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12
Understanding
Quadrilaterals
introdUction
In previous classes, you have learnt about curves, open and closed curves, polygons, quadrilaterals
etc.
In this chapter, we shall review, revise and strengthen these. We shall also learn the properties of
quadrilaterals.
• Angle sum property of a quadrilateral
• Types of quadrilaterals – trapezium, kite, parallelogram, rhombus, rectangle, square
• Properties of parallelogram
• Properties of rhombus
• Properties of rectangle and square
cUrVes
Flat surfaces are known as planes. For example, a page of your notebook, top of a table, floor of a
room etc. are all planes.
If we put the sharp tip of a pencil on a sheet of paper and move from one point to the other, without
lifting the pencil, then the shapes so obtained are called plane curves.
Some plane curves are shown below:
(i)
(v)
(ii)
(vi)
(iii)
(vii)
(iv)
(viii)
Open curve. The curves which have different beginning and end points are called open curves.
Intheabovefigure(i),(vi),(vii)and(viii) are all open curves.
Closed curve. The curves which have same beginning and end points are called closed curves.
Intheabovefigure(ii),(iii),(iv)and(v) are all closed curves.
Simple curve. A curve which does not cross itself at any point is called a simple curve.
Intheabovefigure(i),(ii),(iii),(vi)and(vii) are all simple curves.
Notethatcurves(ii)and(iii) are simple closed curves.
Learning Mathematics–VIII
204
Interior and exterior of a closed curve
A simple closed curve divides the region of the plane into three parts:
(i)The region of the plane that lies inside the curve
is called the interior of the curve.
In the adjoining figure, point A lies in the interior
of the curve.
(ii)The region of the plane that lies outside the
curve is called the exterior of the curve.
In the adjoining figure, point B lies in the
exterior of the curve.
(iii)The collection of points that lie on the curve is
called the boundary of the curve.
In the adjoining figure, point c lies on the boundary of the curve.
A
B
C
Polygons
A simple closed curve made up entirely of line segments is called a polygon.
Look at the following figures:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
In the above figure (i), (ii), (iii) and (iv) are all simple closed curves made up entirely of line segments,
so these are all polygons.
In the above figure (v), (vi), (vii) and (viii) are not polygons (why?)
Sides, vertices and diagonals of a polygon
The line segments forming a polygon are called its sides.
D
In the adjoining figure ABCDE is a polygon. It has five sides AB, BC,
CD, DE and EA.
The meeting point of a pair of sides is called a vertex of the polygon.
It has five vertices A, B, C, D and E.
C
E
The line segments formed by joining non-adjacent vertices are called
diagonals of the polygon.
In the above polygon, the line segments AC, AD, BD, BE and CE are
all diagonals of the polygon.
A
B
Understanding Quadrilaterals
205
Convex and concave polygons
Convex polygon. A polygon in which each interior angle is less than 180° is called a convex
polygon.
Or
Polygons in which no portion of any of their diagonals lies exterior to the polygon are called convex
polygons.
Convex polygons
Concave polygon. A polygon in which atleast one interior angle is greater than 180° is called a
concave polygon.
Or
Polygons in which a portion of atleast one diagonal lies exterior to the polygon are called concave
polygons.
diagonal
more than 180°
Concave polygons
Note. In this chapter, by a polygon we would mean a convex polygon only.
Classification of polygons
We can classify the polygons according to the number of sides or vertices they have:
Number of
sides or
vertices
Classification
3
Triangle
4
Quadrilateral
5
Pentagon
6
Hexagon
7
Heptagon
Sample figure
Number of
sides or
vertices
Classification
8
Octagon
9
Nonagon
10
Decagon
n
n-gon
Sample figure
Learning Mathematics–VIII
206
7
Activity
To verify that sum of interior angles of a quadrilateral is 360° by cutting and pasting.
Steps
C
B
1.Take a sheet of paper and draw any quadrilateral ABCD
on it and cut off the four angles i.e. A, B, C and
D as shown in adjoining figure (i).
