Oasis Math 6

AB

Hence, AB is the required line segment with a length of 6 cm.

Steps:
• Mark any point A.
• Place the ruler with its 0 cm mark at point A.
• At the point which corresponds to 6 cm, mark point B.
• Join AB.

Exercise 12.3

1. Measure the length of each line segment.

(a) B (b) (c)

A D E

C

2. Draw the segments of the following length by using the ruler. F

(a) 6 cm (b) 4.9 cm (c) 4.5 cm (d) 5.6 cm

3. Measure the length of given line segment and raw the line segment equal in

length of given line segmen in your copy.

(a) (b) (c)

4. Measure the length of line segments and calculate the perimeter each of the
following figures. [Recall the concept of perimeter].

(a) A (b) P Q (c) A F

BE

B CS R CD

Oasis School Mathematics – 6 195

Answers
Consult your teacher.

Construction of Parallel lines

To construct a line parallel to a given line passing through the given point:
Draw a line AB and mark a point P outside the line. Through P construct CD parallel

to AB.

INCH0 1 2 3 4 5 6
87654321
INCH0 1 2 3 4 5 6
87654321
0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15P CP D CP D
0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
CM 1 2 3 4 5 6 7 8

AB AB

CM 1 2 3 4 5 6 7 8

AB

• Draw AB and mark a point P outside the line AB.
• Place a set square with one arm of its right angle along AB as shown in the figure.
• Holding it straight, place a ruler along the other arm of the right angle.
• Slide the set square along the edge of the ruler until the edge of the set square comes to P.
• Draw a line CD through P along this edge.
• Remove the ruler and set square.

Hence, CD is the required line through P parallel to the line AB.

Construction of Perpendicular Lines:

I. To construct a perpendicular to a given line at a given point on the line.
(Using the protractor):

Draw a line AB and take any point C on it. Construct CD⊥AB at C.

D

60 9070 80 100 110 120 D
50 120 100 80 70
110 B
130 60 130 C

40 50 160
150 20
140 140 30
40
30
150

20 160

AC B 10 170 A
A 170 10

196 Oasis School Mathematics – 6 0 180
180 0
CB

Steps:
• Draw AB and take C as a point on it.
• Place the protractor on the line AB with the centre at C and the 0° – 180° line

along line AB. Hold the protractor securely and mark the point D against the
90° mark.
• Remove the protractor and join D and C. Hence, CD ⊥ AB.

II. To construct a perpendicular to a given line from the given point. (Using
set square):

C

CM 1 2 3 4 5 6 7 8 C

C INCH0 1 2 3 4 5 6 B
A 87654321

AB D0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

AB

D

Steps:

• Draw a line AB and take a point C.
• Place the long edge of the ruler on the line AB.
• Holding the ruler firmly, place a set square with one edge of the 90° angle

against the ruler as shown in the figure.
• Slide the set square firmly till the perpendicular edge is at C.
• Draw a line CD along the perpendicular edge and remove the set square.
Hence, CD ⊥ AB.

Exercise 12.4

1. (a) Draw each of the following figures in your copy and draw a line parallel to the
given line through the given point.

(i) C (ii) Q (iii) X

AB PR Y
Z

(b) Draw each of the following lines in your copy and draw a line parallel to the
given line.

Oasis School Mathematics – 6 197

(i) (ii) (iii)

2. Draw a line as in the given figures in your copy and construct a perpendicular
to the given point on the line (Using protractor)

(a) (b) (c)

Y

A MB P RQ Z

X

3. Draw a line as given in your copy and construct a perpendicular to the given

line from the given point in each of the following. (using set square)

(a) (b) (c)

CQ Y

A BP Q XZ

Answers
Consult your teacher.

Activity A B
C
Take a A4 size paper

• Draw a line segment AB of any measurement.

• Take any point C outside the line AB. B

A E
C
• With the help of set square, draw a line DE||AB D
B
through C. A

• Take any three points P, Q and R on DE.

• With the help of set square draw PM⊥AB, QN⊥AB, DD PP QQ RR EE
RT⊥AB. BB

AA M N T

198 Oasis School Mathematics – 6

Measure the length given line segment.

Line segment PM QN RT

Measurement ...............cm ...................cm ..................cm

Conclusion:................................................................................................................

12.4 C onstruction of Perpendicular Bisector of a Given Line
Segment Using Compass

We shall now learn to make simple constructions, using a straight edge of ruler
and a compass only. The validity of the construction will be verified by actual
measurement.

Draw the perpendicular bisector of given line AB = 6.4cm.

Here we have to draw a line which is perpendicular to the given line and which
bisects the given line.

A figure (i) Steps:

B

• Draw a line AB = 6.4 cm with the

help of ruler. [figure (i)]

B figure (ii) • Taking A as the centre,

A draw an arc with the radius

greater than half of AB.

[figure (ii)]

C • Taking B as the centre,

using the same radius draw

figure (iii) another arc so as to cut the

AB former arc at the points

C and D. [figure (iii)]

D • Join CD and name the

C point where CD meets
AB as E. [figure (iv)]

AE figure (iv) • CD is the perpendicular bisector of
B AB.

D

Observations: BE Remarks
AE 3.2 cm AE = BE

3.2 cm

Oasis School Mathematics – 6 199

∠AEC ∠BEC Remarks
90º 90º CE⊥AB

Since, AE = BE, CD bisects AB.

Again, ∠AEC and ∠BEC both equal to 900, then CE ⊥ AB.

Hence, CD is the perpendicular bisector of AB.
Note

In the given figure, A

AB is the perpendicular bisector of PQ because P Q

PR = RQ and R

∠ARP = ∠ARQ = 90°. B

In the given figures PN = NQ and MN ⊥ PQ, hence MN is the perpendicular
bisector of PQ.

P

MN
Q

Activity
Take a rectangular piece of A4 size colour paper.

Fold it once as shown in the diagram.

Again, fold it as show the diagram.

A P
B

Open the fold and paste it chart paper and the points where R TS
the crease meets its edges as shown.

DQ C

200 Oasis School Mathematics – 6

Observations:

PT TQ RT TS ∠PTS
...................cm ..................cm
..............cm ..............cm ...............cm

Conclusion:................................................................................................................

Exercise 12.5 C

1. In the adjoining figure, if CD is the perpendicular D B
bisector of AB then, A Q
(a) what is the relation of AB with CD?
(b) what is the relation of AD with DB? R

2. From the adjoining figures, identify the perpendicular S
bisector of PQ. Justify your answer. P

3. (a) In the given figure, A
(i) Which line is the perpendicular bisector of BD? Why?
(ii) Is BD the perpendicular bisector of AC? Why? B ED
(b) In the given figure, C
(i) which line is the perpendicular bisector of AC?
(ii) which line is the perpendicular bisector of BD? AD
O

BC

(c) In the given figure, CD is the perpendicular A
bisector of AB,

(i) Is AE = EB? CE D
(ii) Is CE = ED? B
(iii) What is the value of ∠AED?

