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CBSE Questions for Class 9 Maths Constructions Quiz 2 - MCQExams.com
CBSE
Class 9 Maths
Constructions
Quiz 2
Angles to be bisected to obtain an angle of
90
∘
are:
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0%
60
∘
0%
60
∘
and
120
∘
0%
120
∘
and
180
∘
0%
0
∘
and
60
∘
Explanation
Angles to be bisected to obtain an angle of
90
∘
are
60
∘
and
120
∘
as it exactly lies between these two angles.
60
∘
+
120
∘
2
=
90
∘
Hence, option
B
.
Rearrange the following steps of constructing a triangle when the base angle say
∠
B
a
n
d
∠
C
and its perimeter
B
C
+
C
A
+
A
B
is given:
1.
Draw perpendicular bisectors
P
Q
of
A
X
and
R
S
of
A
Y
.
2.
Draw a line segment, say
X
Y
equal to
B
C
+
C
A
+
A
B
.
Let
P
Q
intersect
X
Y
at
B
and
R
S
intersect
X
Y
at
C
. Join
A
−
B
and
A
−
C
.
4.
Make
∠
L
X
Y
equal to
∠
B
and
∠
M
Y
X
equal to
∠
C
.
5.
Bisect
∠
L
X
Y
and
∠
M
Y
X
. Let these bisectors intersect at a point
A
.
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0%
1
→
3
→
5
→
4
→
2
0%
2
→
4
→
5
→
1
→
3
0%
5
→
4
→
3
→
2
→
1
0%
2
→
3
→
5
→
4
→
1
Explanation
Steps of constructing a triangle with given conditions are
i
)
Draw a line segment, say
X
Y
equal to
B
C
+
C
A
+
A
B
.
i
i
)
Make
∠
L
X
Y
equal to
∠
B
and
∠
M
Y
X
equal to
∠
C
.
i
i
i
)
Bisect
∠
L
X
Y
and
∠
M
Y
X
. Let these bisectors intersect at a point
A
.
i
v
)
Draw perpendicular bisectors
P
Q
of
A
X
and
R
S
of
A
Y
.
v
)
Let
P
Q
intersect
X
Y
at
B
and
R
S
intersect
X
Y
at
C
. Join
A
−
B
and
A
−
C
.
So, the correct sequence of given steps is
2→4→5→1→3
Option B is correct.
An angle which can be constructed using a pair of compass and ruler is
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0%
20
∘
0%
80
∘
0%
60
∘
0%
110
∘
Explanation
An angle which can be constructed using a pair of compass and ruler is
60
o
as multiples of
15
o
can be drawn using a compass.
To draw an angle of
150
∘
using a pair of compass and ruler
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0%
bisect
120
∘
and
180
∘
0%
bisect
60
∘
and
120
∘
0%
bisect
0
∘
and
60
∘
0%
None
A
△
A
B
C
in which
A
B
=
5.4
cm
,
∠
C
A
B
=
45
∘
and
A
C
+
B
C
=
9
cm
.
Then, perimeter of
Δ
A
B
C
is
Report Question
0%
14.4
cm
0%
11.4
cm
0%
12.4
cm
0%
15.4
cm
Explanation
⇒
Draw line segment
A
B
=
5.4
c
m
⇒
Take
A
as center and draw angle of
45
o
⇒
Cut off
A
D
=
9
c
m
⇒
Join
B
D
⇒
Now, draw perpendicular bisector of
B
D
which
intersect
A
D
at point
C
.
⇒
Now join
C
B
⇒
Measure the length of
A
C
and
B
C
⇒
We get
A
C
=
5
c
m
and
B
C
=
4
c
m
⇒
Perimeter of
△
A
B
C
=
5.4
+
4
+
5
=
14.4
c
m
The value of
∠
B
A
C
is
Report Question
0%
35
∘
0%
65
∘
0%
75
∘
0%
85
∘
Explanation
⇒
Draw a line segment
P
Q
=
A
B
+
B
C
+
C
A
=
9.8
c
m
.
⇒
Make
∠
L
P
Q
=
45
o
and
∠
M
Q
P
=
60
o
⇒
Now bisect
∠
L
P
Q
and
∠
M
Q
P
such that their bisectors meet at
A
.
⇒
Draw perpendicular bisector of
A
P
and
A
Q
i.e.
X
Y
and
S
T
intersecting
P
Q
at
B
and
C
respectively.
