Lesson: Constructions

Exercise 11.1 (5)

Question: 1

Construct an angle of 90° at the initial point of a

given ray and justify the construction.

Solution:

Steps of construction:

1. Let us draw a ray

OX

.

2. With

O

as a centre and any radius, let us draw an

arc ABC cutting

OX

at

A

.

3. With B as a centre and the same radius, let us

mark a point B on the arc

ABC

. (By construction

= 60AOB

)

4. With B as a centre and the same radius, let us

mark a point

C

on the arc

ABC

. (By construction

= 60BOC

)

5. With B and C as centres, let us draw two arcs

intersecting each other with the same radius at

D.(By construction

=BOD COD

)

6. Let us join

O

and

D

to form a ray

OD

.

7.

AOD

= 90°

Justification for construction:

= 120° = 60 , = 60 :

=

= =

1

= 60

2

= 30

Now, =

= 60

(

30 =

)

(

90

)

AOC AOB BOC

BOC BOD DOC

BOD BOC BOD DOC

BOD

BOD

AOD AOB BOD

Question: 2

Construct an angle of 45° at the initial point of a

given ray and justify the construction.

Solution:

Steps of construction:

1. Let us draw a ray

OX

.

2. With

O

as a centre and any radius, let us draw an

arc ABC cutting

OX

at

A

.

3. With B as a centre and the same radius, let us

mark a point B on the arc

ABC

. (By construction

60AOB

)

4. With

B

as a centre and the same radius, let us

mark a point C on the arc

ABC

. (By construction

60BOC

)

5. With

B

and

C

as centres, let us draw two arcs

intersecting each other with the same radius at

D.(By construction

BOD COD

)

6. Let us join

O

and

D

to form a ray

OD

.

7.

AOD

= 90°

8. With

A

and

D

as centres, let us draw two arcs

intersecting each other with the same radius at

E

.

9.

AOE

= 45°

Justification for construction:

= 120 = 60 , = 60 :

=

1

= =

2

1

BOD = ×60

2

BOD = 30

Now,

= +

= 60 + 30 = 90

1

=

2

= as

( )

is the angle b

(

( se t

)

i c

AOC AOB BOC

BOC BOD DOC

BOD BOC BOD DOC

AOD AOB BOD

AOE AOD

AOE DOE OE

or

1

= of 90

2

)

= 45

Question: 3

Construct the angles of the following measurements:

(i) 30°

(ii) 22.5°

(iii) 15°

Solution

(i) 30°

Steps of construction:

1. Let us draw a ray

AB

.

2. With A as a centre and any radius, let us draw an

arc cutting

AB

at

D

.

3. With

D

as a centre and the same radius, let us

mark a point

E

on the arc.(By construction

= 60BAE

)

4. With D and E as centres, let us draw two arcs

intersecting each other with the same radius at

F(By construction

DAF EAF

)

5. Let us join

A

and

F

to form a ray

AF

that makes

an angle 30° with

AB

(ii) 22.5°

120 60 , 60 :

1

60

2

3

( )

0

( )

AOC AOB BOC

BOC BOD DOC

BOD BOC BOD DOC

BOD

BOD

Now

as is the angle bisector

as is the angle bisector

60 30 90

1

2

1

90 45

2

1

2

( )

1

45 22

(

.5

2

)

AOD AOB BOD

AOE AOD

AOE DOE OE

of

AOG AOE

AOG GOE OG

of

(iii) 15°

Steps of construction:

1. Let us draw a ray AB.

2. With A as a centre and any radius, let us draw an

arc cutting

AB

at

D

.

3. With

D

as a centre and the same radius, let us

mark a point

E

on the arc.(By construction

60BAE

)

4. With

D

and

E

as centres, let us draw two arcs

intersecting each other with the same radius at

F

(By construction

DAF EAF

)

5. Let us join

A

and

F

to form a ray

AF

making an

angle 30° with

AB

.

6. With D and G as centres, let us draw two arcs

intersecting each other with the same radius at

H

(By construction

DAH FAH

) Let us join A

and

G

to form a ray

AG

that makes an angle15°

with

AB

Question: 4

Construct the following angles and verify by

measuring them by a protractor:

(i) 75°

(ii) 105°

(iii) 135°

Solution

(i) 75°

Steps of constructions:

1. Let us draw a ray

OX

.

2. With

O

as a centre and any radius, let us draw an

arc ABC cutting

OX

at

A

.

3. With

B

as a centre and the same radius, let us

mark a point

B

on the arc

ABC

.(By construction

∠AOB = 60°)

4. With B as a centre and the same radius, let us

mark a point C on the arc ABC. (By construction

60BOC

5. With

B

and

C

as centres, let us draw two arcs

intersecting each other with the same radius at

D

.(By construction

BOD COD

6. Let us join

O

and

D

to form a ray

OD

.

7.

AOD

90°

8. With

B

and

D

as centres, let us draw two arcs

intersecting each other with the same radius at

E

.

9.

AOE

=75°

.

(ii) 105°

Steps of constructions:

1. Let us draw a ray

OX

.

