 Lesson: Constructions
Exercise 11.1 (3 Multiple Choice Questions and
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°
(c) 22.5°
(d) 67.5° Solution:
b
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
AB
and
AC
is equal to: (a) 6.9 cm
(b) 5.2 cm
(c) 5.0 cm
(d) 4.0 cm
Solution:
a
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
AB
and
AC
is equal to:
(a) 3.2 cm
(b) 3.1 cm
(c) 3 cm
(d) 2.8 cm
Solution:
d
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
(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
and
12AB BC AC
cm.
Solution:
1. True.
52.5° =
210
4
210 180 30 .
180 30
can be constructed.
2. False.
1
85 .
2
42.5
85° cannot be constructed.
3. False.
As
BC AC
must be greater than
AB
.
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
55
DAG BAG
Question: 2
Draw a line segment
AB
of 4 cm in length. Draw a
line perpendicular to
AB
through
A
and
B
,
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.
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.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 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.
cm
45
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:
As
B
lies on the perpendicular bisector of
AX,
Therefore,
.XB AB
Also,
C
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
(A
s required).
Therefore, the construction is justified.
2.
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
(
AM
is the perpendicular
bisector)
DM MR
((
PM
is the perpendicular bisector)
Therefore, the triangles
PDM
and
PRM
are congruent
by
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. cm
cm
3.5
5.5
AB
BY
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.
In an equilateral triangle all sides are equal and each
internal angle is 60°.
So, at point
D
, we construct a perpendicular. We cut
an arc equal to 3.2 cm on
PD
.
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
AC
and draw its
perpendicular bisector.
We cut an arc of 3 cm from
O
on both the sides of
the perpendicular bisector.
6BD
cm and perpendicular to
AC
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.