4. Draw the line segments of given measurement and hence draw their
perpendicular bisectors.

(a) AB = 7cm (b) PQ = 5.6cm (c) MN = 8.2cm (d) XY = 6.8cm
(e) EF =6.4cm (f) GH = 7.2cm

Oasis School Mathematics – 6 201

5. Draw a line segment AB of measurement 7cm. Draw the perpendicular bisector
CD of AB, which meets AB at E.
(i) Measure the length of AE and BE.
(ii) Measure the angles AEC and BEC.
(iii) Are AE and BE equal?
(iv) Is the measure of ∠AEC and ∠BEC both equal to 90°?

Construction of perpendicular to a line from a point outside it.

Lets be clear with the help of given example.

Draw a line segmentAB of length 6cm. Take any points P outside it and draw a perpendicular
from P to the given line.

A 6 cm B • Draw a line segment AB = 6cm
P B
• Take any point P outside it.
A
P • Taking P as the centre, draw an arc which
cuts AB at C and D
AC DB
• Taking C and D as the centre draw two arcs
P which meet at point Q.

AC DB
Q

P • Join PQ which meets CD at R.
Measure ∠PRC and ∠PRD.
AC R DB
Q ∴ PR⊥AB.

Answers
Consult your teacher.

202 Oasis School Mathematics – 6

Objective Questions

Choose the correct alternatives:

1. How many lines can be drawn from two given points?

(i) infinite (ii) two (iii) one

2. How many rays can be drawn from a given point?

(i) one (ii) two (iii) infinite B
N
3. In the given figure, relation of AB and MN

(i) AB = MN, (ii) AB // MN (iii) AB ⊥ MN A M
M
4. In the given figure, the relation of AB and MN is
(i) AB = MN (ii) AB // MN (iii) AB ⊥ MN B

5. Which of the following statement is not true? N
(i) Distance between two parallel lines be always same. A

(ii) Angle between two parallel line is 900. PR

(iii) Angle between two perpendicular line is 900. QS B
C D
A

6. In the given figure, AB // CD, which of the following

statement is true? C

(i) AB = CD (ii) PQ = AB (iii) PQ = RS

7. In the given figure, CD is the perpendicular bisector of AB then

(i) AD = DB (ii) AD = CD (iii) CD = DB AD B

8. In the given figure, AB = 10cm, CD is the perpendicular bisector of AB then

(i) ∠CDB = 90º, AD = 10cm C

(ii) ∠CDB = 90º, AD = BD = 5cm

(iii) ∠CDB = 90º, DB = 10cm
ADB

9. If PN = 5cm and MN is the perpendicular bisector of PQ, then which of the following

statement is not true? P

(i) PQ=10cm, MN ⊥ PQ (ii) QN = 5 cm, MN ⊥ PQ.

(iii) MN = 10cm, MN// PQ. MN

Q

Oasis School Mathematics – 6 203

Assessment Test Paper

Group A [7 × 1 = 7]

1. Identify whether the following pair of lines are paralel lines or perpendicular
lines.

(a) (b)

2. Answer the following questions. C
(a) What is the angle between two perpendicular lines?
B
(b) How many lines parallel to AB
can be drawn from point C? A

(c) If CD is the perpendicular bisector of AB, AC
then write the relation between D
(i) AB and CD, (ii) AD and DB. B

3. (a) In the given figure, AB // CD, PQ⊥CD and MN⊥CD, AP MB
what is the relation between PQ and MN? CQ ND

Group B [4 × 2 = 8] B

P

4. (a) From the given point P,

draw PM perpendicular to AB. A

P

(b) From the given point P, N
draw APB parallel to MN.

M

5. Draw a line segment of length 7.2 cm.

6. Draw a line segment of 6 cm, and draw the perpendicular bisector of that line.

7. Which line is the perpendicular bisector of AC? A B
Which line is the perpendicular bisector of BD? E

DC

204 Oasis School Mathematics – 6

Unit

13 Angles

13.1 Fundamental Concepts

In the given figure, two rays OA and OB meet at a B
point 'O' to form an angle.

Here, 'O' is the vertex of the angle.

OA

Angle between the arms Angle between the 11 12 1
of the dividers. blades of scissors 10 2

93

84
765

Angle between the
hands of a clock

Note:

• The symbol ‘∠’ is used to represent an angle.

• An angle is represented by three capital letters taken in such a way that the

letter at the middle is always at the vertex of the angle.

For example: A

B C [∠ABC=∠CBA=∠B]

• Angular measurement of a complete turn is 3600. A

In ∠AOB, O is the vertex. OB
OA and OB are the arms.

Measurement of Angles

We use Protractor to measure an angle.
Take the following steps to measure the angle PQR.

90 P70
80 100 110 120
100 80 60 130
60 110 70
50 120
50 160
P 130 150 20
40 140 30
40
140

30
150

20 160

10 170
170 10

SQR 0 180
180 0
S QR

Oasis School Mathematics – 6 205

In the given figure, the arm QP passes through the 60° mark.
Thus, ∠PQR = 60°.

Steps:

• Place the protractor in such a way that its center point is exactly on vertex Q.
• Adjust it so that the 0° – 180° line is along the arm SR.
• Start counting from the zero which is above QR. Read the number of degrees at the

point where the other arm (QP) passes.

Exercise 13.1

1. Copy the given figures in your copy and name the angle, arm and vertex of

each. A (b) P X D
(a)
(c) (d)

Q

B CR Y Z E F

2. Name the angle and size from the given figure.

(a) A (b) 9070 80 100 110
100
9070 80 100 110 P 60 110 80 70 120
100 50 120 60 130
60 110 80 70 120
50 120 60 130 130
40 50 160
130 150 20
40 50 160 140 140 30
150 20 40
140 140 30 30
40 150
30
150 10 20 160

20 160 170 170
10

10 170 CR 0 180 180 Q
170 10 0
(d)
0 180 O
180 0 F E

B 0
180
(c) X
180 10
0 170

170 20
10 160

20 150 30

160 40
140 130 50
9070 80 100 110 30
100 80 70 150
120 50 40 140
60 110 60 130 120
50 120
50 130 120 100 70 60
130 60 110
40 160 110 8090
150 20 70 80 100
140 140 30
40
30 D
150

20 160

10 170
170 10
WZ
Y0 180
180 0

3. Answer the following questions.

(a) What is a point called, where two arms of an angle meet?

(b) What is the angular measurement of one complete turn?

(c) Which geometric instrument is used to measure an angle?

206 Oasis School Mathematics – 6

4. Measure the size of each of the given angles using protractor and name each
of them in two ways.

(a) (b) B (c)

A

D

O O

O BC C

Answers
Consult your teacher.

13.2 Types of Angles

Angles are divided into different kinds according to their size.
Right angle: An angle which is equal to 90° is called a right angle.
In the figure, ∠AOB is a right angle i.e. 90°.

