Lesson: Constructions

Exercise 11.1 (3 Multiple Choice Questions and

Answers)

Question: 1

With the help of a ruler and a compass it is not

possible to construct an angle of:

(a) 37.5°

(b) 40°

(d) 67.5°

Solution:

Using a ruler and a compass, it is possible to

construct 75°. On constructing the angle bisector of

75°, we get the angle of 37.5°.

Using a ruler and a compass, it is possible to

construct

45

. On constructing the angle bisector of

45°, we get the angle of 22.5°.

Using a ruler and a compass, it is possible to

construct 90° and 45°. Thus, 135° can be constructed.

On constructing the angle bisector of 135°, we get

the angle of 67.5°.

We however cannot construct 40° using the same

logic as discussed for 37.5°, 22.5° and 67.5°.

Question: 2

The construction of a triangle

ABC

, given that

6BC 

cm,

45B  

is not possible when the

difference of

and

is equal to:

(a) 6.9 cm

(b) 5.2 cm

(d) 4.0 cm

Solution:

It is not possible to construct a triangle when the

difference of the two sides of a triangle is not less

than or is equal to the third side.

Here the difference of two sides, as given, is 6.9 cm. It

is more than the third side which is 6 cm.

Question: 3

The construction of a triangle

ABC

, given that

cm,3BC 

60C  

is possible when difference of

and

is equal to:

(a) 3.2 cm

(b) 3.1 cm

(d) 2.8 cm

Solution:

It is possible to construct a triangle when the

difference of the two sides of a triangle is less than

the third side. Here the difference of two sides, as

given, is 2.8 cm. It is less than the third side which is

3 cm.

Exercise 11.2

Question: 1

Write True or False in each of the following. Give

reasons for your answer:

(1) An angle of 52.5° can be constructed.

(2) An angle of 42.5° can be constructed.

(3) A triangle

ABC

can be constructed in which

cm,5AB 

45A  

and

cm.5BC AC 

(4) A triangle

ABC

can be constructed in which

cm,6BC 

30C  

and

cm.4AC AB 

(5) A triangle

ABC

can be constructed in which

105 ,B  

90C  

and

10AB BC AC  

cm.

(6) A triangle

ABC

can be constructed in which

60 , 45B C     

and

12AB BC AC  

cm.

Solution:

1. True.

52.5° =

210



210 180 30 .    

180 30  

can be constructed.

2. False.

85 .

42.5   

85° cannot be constructed.

3. False.

BC AC

must be greater than

4. True.

AC AB BC 

.AC AB BC  

So, the triangle

ABC

can be constructed.

5. False.

105 90 195 180 .B C          

The sum of the three angles of a triangle is equal to

180°.

Here, the sum of the two given angles is more than

180°.

So, the triangle

ABC

is not possible.

6. True.

60 45 105 180 .B C          

So, the triangle can be constructed.

Exercise 11.3

Question: 1

Draw an angle of 110° with the help of a protractor

and bisect it. Measure each angle.

Solution:

110

BAD

DAG BAG

  

    

Question: 2

Draw a line segment

of 4 cm in length. Draw a

line perpendicular to

through

and

respectively. Are these lines parallel?

Solution:

Yes, these lines are parallel.

Question: 3

Draw an angle of 80° with the help of a protractor.

Then construct angles of

(i) 40°

(ii) 160°

(iii) 120°

Solution:

80BAC  

(i) To draw 40°, we will construct the angle bisector

of 80°.

is the required angle.DAB

(ii) To draw 160°, we will construct

is the required angle.

80BAC CAD

DAB

    



(iii) To draw 120°, we will construct the angle

bisector of

.DAC

is the required angle.EAB

Question: 4

Construct a triangle whose sides are 3.6 cm, 3.0 cm

and 4.8 cm.

Bisect the smallest angle and measure each part.

Solution:

The angle opposite to the smallest side is the

smallest angle of a traingle.

