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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 $$90^{\circ}$$ are:
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$$60^{\circ}$$
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$$60^{\circ}$$ and $$120^{\circ}$$
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$$120^{\circ}$$ and $$180^{\circ}$$
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$$0^{\circ}$$ and $$60^{\circ}$$
Explanation
Angles to be bisected to obtain an angle of $$ {90}^\circ$$ are $$ {60}^\circ $$ and $$ {120}^\circ $$ as it exactly lies between these two angles.
$$\dfrac{60^\circ+ 120^\circ}2 = 90^\circ$$
Hence, option $$B$$.
Rearrange the following steps of constructing a triangle when the base angle say $$\angle B \,\, and \,\, \angle C$$ and its perimeter $$BC + CA + AB$$ is given:
$$1.$$ Draw perpendicular bisectors $$PQ$$ of $$AX$$ and $$RS$$ of $$AY$$.
$$2.$$ Draw a line segment, say $$XY$$ equal to $$BC + CA +AB$$.
$$$$ Let $$PQ$$ intersect $$XY$$ at $$B$$ and $$RS$$ intersect $$XY$$ at $$C$$. Join $$A-B$$ and $$A-C$$.
$$4.$$ Make $$\angle LXY$$ equal to $$\angle B$$ and $$\angle MYX$$ equal to $$\angle C$$.
$$5.$$ Bisect $$\angle LXY$$ and $$\angle MYX$$. Let these bisectors intersect at a point $$A$$.
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$$1\rightarrow 3\rightarrow 5\rightarrow 4\rightarrow 2$$
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$$2\rightarrow 4\rightarrow 5\rightarrow 1\rightarrow 3$$
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$$5\rightarrow 4\rightarrow 3\rightarrow 2\rightarrow 1$$
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$$2\rightarrow 3\rightarrow 5\rightarrow 4\rightarrow 1$$
Explanation
Steps of constructing a triangle with given conditions are
$$i)$$ Draw a line segment, say $$XY$$ equal to $$BC + CA +AB$$.
$$ii)$$ Make $$\angle LXY$$ equal to $$\angle B$$ and $$\angle MYX$$ equal to $$\angle C$$.
$$iii)$$ Bisect $$\angle LXY$$ and $$\angle MYX$$. Let these bisectors intersect at a point $$A$$.
$$iv)$$ Draw perpendicular bisectors $$PQ$$ of $$AX$$ and $$RS$$ of $$AY$$.
$$v)$$ Let $$PQ$$ intersect $$XY$$ at $$B$$ and $$RS$$ intersect $$XY$$ at $$C$$. Join $$A-B$$ and $$A-C$$.
So, the correct sequence of given steps is
2→4→5→1→3
Option B is correct.
An angle which can be constructed using a pair of compass and ruler is
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$$20^{\circ}$$
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$$80^{\circ}$$
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$$60^{\circ}$$
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$$110^{\circ}$$
Explanation
An angle which can be constructed using a pair of compass and ruler is $$ {60}^{o} $$ as multiples of $${15}^{o} $$ can be drawn using a compass.
To draw an angle of $$150^{\circ}$$ using a pair of compass and ruler
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bisect $$120^{\circ}$$ and $$180^{\circ}$$
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bisect $$60^{\circ}$$ and $$120^{\circ}$$
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bisect $$0^{\circ}$$ and $$60^{\circ}$$
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None
A $$\triangle ABC$$ in which
$$AB= 5.4\ \text{cm}, \angle CAB= 45^{\circ}$$ and $$AC + BC= 9\ \text{cm}$$.
Then, perimeter of $$\Delta ABC$$ is
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$$14.4\ \text{cm}$$
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$$11.4\ \text{cm}$$
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$$12.4\ \text{cm}$$
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$$15.4\ \text{cm}$$
Explanation
$$\Rightarrow$$ Draw line segment $$AB=5.4\,cm$$
$$\Rightarrow$$ Take $$A$$ as center and draw angle of $$45^o$$
$$\Rightarrow$$ Cut off $$AD=9\,cm$$
$$\Rightarrow$$ Join $$BD$$
$$\Rightarrow$$ Now, draw perpendicular bisector of $$BD$$ which
intersect $$AD$$ at point $$C$$.