⇒
Join
A
B
and
A
C
⇒
Now, measure the
∠
B
A
C
⇒
We get,
∠
B
A
C
=
75
o
.
Then, length (in
c
m
) of each of the sides of the triangle is
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0%
3
,
4
,
4
0%
2.7
,
3.7
,
3.4
0%
2.3
,
4.2
,
3.6
0%
4
,
1.8
,
4
Explanation
⇒
Draw a line segment
P
Q
=
A
B
+
B
C
+
C
A
=
9.8
c
m
.
⇒
Make
∠
L
P
Q
=
45
o
and
∠
M
Q
P
=
60
o
⇒
Now bisect
∠
L
P
Q
and
∠
M
Q
P
such that their bisectors meet at
A
.
⇒
Draw perpendicular bisector of
A
P
and
A
Q
i.e.
X
Y
and
S
T
intersecting
P
Q
at
B
and
C
respectively.
⇒
Join
A
B
and
A
C
⇒
Now, measure the length of
△
A
B
C
⇒
We get,
A
B
=
3.4
c
m
,
B
C
=
3.7
c
m
,
A
C
=
2.7
c
m
Given an angle
θ
, which of the following angles cannot be obtained by using the method of construction of angle bisectors?
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0%
θ
2
0%
θ
4
0%
θ
6
0%
θ
8
Explanation
Angle bisector is bisects or divide the angle into two equal parts.
If
θ
is given, we can find
θ
2
by bisecting
θ
into two parts and
θ
2
will be half of
θ
.
By construting angle bisector of angle
θ
2
, we can get
θ
4
which is half of
θ
2
.
And then again bisecting
θ
4
, we can get the angle
θ
8
which is half of
θ
4
.
Hence, by using method of construction of angle
bisectors we can find
θ
2
,
θ
4
,
θ
8
but not
θ
6
as it is half of
θ
3
which can't be determined by angle bisector.
Hence, the answer is
θ
6
.
How many angle bisectors need to be drawn in the steps of construction of an angle
60
∘
?
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0%
0
0%
1
0%
2
0%
3
For constructing a triangle whose perimeter and both base angles are given, the first step is to:
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0%
Draw a base of any length
0%
Draw the base of length
=
perimeter
0%
Draw the base angles from a random line.
0%
Draw a base of length
=
1
3
×
perimeter.
Explanation
Below are the steps for constructing a triangle whose perimeter and base angles are given.
Step 1 : Draw a line segment/base equal to perimeter
Step 2 : From any point X draw ray at one of the given base angles.
From any point Y draw ray at second of the given base angles
Step 3 : draw angle bisector of X and Y, two angle bisectors intersect each other at point A
Step 4 : Draw line bisector of XA and AY respectively these two line bisectors intersect XY at point B and C
Step 5 : Join A to B and A to C
Step 6 : Triangle ABC is the required triangle.
Hence, the answer is 'Draw the base of length
=
perimeter'.
You are asked to "construct" an angle of
90
∘
. Which of the following methods is considered appropriate for construction
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0%
Using a compass and straightedge, copy an angle that appears to be close to
90
∘
from a diagram in your textbook.
0%
Using a compass and straightedge, construct two parallel lines and label on of the angles
90
∘
.
0%
Using a compass and straightedge, construct angles of
60
∘
and
120
∘
and bisect the angle between them.
0%
Using a protractor, draw an angle of
90
∘
Explanation
An angle is drawn most appropriate using a compass.
Now bisector of
60
∘
and
120
∘
gives an angle of
90
∘
So option
C
is correct.
It can also be drawn using protractor but less accurate then compass.
How many angle bisectors need to be drawn in the steps of construction of an angle
45
∘
?
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
Steps for construction:
1
)
Construct a perpendicular line.
2
)
Place a compass on a intersection line.
3
)
Adjust compass width to reach start point
4
)
Draw an arc that intersects perpendicular line.
5
)
Place a ruler on a start point and where arc intersects perpendicular line.
6
)
Draw a 45 degree line
So only
1
angle bisector is need to be drawn in the steps of construction of an angle
45
o
A triangle
ABC
can be constructed in which
∠
B
=
60
o
,
∠
C
=
45
o
and
AB +BC + AC = 11 cm
. Is this Statement true?