AA

80 100 110
100 80 70
60 110Right angle9070 120
120 60 130
50 130 50

40 160
150 20
140 140 30
40
30
150

20 160

10 90° 170
170 10
O
OB 0 180 B
180 0

Acute angle: An angle which is more than 0° and less than 90°, is called an acute
angle.

I n the figure, ∠COD = 50°. It is an acute angle because 50° is more than 0° and less
than 90°.

C 90 C7080 100
OD 60 100 80 110 120
120 110 70
60 130
50
130 50
40 160
150 20
140 140 30
40
30 Acute angle
150 50°

20 160 OD

10 170
170 10

0 180
180 0

O btuse angle: An angle which is more than 90° and less than 180°, is called an
obtuse angle.

Oasis School Mathematics – 6 207

In the figure, ∠PQR = 110°. It is obtuse angle because 110° is more than 90° and

less than 180°. P 9070 80 100 110
100
P 60 110 80 70 120
50 120 60 130
Q 50
40 130 160
150 20
140 140 30
40
30 Obtuse angle
150

20 160

110°10 170
R QR 170 10

0 180
180 0

Straight angle: An angle which is equal to 180°. (i.e. two right angles), is called a
straight angle.

In the given figure, ∠XOY is a straight angle.

∴ ∠XOY = 180°

9070 80 100 110 120
100 80 60 130
60 110 70
50 120
40 50 160
130 150 20
140 30
140 Straight angle 40

30
150

20 160 180°

10 170
170 10

X OY X 0 O 180 Y
180 0

Reflex angle: An angle which is more than 180° and less than 360°, is called a
reflex angle.

In the given figure, ∠MON = 230°. It is a reflex angle because 230° is more than
180° and less than 360°.

O O M
M N
N

:
208 Oasis School Mathematics – 6

Angles Size Example Figure
Acute angle Less than 90° 30°, 70°, 80° etc.

Obtuse angle More than 90° and 110°, 120°, 175° etc.
less than 180°

Right angle 90° 90°

Straight angle 180° 180°

Reflex angle More than 180° and 220°, 330°, 345° etc.
less than 360°

Exercise 13.2

1. Name the type of given angles without measurement.

(a) A (b) D F (c) P (d) M

BC E QO O N

(e) (f) (g) (h) K

I H G A B
X
OY

JL
C

2. Classify the given angles into acute, obtuse, right, straight and reflex angles.

(a) 27° (b) 90° (c) 124° (d) 234°

(e) 180° (f) 332° (g) 345° (h) 165°

(i) 135° (j) 76° (k) 57° (l) 45°

(m) 145° (n) 270°

Oasis School Mathematics – 6 209

3. Measure the following angles with protractor and hence classify them into
acute, obtuse, right, reflex and straight angles.

(a) (b) (c)

(d) (e) (f)



4. (a) In the given figure,
(i) name the angles which are acute.
(ii) name the angle which is a right angle.
(iii) name the angle which is obtuse.

(b) In the given figure, A D
(i) name the angles which are acute. BO C
(ii) name the angles which are obtuse.

5. Answer the following questions:

(a) What is an angle called whose measurement is exactly 90°?

(b) What is an angle called whose measurement is more than 90° and less than 180°?

(c) Is 72° an acute angle ? Give reason.
(d) Is 110° an obtuse angle? Give reason.

6. Find the number of acute, obtuse and right angles in each of the given figure.
(a) (b) (c) (d)

Answers
Consult your teacher.

210 Oasis School Mathematics – 6

Project workCM 1 2 3 4 5 6 7 8 987654321
In a chart paper, paste matchstick and show an acute angle, an obtuse angle a
right angle, a straight angle and a reflex angle.

13.3 Construction of an Angle of given Measurement

Construction of 30°, 60°, and 45° with the help of set square

There are two types of set squares.
A set square contains angles 90°, 45°, and 45° as shown in figure (i)
Another set square contains angles 90°, 30° and 60° as in the figure (ii)

45° 60°

90° 45° 30° CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 90°
Figure (i)
B
Construction of 30°
87654321

Figure (ii)

Step I A
• Draw a line segment AB.
A9 CM 1 B
Step II 8
• Fix the set square as shown. 7 2
6
5 3
4
3 4
2
1 5

6

7

8

9

10
11
12
13
14
15

0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

INCH0 1 2 3 4 5 6
INCH0 1 2 3 4 5 6
Step III 9 A CM 1 B
• From B, draw a line BC as shown 8
7 2
in the figure. 6
5 3
C 4
3 4
2
1 5

6

7

8

9

10
11
12
13
14
15

0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Step IV

• R emove ruler and set square, 30°
B
then ∠ABC = 30°. A
Oasis School Mathematics – 6 211

Construction of 45° A B

Step I
• Draw a line segment AB.

Step II 8 8
• Fix the ruler and set square as 7 7
6 6
shown in the figure. 5 5
4 4
3 3
2 2
1 1

CM
A B0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Step III INCH0 1 2 3 4 5 6 C8 8
INCH0 1 2 3 4 5 6A B0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 157 7
• From A, draw a line AC as shown 6 6
in the figure. INCH0 1 2 3 4 5 65 5
INCH0 1 2 3 4 5 6 4 4
3 3
2 2
1 1

CM

Step IV C

• Remove the ruler and the set square,

then ∠CAB = 45° 45°
A
B

Construction of 60°

Step I AB
• Draw a line segment AB.
A 9 CM 1
Step II 8
• Fix the ruler and set square as 7 2
6
shown in the figure. 5 3
4
3 4
2
1 5

6

7

8

9

10
11
12
13

B14
15

0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

C

Step III 9 CM 1
8 2
7 3
6
5 4
4
3 5
2
• From the point A, draw a line 1 6
7
8

9

AC as shown in the figure. 10
11

12

13

14

15

AB
0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Step IV C

• Remove the ruler and the set 60°
square, then ∠CAB = 60° A

B

212 Oasis School Mathematics – 6

Construction of 90° AB

Step I CM 1 2 3 4 5 6 7 8
• Draw a line AB.

Step II AB
• Fix the ruler and set square on the
0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
line AB as shown in the figure.
C
87654321
Step III CM 1 2 3 4 5 6 7 8
• Draw a line AC from A as shown INCH0 1 2 3 4 5 6
87654321A B0 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
in the figure.
INCH0 1 2 3 4 5 6
• Remove the ruler and the set square, then ∠CAB = 90° 87654321

Construction of other angles using two set squares:

Look at the given figures and get the idea of construction of some angles using two
set square.

I. 75º II. 120º 9 8 7 6 5 4 3 2 1
CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
CM 1 87654321 CM 1 2 3 4 5 6 7 8
2
9 3 CM 1 2 3 4 5 6 7 8
8 4
7
6 5
5
4 6
3
2 7
1
8

9

10 120°
11
It is the combination of 30º and 90º
12
13
14
15

75°

It is the combination of 30º and 45º.

Oasis School Mathematics – 6 213

III. 105º CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Now! I can construct 135°
and 150° using two set
87654321 squares.