$$\Rightarrow$$ Now join $$CB$$
$$\Rightarrow$$ Measure the length of $$AC$$ and $$BC$$
$$\Rightarrow$$ We get $$AC=5\,cm$$ and $$BC=4\,cm$$
$$\Rightarrow$$ Perimeter of $$\triangle ABC=5.4+4+5=14.4\,cm$$
The value of $$\angle BAC$$ is
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$$35^\circ$$
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$$65^\circ$$
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$$75^\circ$$
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$$85^\circ$$
Explanation
$$\Rightarrow$$ Draw a line segment $$PQ = AB + BC + CA =9.8\,cm.$$
$$\Rightarrow$$ Make $$\angle LPQ= 45^o$$ and $$\angle MQP = 60^o$$
$$\Rightarrow$$ Now bisect $$\angle LPQ$$ and $$\angle MQP$$ such that their bisectors meet at $$A.$$
$$\Rightarrow$$ Draw perpendicular bisector of $$AP$$ and $$AQ$$ i.e.
$$XY$$ and $$ST$$ intersecting $$PQ$$ at $$B$$ and $$C$$ respectively.
$$\Rightarrow$$ Join $$AB$$ and $$AC$$
$$\Rightarrow$$ Now, measure the $$\angle BAC$$
$$\Rightarrow$$ We get, $$\angle BAC=75^o$$.
Then, length (in $$cm$$) of each of the sides of the triangle is
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$$3, 4, 4$$
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$$2.7, 3.7, 3.4$$
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$$2.3, 4.2, 3.6$$
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$$4, 1.8, 4$$
Explanation
$$\Rightarrow$$ Draw a line segment $$PQ = AB + BC + CA =9.8\,cm.$$
$$\Rightarrow$$ Make $$\angle LPQ= 45^o$$ and $$\angle MQP = 60^o$$
$$\Rightarrow$$ Now bisect $$\angle LPQ$$ and $$\angle MQP$$ such that their bisectors meet at $$A.$$
$$\Rightarrow$$ Draw perpendicular bisector of $$AP$$ and $$AQ$$ i.e.
$$XY$$ and $$ST$$ intersecting $$PQ$$ at $$B$$ and $$C$$ respectively.
$$\Rightarrow$$ Join $$AB$$ and $$AC$$
$$\Rightarrow$$ Now, measure the length of $$\triangle ABC$$
$$\Rightarrow$$ We get, $$AB=3.4\,cm,\,BC=3.7\,cm,\,AC=2.7\,cm$$
Given an angle $$\theta$$, which of the following angles cannot be obtained by using the method of construction of angle bisectors?
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$$\dfrac{\theta}2$$
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$$\dfrac{\theta}4$$
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$$\dfrac{\theta}6$$
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$$\dfrac{\theta}8$$
Explanation
Angle bisector is bisects or divide the angle into two equal parts.
If $$\theta$$ is given, we can find $$\dfrac{\theta}{2}$$ by bisecting $$\theta$$ into two parts and $$\dfrac{\theta}{2}$$ will be half of $$\theta$$.
By construting angle bisector of angle $$\dfrac{\theta}{2}$$, we can get $$\dfrac{\theta}{4}$$ which is half of
$$\dfrac{\theta}{2}$$.
And then again bisecting
$$\dfrac{\theta}{4}$$, we can get the angle
$$\dfrac{\theta}{8}$$ which is half of
$$\dfrac{\theta}{4}$$.
Hence, by using method of construction of angle
bisectors we can find $$\dfrac { \theta }{ 2 } , \dfrac { \theta }{ 4 } ,\dfrac { \theta }{8} $$
but not $$\dfrac { \theta }{ 6 } $$ as it is half of
$$\dfrac{\theta}{3}$$ which can't be determined by angle bisector.
Hence, the answer is $$\dfrac{\theta}{6}$$.
How many angle bisectors need to be drawn in the steps of construction of an angle $$60^\circ$$?
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$$0$$
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$$1$$
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$$2$$
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$$3$$
For constructing a triangle whose perimeter and both base angles are given, the first step is to:
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Draw a base of any length
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Draw the base of length $$=$$ perimeter
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Draw the base angles from a random line.