A
D
Fig. (i)
2.Mark any point O on the copy.
3.Paste all the four cut out angles in such a way that the
vertices of these angles are at the marked point and they
form adjacent angles as shown in figure (ii).
These angles forms a complete angle at the point O.
∴A + B + C + D = 360°.
Thus, we have verified:
The sum of all the interior angles of quadrilaterals is 360°.
C
B
D
O
A
Fig. (ii)
Regular and irregular polygons
A polygon which has all its sides are of equal length and
all its interior angles are of equal measure is called a
regular polygon.
For example:
Equilateral triangle and square are regular
polygons.
Equilateral triangle
Square
A polygon which is not regular is called
irregular polygon.
For example:
Rectangle
Rhombus
(i)Rectangle is equiangular but all of its sides are not equal, so it is an irregular polygon.
(ii)Rhombus has all its sides equal but it is not equiangular, so it is also an irregular polygon.
Thus, in a regular polygon:
• All sides are equal in length.
• All interior angles are of equal measure.
• All exterior angles are of equal measure.
All regular polygons are convex.
Understanding Quadrilaterals
207
Angle sum property of a polygon
In class VII, you have learnt that sum of all interior angles of a triangle is 180°.
Can you tell that what is the sum of all interior angles of a quadrilateral, a pentagon, a hexagon and
so on?
To find this, let us try this method:
D
Take a quadrilateral ABCD. Divide it into two triangles ABC
5
and ACD by drawing a diagonal AC. We get six angles
C
4
1, 2, 3, 4, 5 and 6.
3
By angle sum property of a triangle
6
In ABC, we have 1 + 2 + 3 = 180°
…(i)
2
1
A
In ACD, we have 4 + 5 + 6 = 180°
…(ii)
B
Now in quadrilateral ABCD
A + B + C + D = 1 + 6 + 2 + 3 + 4 + 5
or
A + B + C + D = 1 + 2 + 3 + 4 + 5 + 6
…(iii)
From equations (i), (ii) and (iii), we have
A + B + C + D = 180° + 180° = 360°.
Thus, the sum of measures of all the interior angles of a quadrilateral is 360°.
We can extend this idea to other polygons.
Triangle
Figure
Quadrilateral
i
i
Pentagon
i
ii
Hexagon
iii
i
iii
ii
ii
Number of
sides
Number of
triangles
Sum of
interior angles
iv
3
4
5
6
1
2
3
4
180°
180° × 2 = 360°
180° × 3 = 540°
or
or
(4 – 2) × 180°
(5 – 2) × 180°
= 360°
= 540°
Hence, from the above table we can conclude that:
Sum of measures of all interior angles of a n-sided polygon = (n – 2) × 180°.
Each interior angle of a n-sided regular polygon =
180° × 4 = 720°
or
(6 – 2) × 180°
= 720°
(n − 2) # 180c
.
n
Remark
If a polygon has n sides, then the number of diagonals of the polygon =
n (n − 3)
.
2
Example 1. Some figures are given below:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
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208
Classify each of them on the basis of
(a)simple curve (b) simple closed curves
(d)convex polygon (e) concave polygon.
(c) polygon
Solution.
(a)figures (i), (ii), (iii), (iv) and (vii) are simple curves
(b)figures (i), (iv) are simple closed curves
(c)figures (i) and (iv) are polygons
(d)figure (i) is convex polygon
(e)figure (iv) is concave polygon.
Example 2. Find the sum of measures of all the interior angles of a polygon with
(i)7 sides
(ii) 9 sides.
Solution. We know that:
Sum of measures of all interior angles of a n-sided polygon = (n – 2) × 180°.
(i)Sum of measures of all interior angles of a 7-sided polygon= (7 – 2) × 180°
= 5 × 180° = 900°
(ii)Sum of measures of all interior angles of a 9-sided polygon= (9 – 2) × 180°
= 7 × 180° = 1260°.
Example 3. How many sides does a regular polygon have if each of its interior angles is 120°?