CM 1

2

345678
105°
987654321
It is combination of 60º and 45º.
9
8
Exercise 13.3 7
6
1. How many types of set square are there? 5
2. What are the values of the 4
angles in the given set square? 3
2
CM 1 1
123456789
3. What are the values of three 2 CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
angles in the given set square? 3

4

5

6

7

8

9

10
11

12
13
14
15

987654321

4. (a) What is the value of mentioned angles formed by
combining two set squares in each of the given figure.
CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
8 7 6 5 4 3 2 1 MC 12345678 1 MC
2 9
3 8
4 7
6
5
4
3
2
1
(b) 5 123456789

6 CM 1
2
7 3
4
8 5
6
9 7
8
01 9 87654321
11 10
11
21 12
31 13
41 14
51 15

CM 1 2 3 4 5 6 7 8
(c) 9
214 Oasis School Mathematics – 6 8
7
6
5
4
3
2
1

CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

4. Using ruler and set square draw the following angles.

(a) 30° (b) 45° (c) 60° (d) 90°

5. Using two set squares draw the following.

(a) 105° (b) 120° (c) 135° (d) 75° (e) 150°

To construct a given angle by using compass:
Let us construct the following angles by using compass.

I. To construct an angle of 60°.

D

EE

A B AC B AC B

Steps:

• Draw a line AB of suitable length.
• At A, draw an arc of any size to cut AB at C.

• With C as center, draw the same size arc, to cut the first arc at E.
• Join AE and produce it to D.

Hence, ∠DAB = 60° is the required angle.

Note: Measure the ∠DAB (By protractor) and check.

II. To construct an angle of 120°. P

ED ED

A B AC B 120° B
AC

Steps:

• Draw a line AB of suitable length.
• At A, draw an arc of any size to cut AB at C.
• With C as center, draw the same size arc, to cut the first arc at D.
• With D as centre, draw one more arc of the same size which cuts the first

arc at E.
• Join AE and produce it to P.

Hence, ∠PAB = 120° is the required angle.

Oasis School Mathematics – 6 215

III. To draw a bisector of angle

In the given figure, ∠BOC = 300 and ∠AOC = 300. OC divides ∠AOB into two equal

parts. Then, OC is the bisector of ∠AOB. A
How to bisect the given angle?

Let us take an ∠AOB. 300 C
To bisect this angle, lets follow the following steps: B
O 300

Steps:

• Taking centre 'O' take an arc of suitable length which meets OA and OB
at C and D respectively.

• TakingCandDascentres,drawtwoarcsofequalradiitointersecteachother
at E.

• Join OE.

OE is the bisector of ∠AOB. A
Verification, CE
Lets measure ∠AOE and ∠BOE.
Measurement of both angles should be the same. O DB

IV. To construct an angle of 30°.

D DE

A B AC B A 30° B

Steps: C

• Draw a line AB of suitable length.
• At A, draw an arc of any size to cut AB at C.
• With C as center, draw the same size arc, to cut the first arc at D.
• Taking C and D as centers, draw arcs of equal radii to intersect each other

at point E.
• Join AE.

Hence, ∠EAB = 30° is the required angle.

216 Oasis School Mathematics – 6

V. To construct an angle of 90°:
Let AB be a line and at point A, an angle of 90° is to be drawn.

A ED F B
B AC ED
Steps: B AC

• Draw a line AB of suitable length.
• At A, draw an arc of any size to cut AB at C.
• With C as centre, draw the same size arc to cut the first arc at D.
• With D as centre, draw one more arc of the same size which cuts the first

arc at E.
• Taking D and E as centers, draw arcs of equal radii to intersect each other

at point F.
• Join AF.

Hence, ∠FAB = 90° is the required angle.

VI. To Construct angle of 45°. F
EG D
F
H
EG D

A B AC B AC B

Steps:

• Draw a line segment AB of suitable length.
• At A, draw an arc of any size to cut AB at C.
• With C as the centre use the arc of the same size to cut the first arc at D.
• With D as the centre use the arc of the same size to cut the first arc at E.
• A gain with D and E as the centre, use the arc of the same size to cut at the point F.
• Join F and A which cuts the first on G. Then ∠FAB = 90°
• Bisect the ∠FAB. i.e. taking C and G as the centre, cut at the point H and join AH.

Hence, ∠HAB = 45° is a required angle.

Oasis School Mathematics – 6 217

Exercise 13.4

1. Use protractor or set square to draw the following angles and bisect them.

(a) 60° (b) 30° (c) 45° (d) 120°

2. Construct the angles of following size using a ruler and compasse.

(a) 60° (b) 120° (c) 90° (d) 30° (e) 15°
(f) 45° (g) 75° (h) 105° (i) 135° (j) 150°

3. (a) Draw an angle 700 with the help of protractor and bisect it.

(b) Draw a line segment OP of any measurement using a protractor and draw an
angle QOP = 50°. Draw a bisector OR of ∠POQ. Find the measure of ∠ POR and
∠QOR.

Answers
Consult your teacher.

Activity

Construction of angular bisector using paper folding method:
• Take a sheet of paper

• Draw an ∠ AOB

• Cut the ∠AOB A A
A
A

C

O BO BO BO B

• Fold the paper in such a way that OA falls on OB. Draw OC on the crease

OC is the bisector of ∠AOB.

13.4 Pair of Angles C A
O B
There are some pairs of angles. Let us discuss them.
Adjacent angles:

In the given figure, ∠AOB and ∠AOC both have the
common vertex O and common arm OA.

∴ ∠AOB and ∠AOC are adjacent angles.

Hence, two angle having a common vertex and a common
arm are adjacent angles.

218 Oasis School Mathematics – 6

Linear pair: C
OB
In the given figure, ∠BOC and ∠AOC are adjacent angles

and their sum is 180°. Therefore ∠AOC and ∠BOC are the

linear pairs. A

i.e. "linear pair are the adjacent angles whose sum is 180°".

Vertically Opposite angles:

In the given figure, lines AB and CD interest at O. Among four angles formed at O,

∠AOC and ∠BOD lie on the opposite side of common vertex O. Similarly, ∠AOD

and ∠BOC lie on the opposite side of common vertex O. A C
∴ ∠AOC and ∠BOD are vertically opposite angles

Similarly, ∠AOD and ∠BOC are vertically opposite angles. D O
Vertically opposite angles are equal. B
∴ ∠AOC = ∠BOD and ∠AOD = ∠BOC

Complementary angles:

In the given figure, ∠AOB = 90°. Line OC divides ∠AOB B C
into two angles ∠AOC and ∠BOC.

i.e. ∠AOB = ∠AOC + ∠BOC

or, ∠AOC + ∠BOC = 90° O A
then, ∠AOC and ∠BOC are the complementary angles.

Hence, two angles are said to be complementary if their sum
is 90°.

If x + y = 90°, then x and y are complementary angles and we say, x is the complement
of y and y is the complement of x.

Remember: Complement of x is always 90° – x0.

Supplementary angles: C

In the given figure, ∠AOB is a straight angle i.e.