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0%
True
0%
False
Explanation
∵
Sum of angles of a triangle=
180
∘
⟹
∠
B
A
C
=
180
−
(
60
+
45
)
=
75
∘
Now, to check the statement.
i)Draw a line segment XY=11cm.
ii)Make an angle equal to
∠
B
=
60
from point X and mark L on this line.
iii)Make an angle equal to
∠
C
=
45
∘
from point Y and mark M on this line.
iv)Bisect
∠
L
X
Y
a
n
d
∠
M
Y
X
,l
et the bisectors meet at point A.
v)Make perpendicular bisector of AX,let it intersect XY at B.
vi)Make perpendicular bisector of AY,let it intersect XY at point C.
vii)Join AB & AC ,Then measure AB+BC+AC.
If that is equal to 11cm.
The statement is true and ABC is the required triangle.
The diagram represents the construction of
triangle ABC with which of the following dimensions?
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0%
B
C
=
4.8
c
m
,
∠
B
=
60
∘
and
A
B
−
A
C
=
12
c
m
.
0%
B
C
=
4.8
c
m
,
∠
B
=
30
∘
and
A
B
−
A
C
=
15
c
m
.
0%
B
C
=
4.8
c
m
,
∠
B
=
30
∘
and
A
B
−
A
C
=
12
c
m
.
0%
None of these
Construct a
△
A
B
C
in which:
A
B
=
5.4
c
m
,
∠
C
A
B
=
45
0
and
A
C
+
B
C
=
9
c
m
.
Then the length of
A
C
(in
c
m
.
) is:
Report Question
0%
4
0%
7
0%
5
0%
None of these
In a
△
A
B
C
in which
A
C
=
5
c
m
and
∠
B
A
C
=
60
∘
and
B
C
−
A
B
=
1.2
c
m
. The,
A
B
is
Report Question
0%
3.18
0%
4.32
0%
5.12
0%
None of these
Explanation
Given :
A
C
=
5
c
m
\angle BAC=60°
BC-AB=1.2cm
BC=1.2+AB
\Longrightarrow \cos 60°=\cfrac{{AC}^{2}+{AB}^{2}-{BC}^{2}}{2AC\times AB}
\Longrightarrow \cfrac{1}{2}=\cfrac{25+{x}^{2}-{(1.2+x)}^{2}}{2\times 5\times x}
\Longrightarrow 5x=25+{x}^{2}-1.44-{x}^{2}-2.4x
\Longrightarrow 7.4x=23.56
\Longrightarrow x=3.18
\therefore AB=3.18, BC=4.38
The steps of construction of an
\angle AOB=45^{o}
is given in jumbled order below:
1. Place compass on intersection point.
Place ruler on start point and where arc intersects perpendicular line.
Adjust compass width to reach start point.
Construct a perpendicular line.
Draw
45
degree line.
Draw an arc that intersects perpendicular line.
Which step comes last ?
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0%
2
0%
3
0%
4
0%
5
Explanation
Correct sequence is :
1. Construct a perpendicular line .
2. Draw an arc that intersect the perpendicular line.
3. Adjust the compass width to reach the start point .
4.Place compass on intersection point.
5. Place ruler on start point and where the arc intersects the perpendicular line.
6. Draw
45
degree line.
So the last step is
5
Option
D
is correct.
To construct a
\triangle
ABC in which BC =
10
cm and B=
60^o
and AB +
AC =14 cm, then the length of BD used for construction
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0%
7
cm
0%
14
cm
0%
20
cm
0%
10
cm
Explanation
Construction:
A]. Draw the base BC and at the point, B makes an angle, say
XBC
equal to the given angle.
B]. Cut a line segment
BD
equal to
AB + AC
from the ray
BX
.
C]. Join
DC
and make an angle
DCY
equal to
BDC
.
D]. Let
CY
intersect
BX
at
A
E].
ABC
is the required triangle.
F]. In triangle
\Delta ACD
,
∠ACD = ∠ ADC
.
G]. So,
AB = BD – AD = BD – AC
H].
AB + AC = BD
From this we get,
BD=14cm
.
Write True or False in each of the following . Give reason for your answer:
An angle of
42.5^{\circ}
can be constructed .
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0%
True
0%
False
Explanation
Since ,
42.5^{\circ} = \dfrac{1}{2} \times 85^{\circ}
and
85^{\circ}
cannot be constructed by using ruler and compass . Also
42.5^{\circ}
is not multiple of
3
. Therefore , an angle of
42.5^{\circ}
can not be constructed.
The bisector of
90^0
is
40^0
. True or false
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0%
True
0%
False
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