We have bisected the smalest angle

.DAB

Question: 5

Construct a triangle

ABC

in which

cm,5BC 

60B  

and

cm.7.5AC AB 

Solution:

60B  

5BC 

cm.5AD 

cm.7.5AC AB 

Question: 6

Construct a square of side 3 cm.

Solution:

cm3AB BC CD AC   

Question: 7

Construct a rectangle whose adjacent sides are of

lengths 5 cm and 3.5 cm.

Solution:

cm3.5AB CD 

cm5BC AD 

Question: 8

Construct a rhombus whose side is of length 3.4 cm

and one of its angles is 45°.

Solution:

3.4

is a rhombus.

ABCD

A  

   AB BC CD DA

Exercise 11.4

Question: 1

Construct each of the following and give justification:

(1) A triangle if its perimeter is 10.4 cm and two

angles are 45° and 120°.

(2) A triangle

PQR

given that

cm,3QR

45PQR  

and

2QP PR 

cm.

(3) A right triangle when one side is 3.5 cm and sum

of other sides and the hypotenuse is 5.5 cm.

(4) An equilateral triangle if its altitude is 3.2 cm.

(5) A rhombus whose diagonals are 4 cm and 6 cm in

lengths.

Solution:

1. A triangle if its perimeter is 10.4 cm and two

angles are 45° and 120°.

cm10XY 

45LXY  

120MYX  

Proof that the given method is justified:

lies on the perpendicular bisector of

AX,

Therefore,

.XB AB

Also,

lies on the perpendicular bisector of

AY.

Therefore,

.CY AC

Now,

. (BC CA AB BC XB CY XY As XB AB      

and

CY AC

)

Again,

 

As in ,BAX AXB AXB AB XB   

ABC BAX AXB 

(Exterior angle of a triangle is

equal to the sum of its interior opposite angles)

2AXB

LXY

Similarly

ACB MYX 

s required).

Therefore, the construction is justified.

cm, 3 45QR PQR   

and

cm2QP PR QD  

Proof that this method gives the required

triangle:

In triangles,

PDM

and

PRM

PM PM

(Common side)



90PMD AMR  

(

is the perpendicular

bisector)

DM MR

((

is the perpendicular bisector)

Therefore, the triangles

PDM

and

PRM

are congruent

SAS

congruency rule.

So,

 

by CPCTPD PR

Now,

.QD PD PQ PQ PR   

Thus, we can justify the construction.

3. A right triangle when one side is 3.5 cm and sum

of other sides and the hypotenuse is 5.5 cm.

3.5

5.5



Proof that this method gives the required

triangle:

In triangle

ACY

ACY AYC

(By construction)

Therefore,

AC AY

(Sides opposite to equal angles

are equal)

Now,

BC BY CY 

 

As BY AC AC CY  

BC AC BY 

4. An equilateral triangle if its altitude is 3.2 cm.

cm3.2AD 

In an equilateral triangle all sides are equal and each

internal angle is 60°.

So, at point

, we construct a perpendicular. We cut

an arc equal to 3.2 cm on

At point A, we draw two angles

DAC

and

BAB

such

that

30 .DAB DAC    

ABC is the required triangle.

( 90 60 , 90 6 )0DBA DAB DCA DAC         

5. A rhombus whose diagonals are 4 cm and 6 cm in

lengths.

cm6BD 

cm4AC 

The construction is done based on the property that

all sides of a rhombus are equal and the diagonals of

a rhombus are perpendicular bisector of one another.

So, we take one of the diagonals as

and draw its

perpendicular bisector.

We cut an arc of 3 cm from

on both the sides of

the perpendicular bisector.

6BD 

cm and perpendicular to

is the other

diagonal.

By SAS rule; ,

( Common angle)

ABO CBO

AO OC AOB BOC OB OB

  

    

 

 

 

In the same way, it can be proved that the triangles

AOB, BOC, COD

and

DOC

are congruent to each

other.

Therefore,

.AB BC CD DA  

Thus, ABCD is the required rhombus.