∠AOB = 180°. OC divides ∠AOB into two parts ∠BOC A B

and ∠AOC then ∠AOC + ∠BOC = 180°. O

Here, ∠AOC and ∠BOC are the supplementary angles.

Hence, two angles are said to be supplementary if their sum is 180°. If x + y = 180°,
then, 'x' and 'y' are supplementary angles. Here, x is the supplement of y and 'y' is
the supplement of x.

Remember ! Supplement of x0 is always 1800 – x.

Oasis School Mathematics – 6 219

Angle at a line and Angle around a point b
ca
In the figure, a, b and c are angle at a line. Their sum is 180º.
a + b + c = 180º·

In the given figure, p, q, r and s are angles around a point q
which make a complete turn. ∴ p+ q + r + s = 360º. rp

s

Remember!

• Sum of linear pairs is 180º • Vertically opposite angles are equal
• Two angles are said to be supplementary if their sum is 180º
• Two angles are said to be complementary if their sum is 900

Worked Out Examples

Example: 1

Find the value of x in the given figure.

(a) (b) x°

30° 50°

x° 60°

Solution: Solution:
x° +60° = 180°
[∵ Straight angle = 180°] 30°+x°+50° = 180°
or, x = 180 – 60°
∴ x = 120° [∵ Straight angle = 180°]

or, x+80° = 180°

or, x = 180°–80°

Example: 2 ∴ x = 100°
Find the value of x.

Solution:

x° + 2x° + 12x° = 360° [∵ Angle around or point]

or, 15x° = 360° x°
∴ x° = 360o
15 = 24° 2x° 12x°

Example: 3

If (x +20°) and (2x + 10°) are the supplementary angles, find the size of

each angle.

220 Oasis School Mathematics – 6

Solution:
Here, (x + 20°) + (2x + 10°) = 180° [∵ Sum of supplementary angles]
or, 3x + 30° = 180°
or, 3x = 180° – 30°

or, 3x = 150°
150o
or, x = 3

or, x = 50°.

Hence two angles are,

x + 20° = 50° + 20° = 70° and (2x + 10)° = 2×50° + 10° = 110°.

Example: 4
In the given figure, find the value of w°, x°, y°, and z°.
Solution:

w° = 90°, y° = 60° [∵ Vertically opposite angles] y° 90°
Again, x°

x° + y° + 90° = 180° [∵ Straight angles] w z°
or, x° + 60° + 90° = 180° 60°

or, x° = 180° – 150° = 30°

Again, x° = z° [∵ Vertically opposite angles]

∴ z° = 30°.

Exercise 13.5 A B
C
1. (a) Name the adjacent angle of ∠AOB. D

O BC

(b) Name two angles which are linear pair? A
What is their relation?

(c) In the given figure,

(i) which is the vertically opposite angle of ∠AOC? A D
OB
(ii) which is the vertically opposite angle of ∠AOD? C
(iii) which is the linear pair of ∠AOD? a
db
(d) Look at the given figure and fill in the blanks.
c
(i) a = ......... (ii) b = ...............

(iii) a + ....... = 1800 (iv) a + ........ = 1800

(v) c + ........ = 1800 (vi) c + ....... = 1800

Oasis School Mathematics – 6 221

2. Find the value of x in each of the following figures. (d)
(a) (b) (c)

x 60° x 90° 45° x x 85°

3. Find the value of unknown angles.

(a) (b) (c)

x 60° x 50°
20° x 70°
60° 40°

(d) (e) 20° (f)

90° 26° x 65° x
x 30° 40°

4. Find the value of x in each of the following figures.

(a) (b) (c)

6x 4x 3x 2x x
2x x

(d) (e)

(3x + 20)° x° 60°
(2x + 20)°
2

5. Find the value of x . 140° (c) 150° (d)
(a) (b) 160°
x x 140°
x
2x° 60°
280° 40°

6. Fill in the blanks. (b) Supplementary angle of y° is ……
(a) Complementary angle of x° is …… (d) Sum of the linear pairs is ……
(c) Vertically opposite angles are ……
(e) Complement of 60° is ……… C
E
7. In the given figure write,
A B
(a) Vertically opposite angle of ∠CFE. F

(b) Complementary angle of ∠CFE. G
D
(c) Vertically opposite angle of ∠AFC.
(d) Adjacent angles of ∠CFE.

222 Oasis School Mathematics – 6

8. Find the value of unknown angles in the given figures.

(a) (b) (c)

500 x 860 500 1300
x xy

9. Find the value of unknown angles in each of the following figures.

(a) (b) (c) z° x°
3x°
x b° a°
35°
y 95°
z c°

(d) (e) (f)

x° a 2a a 3x x
x° yz c
2x a b
z° y x

10. Find the complement of following angles.

(a) 250 (b) 420 (c) 620 (d) 700

11. Find the supplement of following angles.

(a) 1000 (b) 920 (c) 1050 (d) 50

12. (a) Are 800 and 100 complementary angles? Justify your answer.

(b) Are 1300 and 700 supplementary angles? Justify your answer.

13. (a) If 2x and 3x are the supplementary angles, find the value of each angle.
(b) If y and 2y are the complementary angles, find the value of each angle.
(c) If 2x° and (x + 60)° are the vertically opposite angles, find their values.
(d) If (2x + 30)° and (3x + 50)° are the linear pairs, find their values.

Answers

1. Consult your teacher

2. a) 120º b) 90º c) 135º d) 95º 3. a) 80º b) 100º c) 60º d) 60º

e) 134º f) 75º 4. a) 18º b) 45º c) 45º d) 28º e) 60º

5. a) 80º b) 220º c) 50º d) 60º 6. Consult your teacher. 7. Consult your teacher.

8. a) x = 500 b) a = 860 c) x = 1300, y = 500 9. a) x = 85º, y = 95º, z = 85º b) a = 145º, b = 35º,

c = 145º c) x = 45º, y = 45º, z = 135º d) x = 45º, y = 45º, z = 90º e) a = 45º, x = 90º, y = 45º,

z - 45º f) x = 30º, a = 30º, c = 60º 10. Consult your teacher

11. Consult your teacher 12. Consult your teacher

13. a) 72º, 108º b) 30º, 60º c) 120º d) 70º, 110º

Oasis School Mathematics – 6 223

Objective Questions

Choose the correct alternatives: P
1. Which one of the following is not the correct name of given angle?

(i) ∠PQR (ii) ∠QPR (iii) ∠RQP

2. Instrument which is used to measure the given angle is Q R

(i) set square (ii) ruler (iii) protractor

3. I am an angle, my value is more than 900 but less than 1800, I am

(i) an obtuse angle (ii) an acute angle (iii) a reflex angle

4. If x + y = 1800 then x and y are

(i) complementary angles (ii) supplementary angles (iii) vertically opposite angles

5. Complement of 200 is

(i) 1600 (ii) 200 (iii) 700

6. Supplement of 600 is

(i) 600 (ii) 1200 (iii) 300

7. Which one of the following statement is not true?

(i) Vertically opposite angles are equal.