Solution. We know that:
Each interior angle of a n-sided regular polygon =
(n − 2) # 180c
.
n
Given each interior angle= 120°,
∴
(n − 2) # 180c
=
n
⇒
(n – 2) × 3 = 2n
⇒
3n – 6 = 2n
120°
3n – 2n = 6
⇒
⇒
n = 6.
∴ Number of sides of regular polygon = 6.
Example 4. Find the angle measure x in the following figures:
120°
x
x
140°
60°
150°
70°
50°
x
(i)
(iii)
(iii)
Solution.
(i)As the sum of all interior angles of a quadrilateral is 360°,
120° + 60° + 50° + x = 360°
⇒
230° + x = 360°
120°
60°
x
50°
Understanding Quadrilaterals
209
B
x = 360° – 230° = 130°
⇒
(ii)As DAE is a straight line
∴
150°
140° + BAD = 180°
140°
⇒
BAD = 180° – 140°
⇒
BAD = 40° ...(i)
C
x
E
70°
D
A
As the sum of all interior angles of a quadrilateral is 360°,
x + 150° + 70° + BAD = 360°
∴
x + 150° + 70° + 40° = 360°
⇒
x + 260° = 360°
(from (i))
x = 360° – 260° = 100°
⇒
(iii)As the given figure is a regular hexagon,
each interior angle of a regular hexagon=
∴ x = 120°.
(6 − 2) # 180c
6
= 4 # 180c
6
= 4 × 30° = 120°
Example 5. If the angles of a quadrilateral are in the ratio 5 : 8 : 11 : 12, find the angles.
Solution. Since the angles of quadrilateral are in the ratio 5 : 8 : 11 : 12, let these angles be 5x, 8x,
11x and 12x.
As the sum of measures of all interior angles of a quadrilateral is 360°
∴ 5x + 8x + 11x + 12x = 360°
⇒
⇒
36x = 360°
x = 360c = 10°.
36c
∴ The angles of the quadrilaterals are 5 × 10° = 50°, 8 × 10° = 80°,
11 × 10° = 110° and 12 × 10° = 120°
Hence, the angles of the quadrilateral are 50°, 80°, 110°, 120°.
Example 6. From the adjoining diagram, find x + y + z.
D
Solution. As an exterior angle of a triangle
= sum of two opposite interior angles
∴
z = 70° + 60°
z = 130°
70°
 DAC is a straight line,
∴
x + 70° = 180° ⇒ x = 110°
...(ii)
Again BCF is a straight line,
∴
60° + y = 180° ⇒ y = 120°
∴ From (i), (ii) and (iii), we have
x + y + z = 110° + 120° + 130° = 360°.
A
x
...(i)
...(iii)
E
y
60°
B
z
C
F
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210
8
Activity
Sum of the measures of the exterior angles of a polygon.
Steps
1. Draw a polygon (a pentagon ABCDE as shown in adjoining
figure) on the floor, using a piece of chalk.
2. Start walking from A. Walk along AB, on reaching B, you
need to turn through an angle of m 1 to walk along BC. On
reaching at C, you need to turn through an angle of m 2 to
walk along CD.
3 D
E
4
2
C
1
3. Continue moving in the same manner, until you return to the
A 5
B
side AB.
4. You will find that you have made one complete turn.
Therefore, m 1 + m 2 + m 3 + m 4 + m 5 = 360°.
Hence, the sum of the measures of the exterior angles of a pentagon is 360°. This is true for every
polygon.
Thus, the sum of measures of all the exterior angles of a polygon = 360°.
Example 7. Find the measure of angle x in the following figures:
F
80°
C
E
D
A
x
B
D
C
150°
F
B
120°
(i)
160°
150°
x
A
U
T
V
20°
S
R
80°
P
Q
x
(ii)
(iii)
Solution.
