
3 |
2 |
- |
5 |
|
7 |
2(adjacent) |
- |
3 |
5 |
General method for
construction of quadrilateral:
(The method is more or less the same for construction of
quadrilateral, for any combinations of the data given in the above table)
Notes:
a) Step for drawing a side:
First draw a line. Mark a point on this line
From the above point, draw an arc of given radius (length
of the side) to cut this line to get the required side.
b) Use protractor to construct angles of required measure,
wherever required.
c) For drawing of any figure follow the procedure:
Step 1: Draw a rough diagram
Step 2: Follow the steps described above (a,b), to construct sides/angles of
required measure.
1. When two sides,
two diagonals and one angle are given
6.7.1
Problem 3: Construct
a quadrilateral ABCD with AB=4cm, BC=2cm, AC=5cm and BD=4cm and 
DAB= 600
First draw a rough diagram
|
Step |
Construction |
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1 |
Mark a point A and draw a line
though A |
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2 |
With A as center, draw an arc of radius 4cm to cut above
line at B (AB=4cm) |
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3 |
From B, dawn an arc of radius 2cm above AB |
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4 |
From A, draw an arc of radius 5cm to cut the above arc
at C (AC=5cm,BC=2cm) |
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|
5 |
From A, draw a line at an angle 600 with AB |
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6 |
From B, draw an arc of radius 4cm to cut the above line
at D ( |
ABCD is the required quadrilateral.
2. When two sides,
one diagonal and two angles are given
6.7.1
Problem 4: Construct a quadrilateral ABCD with AB=4cm, BC=3cm, BD=5cm and ABC= 600, BCD= 650
First draw a rough diagram.
|
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Construction |
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|
1 |
Mark a point A
and draw a line though A |
|
|
2 |
Cut the above
line by an arc of radius 4cm to cut at B (AB=4cm) |
|
|
3 |
From B, draw a
line at an angle 600 with AB |
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|
4 |
From B, dawn an
arc of radius 3cm to cut the above line at C (BC=3cm, |
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|
5 |
From C, draw a
line at an angle 650 with AB |
|
|
6 |
From B, draw an
arc of radius 5cm to cut the above line at D (BD=5cm) |
ABCD is the required quadrilateral
3. When four sides
and one angle are given
6.7.1
Problem 5: Construct
a quadrilateral PQRS with PQ=4cm, QR=3cm, RS=2.5cm and PS=3.5cm and SPQ= 500
First draw a rough diagram
|
Step |
Construction |
|
|
1 |
Mark a point P
and draw a line though P |
|
|
2 |
Cut the above
line by an arc of radius 4cm to cut at Q
(PQ=4cm) |
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|
3 |
From P, draw a
line at an angle 500 with
PQ |
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4 |
From P, dawn an
arc of radius 3.5cm to cut the above line at S(PS=3.5cm, |
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5 |
From S, dawn an
arc of radius 2.5cm. |
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6 |
From Q, dawn an
arc of radius 3cm to cut the above arc at R(SR=2.5cm,QR=3cm).
Join SR and QR. |
PQRS is the
required quadrilateral
6.7. 1
Exercise: In Table 6.7.1 we have listed seven combinations of
data required for construction of a quadrilateral, out of this we constructed
quadrilateral of three combinations (green color). Construct quadrilaterals with the four remaining
combinations of data (yellow
color).
6.7.2 Area of a Triangle
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In the adjacent figure of triangle ABC, AD, BF, CE are
altitudes to the bases BC, AC and AB respectively from the opposite vertices
A,B and C respectively. Area of a triangle =(1/2)Base*height(altitude) Area of We are going to prove this formula for
calculation of area of triangle in section 6.8.7. |
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6.7.2
Problem 1: The
base of a triangular field is three times its height. If the cost of
cultivating field at the rate Rs36.72 per 100 sq mts
is Rs 49,572, find its base and height
Solution:
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Total cost of cultivating the field = 49,572 Area of the fileld =
49572*100/36.72 = 135000 Let x be the height of the field
Thus height of the triangle is 300m and base is 900m |
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Verification:
Area of triangle = 1/2*900*300 = 900*150 = 135000
Cost of cultivation = 135000*36.72/100 = 49572 which is as
given in the problem
6.7.3 Area of a quadrilateral
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We have seen that a diagonal of a quadrilateral cuts the
quadrilateral in to 2 triangles. We shall use this Property to calculate the area of a quadrilateral,
as we already know how to calculate the area of a triangle. Let PQRS be the quadrilateral. Draw the diagonal PR. Draw a
perpendicular (QA= h1) to PR
from vertex Q. Draw another perpendicular (SB=h2) to PR from
vertex S. h1 and h2 are altitudes of Area of Area of
= ½(PR*
h1)+ ½(PR* h1) = 1/2*PR* (h1+h2)
sq units
1/2 * diagonal * sum of altitudes of the two triangles with diagonal as base |
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6.7.4 Types of quadrilaterals
Depending on the shape of quadrilateral, they are
classified as follows:
|
Type |
Main(Basic) Property of quadrilateral |
Figure |
Relationship between Sides |
Relationship between Angles |
Relationship between Diagonals |
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Parallelogram |
Both pairs of opposite
sides are parallel |
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1.Both pairs of opposite sides are parallel 2.Both pairs of opposite sides are equal |
1.Oppopsite angles are
equal. 2. Sum of any two consecutive angles = 1800 |
1.Diagonals divide the parallelogram in to two congruent triangles 2. Diagonals bisect each other |
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Trapezium |
Only one pair of
opposite sides are parallel |
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A
pair of opposite sides are parallel |
Pairs of consecutive angles at the end points of the two
non parallel sides are supplimentary |
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Isosceles
Trapezium |
One pair of
opposite sides are parallel and non parallel sides are equal |
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1.A
pair of opposite sides are parallel 2.