(ii) Supplementary angles are equal.

(iii) Sum of complementary angles is 900. P O S
8. In the given figure, vertically opposite angle of ∠POR is

(i) ∠POS (ii) ∠QOS (iii) ∠ROQ R Q

9. Adjacent angle of ∠ABD is AD
C
(i) ∠CBD (ii) ∠BAD (iii) ∠CDB
B
10. 1600 is
(i) reflex angle (ii) obtuse angle (iii) actute angle.

11. In the given figure, values of 'a' and 'b' a
(i) a = 600 b= 1200 (ii) a = 1200, b = 600 b 600
(iii) a = 600, b = 1800

12. Which of the following statement is not true?
(i) Adjacent angles are equal. (ii) Angle around a point is 3600.
(iii) Sum of linear pairs is 1800.

224 Oasis School Mathematics – 6

13. Which of the following statement is true?
(i) 500 and 400 are complementary angles.
(ii) 1500 and 500 are linear pairs.
(iii) 1200 and 1500 are supplementary angles.

Project work
Preparation of paper protractor.

1. Take a rectangular sheet of paper.

2. Fold the paper as shown in the figure and unfold it and draw a line on the
crease.

900

3. Again, fold it in the same way. Again make next fold as shown in the figure
and unfold it.

1350
450 900

4. Again, unfold the paper and fold it as shown in the figure and unfold it.

1200 900 600
1350 450

Again, we can further fold it to get 300. 1800 00

Oasis School Mathematics – 6 225

Assessment Test Paper

Attempt all the questions.

Group A (5 × 1= 5]

1. (a) What is an angle called whose value is more than 1800 and less than

3600? P

(b) Measure the given angle O Q
and identify its type. A

B

2. (a) In the given figure, write the OC
adjacent angle of BOC.

M Q
O
(b) Write the vertically opposite angle of
(i) ∠MOQ (ii) ∠QON. PN

(c) Is 1300 an obtuse angle? Give reason.

Group B (5 × 2 = 10]

3. Using compass draw an angle of 1200.

4. Draw a line segment of length 7cm. Draw an angle 600 and bisect the
angle 600.

5. Find the value of x. 1250
6. Find the value of x.
x

250
1200
x

7. From the given figure, 400 z
find the values of x, y and z. xy

226 Oasis School Mathematics – 6

Unit Triangle and
Quadrilateral
14

14.1 Triangle

Introduction

Let's observe the objects alongside:

Let's find the common feature of these two figures.
Both figures are bounded by three straight lines.

Hence, a triangle is a closed figure bounded by
three straight lines.

These two figures are triangles.

Given figure is a triangle. It has three A CA
sides, three angles and three vertices. BC

Its three sides are AB, BC and AC.

Its three angles are ∠A, ∠B and ∠C. B

Its three vertices are A, B and C.

Types of triangles on the basis of its sides

Scalene triangle: A
In the figure, AB = 4 cm, BC = 5 cm, CA = 6 cm.
All of its sides are not equal in length. 4cm
∴ ∆ABC is a scalene triangle. 6cm

If all three sides of a triangle are unequal, it is called a B 5cm C
scalene triangle.
A
Isosceles triangle:
In the figure, AB = AC = 2 cm. and BC = 3cm.
Its two sides are equal in length.
∴ ∆ABC is an isosceles triangle.

If any two sides of a triangle are equal, then it is called B C
an isosceles triangle.

Oasis School Mathematics – 6 227

Note:

• If two sides of a triangle are equal, the angles opposite A
to them are equal. C

In the given figure, if AB = AC then ∠B = ∠C.

• If two angles of a triangle are equal, the sides opposite

to them are also equal. B
In the given figure, if ∠B = ∠C then AB = AC.

Equilateral triangle:

In the figure, AB = BC = AC = 2cm. A
Its all sides are equal in length.

∴ ∆ABC is an equilateral triangle.

If all three sides of a triangle are equal, it is an equilateral

triangle. B 2 cm C

Note: Each angle of an equilateral triangle is 600.

Types of triangles on the basis of its angles

Right angled triangle: In ∆ABC, ∠B = 90°, so, ABC is A
a right angled triangle. If one angle of a triangle is a right
angle, it is called a right angled triangle.

Note: Two angles of a right angled triangle are acute. B C

Obtuse angled triangle: In the given figure, A
∠B = 130º, which is obtuse angle.
1300 C
If one angle of a triangle is obtuse, it is called an obtuse
angled triangle. B
Note: Two angles of an obtuse angled triangle are acute.

Acute angled triangle: A
In ∆ABC, ∠A= 60º, ∠B = 50º and ∠C = 70º are acute angles.
∴ ∆ ABC is an acute angled triangle. 600
All the angles are acute angle.
B 500 700 C

Hence, if all three angles of a triangle are acute, then it is an acute angled triangle.

228 Oasis School Mathematics – 6

Exercise 14.1

1. Mention the type of following triangles on the basis of their sides.

(a) A (b) X (c) L (d) P

B 4 cm C Y 3 cm ZM N Q 4 cm R

2 cm

2. Identify the types of triangle if the measurement of its three sides of triangle
are as shown.

(a) 4 cm, 5 cm and 6 cm (b) 4.5 cm, 6cm and 6 cm
(c) 7cm, 7cm and 7 cm (d) 8cm, 6.5 cm and 6.5 cm

3. Measure the lengths of each side of the following triangles and distinguish

them according to their sides. X

(a) A (b) P (c)

Y

B CQ R

4. Distinguish the following triangles according to their angles. Z

(a) (b) P (c) C

30° 60° 60°

120° 30° Q 50° 70° R D 90° 30° E

5. Identify the types of triangle (on the basis of angles) if the measurement of

three angles are as shown.

(a) 600, 700 and 500 (b) 300, 600 and 300 (c) 200, 300 and 1300

6. In each of the triangle, measure the size of each angle and distinguish according

to their angles. (c) P

(a) A (b) X

B CY ZQ R

Oasis School Mathematics – 6 229

7. Write true (T) or false (F) in the given statements.

(a) All three sides of an equilateral triangle are equal.
(b) Any two sides of an isosceles triangle are equal.
(c) All the sides of a scalene triangle are equal.
(d) Every equilateral triangle is also an isosceles triangle.
(e) One angle of right angled triangle is acute.
(f) Two angles of an obtuse angled triangle are acute.
(g) Each angle of an equilateral triangle is 60°.
(h) Two angles of an isosceles triangle are equal.
(i) All angles of an acute angled triangle are acute.
8. Answer the following questions:

(a) What is a triangle called, whose all three sides are equal?
(b) What is a triangle called, whose none of the sides are equal?
(c) If the length of all three sides of a triangle are different, then what type of triangle is this?
(d) In ∆ABC, if ∠A = 62°, ∠B = 35°, ∠C = 83°, which type of triangle is this?
(e) In ∆ABC, if ∠A = 30°, ∠B = 96°, ∠C = 54°, which type of triangle is this?

(f) What is the value of each angle of an equilateral triangle?