(i)Since CAE is a straight line,
⇒
BAE + 90° = 180°
( CAB = 90°, given)
BAE = 180° – 90° = 90°
As the sum of measures of all the exterior angles of a polygon = 360°
⇒
x + 90° + 150° = 360°
x + 240° = 360° ⇒ x = 360° – 240° = 120°
(ii) BCF is a straight line,
∴
⇒
FCD + 90° = 180°
FCD = 180° – 90° = 90°
As the sum of measures of all exterior angles of a polygon = 360°,
∴
x + 90° + 20° + 120° = 360°
⇒
x + 230° = 360°
⇒
x = 360° – 230° = 130°
(Given BCD = 90°)
Understanding QUadriLateraLs
211
(iii)  QRU is a straight line,
∴
150°+URS =180°⇒ URS=30°
 STV is a straight line,
∴
160°+VTP =180°⇒ VTP=20°
Asthesumofmeasuresofallexterioranglesofapolygon=360°,
∴ 80°+30°+80°+20°+x =360°
⇒
210°+x =360°⇒ x=150°.
Example 8. Find the number of sides of a regular polygon whose each exterior angle has a measure
of45°.
Solution.Asthesumofmeasuresofallexterioranglesofapolygon=360°andgiveneachexterior
angle=45°.
∴Thenumberofexteriorangles= 360c =8.
45c
Hence, the polygon has 8 sides.
Example 9.Isitpossibletohavearegularpolygonwithmeasureofeachexteriorangleas22°?
Solution.As sum of measures of all exterior angles of a polygon = 360° and given each exterior
angle=22°,
∴thenumberofsides= 360c =16 4 , which is not a natural number.
22c
11
Hence,itisnotpossibletohavearegularpolygonwithmeasureofeachexteriorangleas22°.
Example 10.Whatistheminimuminterioranglepossibleforaregularpolygon?Why?
Solution. We know that as number of sides of a regular polygon increases, each interior angle also
increases.
So, the minimum interior angle is possible in a polygon of least number of sides i.e. in an equilateral
triangle.
Eachangleofanequilateraltriangle= 180c =60°.
3
So,theminimuminteriorangle=60°.
Example 11.Whatisthemaximumexterioranglepossibleforaregularpolygon?
Solution.Asthesumofmeasuresofinteriorangleandcorrespondingexteriorangleis180°(constant),
so for a regular polygon each exterior angle will be maximum when each interior angle is minimum.
Fromexample10,weknowthattheminimuminteriorangle=60°.
∴Maximumexteriorangleofaregularpolygon=180°–60°=120°.
exercise 12.1
1. Somefiguresaregivenbelow.
(i)
(ii)
(iii)
(iv)
(v)
classify each of them on the basis of the following:
(a) Simplecurve
(b) Simpleclosedcurve (c) Polygon
(d) Convexpolygon (e) concave polygon
(vi)
(vii)
(viii)
Learning MatheMatics–Viii
212
2.
3.
4.
5.
6.
7.
Howmanydiagonalsdoeseachofthefollowinghave?
(a) Aconvexquadrilateral (b) A regular hexagon
(c) A triangle.
Find the sum of measures of all interior angles of a polygon with number of sides:
(i) 6
(ii) 8
(iii) 10
(iv) 12
Find the number of sides of a regular polygon if each of its interior angle is
(i) 60°
(ii) 90°
(iii) 108°
(iv) 135°
(v) 165°
Iftheanglesofaquadrilateralareintheratio2:3:4:6,findtheangles.
Iftheanglesofapentagonareintheratio7:8:11:13:15,findtheangles.
InaquadrilateralABCD,AB||DC.IfA : D=2:3andB:C=7:8,findthemeasure
of each angle.
D
8. Fromtheadjoiningfigure,find
(i) x
(ii) DAB
(iii) ADB
(3
x+
10
)°
(5x
C
+8
(3x + 4)°
)°
50°
A
x°
B
9. Find the angle measure xinthefollowingfigures:
30°
50°
x
110°
x
x
x
130°
120°
(i)
60°
70°
60°
(ii)
(iii)
83°
x
86°
110°
x
x
(iv)
75°
102°
(v)
(vi)
x
10. (i)Intheadjoiningfigure,findx + y + z.
90°
z
30°
y