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Draw a base of length $$=\dfrac13 \times$$ perimeter.
Explanation
Below are the steps for constructing a triangle whose perimeter and base angles are given.
Step 1 : Draw a line segment/base equal to perimeter
Step 2 : From any point X draw ray at one of the given base angles.
From any point Y draw ray at second of the given base angles
Step 3 : draw angle bisector of X and Y, two angle bisectors intersect each other at point A
Step 4 : Draw line bisector of XA and AY respectively these two line bisectors intersect XY at point B and C
Step 5 : Join A to B and A to C
Step 6 : Triangle ABC is the required triangle.
Hence, the answer is 'Draw the base of length $$=$$ perimeter'.
You are asked to "construct" an angle of $$90^\circ$$. Which of the following methods is considered appropriate for construction
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Using a compass and straightedge, copy an angle that appears to be close to $$90^\circ$$ from a diagram in your textbook.
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Using a compass and straightedge, construct two parallel lines and label on of the angles $$90^\circ$$.
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Using a compass and straightedge, construct angles of $$60^\circ$$ and $$120^\circ$$ and bisect the angle between them.
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Using a protractor, draw an angle of $$90^\circ$$
Explanation
An angle is drawn most appropriate using a compass.
Now bisector of $$60^{\circ}$$ and $$120^{\circ}$$ gives an angle of $$90^{\circ}$$
So option $$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^\circ$$?
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$$0$$
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$$1$$
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$$2$$
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$$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$$
A triangle
ABC
can be constructed in which$$\angle B =60^{o}$$, $$\angle C= 45^{o}$$ and
AB +BC + AC = 11 cm
. Is this Statement true?
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True
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False
Explanation
$$\because$$ Sum of angles of a triangle=$${ 180 }^{ \circ }$$
$$\implies \angle BAC=180-(60+45)={ 75 }^{ \circ }$$
Now, to check the statement.
i)Draw a line segment XY=11cm.
ii)Make an angle equal to $$\angle B={ 60 }$$ from point X and mark L on this line.
iii)Make an angle equal to $$\angle C={ 45 }^{ \circ }$$ from point Y and mark M on this line.
iv)Bisect $$\angle LXY and \angle MYX $$,l
et the bisectors meet at point A.
v)Make perpendicular bisector of AX,let it intersect XY at B.
vi)Make perpendicular bisector of AY,let it intersect XY at point C.
vii)Join AB & AC ,Then measure AB+BC+AC.
If that is equal to 11cm.
The statement is true and ABC is the required triangle.
The diagram represents the construction of
triangle ABC with which of the following dimensions?
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$$ \displaystyle BC=4.8 $$ $$cm$$, $$ \displaystyle \angle B=60^{\circ} $$ and $$ \displaystyle AB-AC=12\ cm $$.
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$$ \displaystyle BC=4.8 $$ $$cm$$, $$ \displaystyle \angle B=30^{\circ} $$ and $$ \displaystyle AB-AC=15\ cm $$.
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$$ \displaystyle BC=4.8 $$ $$cm$$, $$ \displaystyle \angle B=30^{\circ} $$ and $$ \displaystyle AB-AC=12\ cm $$.
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None of these
Construct a $$\triangle ABC$$ in which:
$$AB= 5.4\ cm$$, $$\angle CAB= 45^{0}$$ and $$AC\, +\, BC= 9\ cm$$.
Then the length of $$AC$$ (in $$cm.$$) is:
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$$4$$
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$$7$$
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$$5$$
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None of these
In a $$\triangle ABC$$ in which $$AC= 5\ cm$$ and $$\angle BAC= 60^{\circ}$$ and $$BC - AB= 1.2\ cm$$. The, $$AB$$ is
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$$3.18$$
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$$4.32$$
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$$5.12$$
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None of these
Explanation
Given : $$AC=5cm$$
$$\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$$ degree line.
Draw an arc that intersects perpendicular line.
Which step comes last ?
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$$2$$
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$$3$$
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$$4$$
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$$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 $$and AB +
AC =14 cm, then the length of BD used for construction
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$$7$$ cm
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$$14$$ cm
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$$20$$ cm
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$$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} $$ can be constructed .
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True
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False
Explanation
Since , $$ 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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True
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False
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