(g) How many angles of an obtuse angled triangle are acute?

9. (a) In ∆ABC, if AB = AC, then A
which two angles are equal?

P BC

(b) In ∆PQR, which two angles are
equal if PR = QR?

Q R

X

10. (a) In ∆XYZ, if ∠XYZ = ∠XZY,
which two sides are equal?

YZ

PQ

(b) In ∆PQR, if ∠QPR = ∠PQR,
which two sides are equal?

R

Answers
Consult your teacher.



230 Oasis School Mathematics – 6

Sum of the angles of a triangle

Let's draw two triangles ABC of different measurements and measure all three

angles. A A

BC B C
Fig (i) Fig (ii)

Observations:

Figure ∠A ∠B ∠C ∠A + ∠B + ∠C
55°+60°+65° = 180°
(i) 55° 60° 65°

(ii) ...... ...... ...... .......

We will see that, sum of three angles of triangle is 180°

Remember! A
• In ABC, ∠A + ∠B +∠C = 180º. BC

Worked Out Examples

Example: 1 x
50° 60°
Find the value of x.

Solution:

x + 50° + 60°= 180° [∵ Sum of the angles of a triangle]

or, x + 110° = 180°

or, x = 180° – 110°

∴ x = 70°.

Example : 2 A

Find the value of 'x' in the following figure. 3x

Solution: In ∆ABE, B 600 2x

3x + 2x + 60º = 180º

[Sum of three angles of a triangle] C

or, 5x + 60º = 180º

or, 5x = 180º – 60º

or, 5x = 120º
or,
∴ x= 1200
5

x = 24º.

Oasis School Mathematics – 6 231

Example : 3 A

Find the value of 'x' in the following figure. B 580 x C
Solution:

Given, AB = AC
∠B = ∠C [Sum of three angles of a triangle]

or, 58º = xº
∴ x = 58º.

Exercise 14.2

1. Find the value of x: (b) P (c) X
(a) A 60°
50° Z
80°

B 50º x° C Q x 50° R Y x°

(d) D (e) G

x 25º

E 30º 70° F x 25º I
H


2. a) In ∆ABC, ∠A = 60º, ∠B = 80º, find the value of ∠C.

b) In ∆PQR, ∠P = 45º, ∠Q = 65º, find the value of ∠R.

c) In ∆XYZ, ∠X = 100º, ∠Y = 25º find the value of ∠Z.

3. Find the value of 'x' in the following figures:

(a) D (b) (c) X
4x
90° A 40° C Y 3x
E 2x° 2x Z
70° A

x° F B 2x 7x

(d) G (e) J (f) 6x

68º 30º

H 3x xI K 7x B 5x

8x L C

232 Oasis School Mathematics – 6

4. Find the value of 'x' and 'y' from the given figures:

(a) A (b) D (c) Y

400

B 700 x C Ex 250 F X x

Answers Z

1. a) 50º b) 70º c) 40º d) 80º e) 130º 2. a) 40º b) 70º c) 55º
d) 28º e) 10º f) 10º
3. a) 30º b) 35º c) 20º
4. a) 70º b) 25º c) 40º

14.2 Quadrilateral

A quadrilateral is a closed figure bounded by four line A
B
segments. A quadrilateral has four sides and four angles. In
C
the given figure, ABCD is a quadrilateral. AB, BC, CD and
AD are its four sides and ∠A, ∠B, ∠C and ∠D are its four D

angles.

Sum of the four angles of a quadrilateral

Let's draw two quadrilaterals ABCD of different shape. Measure their angles and

tabulated in the given table. A

A

D B
D
B

figure (i) figure (ii)
C C

Observations:

Figure ∠A ∠B ∠C ∠D ∠A +∠ B + ∠C + ∠D

(i) .......º .......º .......º .......º 360º

(ii) .......º .......º .......º .......º 360º

Conclusion : Hence, the sum of angles of a quadrilateral is 3600.

Oasis School Mathematics – 6 233

Worked Out Examples A B
92° 150°
Example : 1
x 63° C
Find the value of x, from the given figure. D
Solution :
Here, x+92°+150°+63° = 360°
[Sum of four angles of a quadrilateral]
or, x+305° = 360°
or, x = 360°–305°
∴ x = 55°

Example : 2

Find the value of x, y and z, from the given figure. 72°

Solution : 127° x
Here, x = 72° [Vertically opposite angles] y z

Again,y +82° = 180° [Being a linear pair]

or, y = 180°–82° 82°

∴ y = 98°

Again, x+y+z+127° = 360° [Sum of four angles of a quadrilateral]

or, 72°+98°+z+127° = 360°

or, z+297° = 360°

or, z = 360°–297°

∴ z = 63°
Example : 3 AB
2x 2x+300
Find the value of x and the value of unknown angles

in the given figure.

Solution : 2x–200 x
Given, ABCD is a quadrilateral.
D C

2x + 2x – 200 + 2x + 300 + x = 3600

[Sum of four angles of a quadrilateral.]

or, 7x + 100 = 3600

or, 7x = 3600 – 100

or, 7x = 3500
or,
x= 3500
7

234 Oasis School Mathematics – 6

∴ x = 50°.
Now, ∠A = 2x = 2 × 50° = 100°
∠B = 2x + 30° = 2 × 500 + 300 = 130°
∠C = x = 500
∠D = 2x – 20° = 2 × 50° – 20° = 80°.

Exercise 14.3

1. Find the value of unknown angles in each of the following figures.

(a) (b) W (c) (d)

A S M
140°
130° x 90° P 95° xN

B 85° D X 150° 65° Z 100°
70°
a Q 90° x P 65°
C R 110°

YO

2. (a) If xº, 45º, 60º and 140º are four angles of a quadrilateral, find the value of x.

(b) In a quadrilateral ABCD, If ∠A = 70º, ∠B = 65º, ∠C = 85º, find the value of ∠D.

(c) In a quadrilateral PQRS if ∠P = 65º, ∠Q = 85º, ∠R = 95º, find the value of S.

3. Find the value of unknown angles. 52° (c)
(a) (b)
x 82° y z
110° 100°
70° x 35° 96° 100° x
106°
yy

4. Find the value of x and the value of unknown angles in the given figures:

(a) A (b) P 3x x Q (c) W 2x 2x+30ºX

(x+10)º

x+30ºB 2x D 4x Z 2x–20º

(x–20)º 2x R xY

S

C c) 115º

Answers
1. a) 75º b) 55º c) 75º d) 45º 2. a) 115º b) 140º

3. a) x = 35º, y = 145º b) x = 52º, y = 112º c) x = 740, y = 820, z = 1040

4. a) x = 68º ∠A = 78º, ∠B = 98º ∠C = 48º, ∠D = 136º

b) x = 36º ∠P = 108º, ∠Q = 36º, ∠R = 144º, ∠S = 72º

Oasis School Mathematics – 6 235

Types of quadrilateral D C
B
Parallelogram:
C
In the given figure, ABCD is a quadrilateral. Where, A B
AB = DC, AB||DC, AD = BC AD||BC. D
C
Then, ABCD is a parallelogram. Hence, a parallelogram is a B
quadrilateral whose opposite sides are equal. C
Rectangle: B

In the given figure, ABCD is a parallelogram. Where, B
C
AB = DC, AB || DC, AD = BC, AD || BC, A
∠A = ∠B = ∠C = ∠D = 90°.