Non parallel sides are equal. |
1.Pairs of consecutive angles at the end points of the two
non parallel sides are suplementary 2.Pairs of
consecutive angles at the end points of the two parallel sides are equal. |
Diagonals are equal |
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Rectangle |
Both pairs of opposite
sides are parallel and all angles are
right angle |
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1.Opposite
sides are equal 2.
Both pairs of
opposite sides are parallel |
All angles are equal and are right angles |
1.Diagonals divide the rectangle in to two Equal triangles 2. Diagonals are equal 3 Diagonals bisect each
other |
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Rhombus |
All sides are
equal and both pairs of opposite
sides parallel |
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1.
All sides are equal 2.
Both pairs of opposite sides are parallel |
1.Oppopsite angles are
equal. 2.Sum of any two consecytive angles = 1800 |
1.Diagonals divide the rhombus in to two congruent
triangles 2. Diagonals bisect each other. 3. Diagonals are | to each other. |
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Square |
All sides are
equal and all angles are
right angles |
|
1
All sides are equal 2.Both pairs of opposite sides are parallel |
All angles are equal
and right angles |
1.Diagonals
divide the sqaure in to two congruent triangles 2. Diagonals
are equal. 3. Diagonals bisect each other. 4. Diagonals are | to each other. |
Note : Why mothers and grandmothers cut burfies
in the shape of parallelogram and not in rectangles?
Because they could cut more pieces from the same spread of sweet
preparation (Did they study Geometry?)
The reason we will learn later (Section6.8.2)
is that the area of a rectangle (base*another side) is greater than the area of
parallelogram (base*height) when their sides are of same measures.
The hierarchy of different types of quadrilaterals can be
represented by the following chart:
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6.7.4
Problem 1: Two consecutive angles of a parallelogram have
measures, (x+30) and (2x-60) respectively. Find the measures of all angles.
Solution:
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We know that in a parallelogram, sum of any two consecutive angles = 1800.
Hence the angles are 100,80,100 and 80 |
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6.7.4
Problem 2: In a parallelogram ABCD, DAB= 700, DBC = 800 Find CDB and ADB.
Solution:
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We know that in a parallelogram, sum of any two consecutive angles 1800.
i.e. 700+
Since BA || CD, corresponding angles are equal
Since in a parallelogram opposite angles are equal, So the angles are 700, 1100, 700
and 1100. |
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6.7.4
Problem 3: The
ratio of two sides of a parallelogram is 3:5 and the perimeter is 48cm. Find
the sides of the parallelogram.
Solution:
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We know that in a parallelogram Perimeter = sum of four
sides = 2*(sum of any 2 adjacent sides) Since it is given that the perimeter = 48cm
Since the ratio of adjacent sides = 3:5, let the sides
be 3x and 5x
So the lengths of adjacent sides are 9cm and 15cm Hence the sides are 9cm, 15 cm, 9 cm and 15cm. Verification:
Perimeter = sum of all sides = 9+15+9+15 = 48cm, which
is as given in the problem and hence our solution is correct. |
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6.7 Summary of learning
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No |
Points to remember |
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1 |
Area of quadrilateral = 1/2 * diagonal * sum of
altitudes of the 2 triangles with diagonal as base |
Additional
Points:
Heron’s formula for
calculation of area of a triangle:
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If a, b and c are measures of the three sides of a
triangle, then its area can also be calculated by using the formula: Area = Where s = (a+b+c)/2 (semi
perimeter of the circle) Note: Though this has come to be known as Heron’s formula, ‘Bhaskara’ had provided the
proof of this in his book (‘Leelavati’- shloka 169). |
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6.7.2
Problem 2: Find the area of a triangular park
whose sides are 18m, 24m and 30m. Also find the length of the altitude
corresponding to the largest side of the triangle.
Solution:
Let a=18, b=24 and c=30.
s = (a+b+c)/2 = 36
Area =
= 
= 
= 216
Since the largest base is 30, let the height on this base
be h.
We have 216 = 1/2*30*h (
area of a triangle = (1/2)*base*height)
h = 216/15 = 14.4m
Verification:
Area of triangle = 1/2*30*14.4 = 216
Note: Heron’s formula is very helpful if we have to calculate
the area of a quadrilateral given its sides and a diagonal, as in the problem
given below.
6.7.2 Problem 3: A floral design on a
floor is made up of 8 tiles
which
are quadrilaterals. Its sides are 9cm, 28cm, 9cm and 28cm and it’s diagonal is 35cm as shown in the adjoining figure. Find
the cost of polishing the tiles at the rate of 50paisa per sq.cm.
Solution:
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Let us calculate
the area of each tile Area of a tile =
area of quadrilateral = 2*area of the triangle The sides of the
triangle are a=28, b=9 and c=35.
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