2. With O as a centre and any radius, let us draw an

arc ABC cutting

OX

at

A

.

3. With

B

as a centre and the same radius, let us

mark a point B on the arc

ABC

. (By construction

60AOB

)

4. With B as a centre and the same radius, let us

mark a point C on the arc

ABC

. (By construction

60BOC

)

5. With

B

and

C

as centres, let us draw two arcs

intersecting each other with the same radius at

D

.(By construction

BOD COD

)

6. Let us join

O

and

D

to form a ray

OD

.

7.

AOD=

90°

8. With

B

and

C

as centres, let us draw two arcs

intersecting each other with the same radius at

E

.

9.

AOE

=105°

(iii) 135°

Steps of constructions:

1. Let us draw a ray

OX

.

2. With

O

as a centre and any radius, let us draw

an arc ABC cutting

OX

at

A

and

E

.

3. With

B

as a centre and the same radius, let us

mark a point

B

on the arc

ABC

. (By construction

60AOB

)

4. With

B

as a centre and the same radius, let us

mark a point C on the arc

ABC

. (By

construction

60BOC

)

5. With

B

and

C

as centres, let us draw two arcs

intersecting each other with the same radius at

D

.(By construction

BOD COD

)

6. Let us join

O

and

D

to form a ray

OD

.

7.

AOD

= 90°

8. With

D

and

E

as centres, let us draw two arcs

intersecting each other with the same radius at

F

.

9.

AOF

=135°

Question: 5

Construct an equilateral triangle, given its side and

justify the construction.

Solution

Let us consider a ray

AB

with initial point

A

, as

shown below.

We want to construct a ray AC such that

60 .CAB

Steps of Construction:

1. Taking

A

as centre and any radius draw an arc

that intersects

AB

, say at point

D

.

2. Taking

D

as centre and with the same radius as

before, draw an arc intersecting the previously

drawn arc, say at point

E

.

3. Draw a ray from

A

passing through

E

. Then

CAB

is the required angle of 60°.

Proof that this method gives the required angle

of 60°:

Join

DE

.

So,

AE

=

AD

=

DE

(Same radius)

This means Δ

EAD

is an equilateral triangle and

therefore

EAD

is equal to 60°. Now

60 .CAB EAD

Hence proved.

Exercise 11.2 (5)

Question: 1

Construct a triangle ABC in which

BC

= 7 cm,

75B

and

AB

+

AC

= 13 cm.

Solution:

Steps of Construction:

1. Let us draw a line segment BC of 7 cm.

2. At point

B

, let’s construct an angle

75 .XBC

3. Let us cut the line segment

BD

13 cm on

.BX AB AC

4. Let us join DC and make

.DCY BDC

5. Let

CY

intersect

BX

at

A

.

Thus, Δ

ABC

is the required triangle.

Question: 2

Construct a triangle

ABC

in which

BC

= 8

cm,

45B

and AB

AC = 3.5 cm.

Solution:

Steps of Construction:

1. Let us draw a line segment

BC

= 8 cm .

2. At point B, let’s make an angle of 45° i.e.

XBC

.

3. Let us cut the line segment

BD

= 3.5 cm (equal to

AB

AC

) on ray

BX

.

4. Let’s join

DC

and draw the perpendicular bisector

of

DC

.

5. Let it intersect

BX

at point

A

.

6. Let us join

AC

.

Thus, Δ

ABC

is the required triangle.

Question: 3

Construct a triangle

PQR

in which

QR

= 6cm,

60Q

and

PR

–

PQ

= 2cm.

Solution:

Steps of Construction:

1. Let’s draw a ray

QX

and cut off a line segment

QR

= 6 cm from it.

2. Let us construct a ray

QX

making an angle of 60º

with QR and produce XQ is produced it to

D

such

that

QD

=

PR – PQ

= 2cm.

3. Let us draw a perpendicular bisector of

DS

intersecting

QX

at a point

P

. and join

PR

is joined.

Thus, Δ

PQR

is the required triangle.

Question: 4

Construct a triangle XYZ in which

30 , 90Y Z

and

XY

+

YZ

+

ZX

= 11 cm.

Solution:

Steps of Construction:

1. Let us draw a line segment

YZ

= 11 cm. (

XY

+

YZ

+

ZX

= 11 cm).

2. We will draw an angle,

30ZYM

at point A

and an angle

90YZN

at point B.

3. We will bisect

XYM

and

YZN

. The bisectors of

these angles intersect each other at point X.

4. We will construct the perpendicular bisectors

PQ

of

XY

and

RS

of

XZ

intersecting the segment

YZ

at

Y

and

Z

respectively.

5. Thus, Δ

XYZ

is the required triangle.

Question: 5

Construct a right triangle whose base is 12 cm and

sum of its hypotenuse and other side is 18 cm.

Solution:

Steps of Construction:

1. We draw a ray BX and a cut off a line segment BC

= 12 cm on it.

2. We construct

90 .BCY

3. We cut off a line segment

CD

= 18 cm on

CY

.

4. We join.

BD

5. We make

ABD

equal to

BDC

Thus, Δ

ABC

is the required triangle.