Hence, a rectangle is a parallelogram whose every angle is

900.
Note: Every rectangle is a parallelogram.

Square: D
Here, ABCD is a square. Where AB = BC = CD = AD and A
∠A = ∠B = ∠C = ∠D = 90°.

Hence, a square is a rectangle whose all four sides are equal.

Rhombus: D
Here, ABCD is a parallelogram.

Where, AB = BC = CD = AD and AB||DC, AD ||BC. Then,

ABCD is rhombus. Hence, a rhombus is a parallelogram A
whose all sides are equal.

Note: Every square is a rhombus. A

Every rhombus is a parallelogram.


Trapezium:

In the given figure, ABCD is a quadrilateral

where AB // CD. It is a trapezium. Hence, a trapezium is D
a quadrilateral whose one pair of opposite sides are parallel.

236 Oasis School Mathematics – 6

Features of different quadrilateral. A B

Parallelogram: D C
• Opposite angles are equal. A
• Opposite a sides are equal. D B
• Opposite sides are parallel. C
Rectangle: A
• Opposite angles are equal. B
• Opposite sides are equal. D
• Opposite sides are parallel. A C
• Each angle is 900. B
Square: D C
• Opposite angles are equal. A B
• All sides are equal. D C
• Opposite sides are parallel.
• Each angle is 900.
Rhombus:
• Opposite angles are equal.
• All sides are equal.
• Opposite sides are parallel.

Trapezium:
• One pair of opposite sides are parallel.

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Activity

Preparation of different geometrical shapes with the help of set squares.
Material required:
Four sets of 450 set squares, four sets of 300 set squares.

450

300

450 600

450 set square 300 set square

I. Square
• Take two sets of 450
set squares.
• Place them as shown
in the figure to get a
square.

II. Equilateral triangle

• Place two sets of 300 set
squares as shown in the
figure to get an equilateral
triangle.

238 Oasis School Mathematics – 6

III. Parallelogram
• Arrange two sets of 450
set squares as shown
in the figure to get a
parallelogram.

IV. Isosceles triangle
• Arrange two sets of 450 set
square as shown in the figure
to get an isosceles triangle.

V. Rhombus
• Arrange four sets
of 450 set squares
as shown in the
figure to get a
rhombus.

VI. Rectangle
• Place two sets of 300 set
squares as shown in the
figure to get a rectangle.

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Worked Out Examples

Example : 1

Find the value of unknown angles from the given figure. z 50°

Solution : Here, y 130° x
x+130° = 180° [Being a linear pair]

or, x = 180°–130°

∴ x = 50°

Again, y = 50° [Opposite angles of a parallelogram]

z = 130° [Opposite angles of a parallelogram]

Example : 2

Find the unknown angles and sides of given rhombus.

Solution : Here, b x
Given, ABCD is a rhombus. z
Its all sides are equal. a
800
1000 y

then, AB = BC = CD = AD 5 cm

∴ x = y = z = 5cm.

Again, b = 1000 [Opposite angles of a rhombus.]

a = 800 [Opposite angles of a rhombus.]

Exercise 14.4

1. Identify the following shapes.

(a) (b) (c)

(d) (e)

240 Oasis School Mathematics – 6

2. Fill in the blanks.
(a) In a parallelogram, ……… sides are equal and ……… angles are equal.
(b) Value of each angle of a rectangle is ……… .
(c) If opposite sides are equal and each angle is 90°in a quadrilateral then it is a
……… .
(d) In a square, ………. angles are equal and ………. sides are equal.
(e) In a rhombus, ……….. angles are equal and ………. sides are equal.
(f) In trapezium .....................sides are parallel.

3. Tick () or cross (×) in the box.

Features Parallelogram Rectangle Rhombus Square

Opposite sides are equal.    

Opposite angles are equal.

All sides are equal

All angles are equal.

Each angle is 90°.

4. Identify the types of quadrilateral having the following properties:
(a) Quadrilateral having opposite sides equal and parallel.
(b) Quadrilateral having all sides equal.
(c) Quadrilateral having all sides equal and each angle 90°.
(d) Quadrilateral having opposite sides equal and each angle 90°.
(e) Quadrilateral having one pair of opposite sides parallel.

5. (a) Write one different property between rhombus and square.
(b) Write one different property between parallelogram and rectangle.
(c) Write one different property between rhombus and parallelogam.
(d) Write one different property between square and rectangle.

Oasis School Mathematics – 6 241

6. From the given information, identify the type of quadrilateral with reason.

(a) S x S (b) (c) x
6cm
S 1200 x
y 8cm
8cm y
600 y

6cm S S S

(d) A B (e) P Q (f) E F

D CS R H G

6. Find the unknown values from the given figure. 50° B
130° C
(a) (b) (c) A
y
x+400 3x–100
Dx
2x 500

(d) E 70° F (e) I x yJ
x
110° L 60° 120° z M
Hy G K

Answers

1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher.
4. Consult your teacher. 5. (a) x = 6cm, y = 8 cm (b) x = 600, y = 1200
6. (a) 400 (b) 200 (c) x = 8cm, y = 6cm (d) x = 50º, y = 130º
e) x = 110º, y = 70º f) x = 120º, y = 60º, z = 60º

242 Oasis School Mathematics – 6

Project work
Tangram

A tangram is a Chinese puzzle made by writing square of thin material into five triangles,
a square and a parallelogram.

These 7 pieces of geometrical shapes can be put together to make different shapes.

How to cut 7 pieces from a square sheet?

• Take a thick square sheet of paper and divide it
into 4 equal parts along its length and breadth as
shown in the figure.

• Draw the solid lines as shown in the figures. 2
1 56

4 7
3

2 6
5
• With the help of scissor cut along the solid lines to get 7
pieces of geometrical shapes. 1
4

37

• Arrange these 7 pieces of paper to get different shapes like.

House Walking Lady Swan

Aeroplane Horse Candle

Try to make some other shapes also.

Oasis School Mathematics – 6 243

Activity

Use of Tangram to prepare different Geometrical shape.
• Make the 7 tangram pieces as the instruction given in the previous page.

34 7
15 6

2

• Make the rectangle, sqaure, parallelogram, trapezium and rhombus.

2 63 7 2
7 56
142
1 54
Trapezium
Rectangle

14.3 Construction of Regular Polygons

Students are advised to draw a rough free hand sketch in each case, before starting
actual construction.

Construction of an equilateral triangle

Example :

Construct an equilateral triangle ABC in which AB = BC = AC = 4 cm.

A

A Rough

BC

B C 4 cm

4 cm

244 Oasis School Mathematics – 6