MCQ Questions for Class 9 Maths Constructions Quiz 2

Angles to be bisected to obtain an angle of

$90^{\circ}$ are:

0%
$60^{\circ}$
0%
$60^{\circ}$ and $120^{\circ}$
0%
$120^{\circ}$ and $180^{\circ}$
0%
$0^{\circ}$ 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

$A$ .

0%
$1\rightarrow 3\rightarrow 5\rightarrow 4\rightarrow 2$
0%
$2\rightarrow 4\rightarrow 5\rightarrow 1\rightarrow 3$
0%
$5\rightarrow 4\rightarrow 3\rightarrow 2\rightarrow 1$
0%
$2\rightarrow 3\rightarrow 5\rightarrow 4\rightarrow 1$

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.

1529653_317208_ans_6715f6c54e874fa3942b6054c0a8f51d.png

An angle which can be constructed using a pair of compass and ruler is

0%
$20^{\circ}$
0%
$80^{\circ}$
0%
$60^{\circ}$
0%
$110^{\circ}$

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

0%
bisect $120^{\circ}$ and $180^{\circ}$
0%
bisect $60^{\circ}$ and $120^{\circ}$
0%
bisect $0^{\circ}$ and $60^{\circ}$
0%
None

$\triangle ABC$ in which

$AB= 5.4\ \text{cm}, \angle CAB= 45^{\circ}$ and

$AC + BC= 9\ \text{cm}$ . Then, perimeter of

$\Delta ABC$ is

0%
$14.4\ \text{cm}$
0%
$11.4\ \text{cm}$
0%
$12.4\ \text{cm}$
0%
$15.4\ \text{cm}$

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

$\triangle ABC=5.4+4+5=14.4\,cm$

1252627_528685_ans_e512701b1d174a0bad961f56b6af8370.png

The value of

$\angle BAC$ is

0%
$35^\circ$
0%
$65^\circ$
0%
$75^\circ$
0%
$85^\circ$

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,

$\angle BAC=75^o$ .

1252626_528738_ans_04c6678278ad4c979e038e418f763715.jpeg

Then, length (in

$cm$ ) of each of the sides of the triangle is

0%
$3, 4, 4$
0%
$2.7, 3.7, 3.4$
0%
$2.3, 4.2, 3.6$
0%
$4, 1.8, 4$

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,

$AB=3.4\,cm,\,BC=3.7\,cm,\,AC=2.7\,cm$

1252625_528734_ans_2bc1997419a445679a376d72ae84e839.jpeg

Given an angle

$\theta$ , which of the following angles cannot be obtained by using the method of construction of angle bisectors?

0%
$\dfrac{\theta}2$
0%
$\dfrac{\theta}4$
0%
$\dfrac{\theta}6$
0%
$\dfrac{\theta}8$

Explanation

Angle bisector is bisects or divide the angle into two equal parts.

$\theta$ 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

$60^\circ$ ?

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:

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 $=\dfrac13 \times$ 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

0%
Using a compass and straightedge, copy an angle that appears to be close to $90^\circ$ from a diagram in your textbook.
0%
Using a compass and straightedge, construct two parallel lines and label on of the angles $90^\circ$ .
0%
Using a compass and straightedge, construct angles of $60^\circ$ and $120^\circ$ and bisect the angle between them.
0%
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

$45^\circ$ ?

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

$\angle B =60^{o}$ ,

$\angle C= 45^{o}$ and AB +BC + AC = 11 cm. Is this Statement true?

0%
True
0%
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

$\angle LXY and \angle MYX$ ,let 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.

1005268_1038238_ans_31dd881cdac9404186752da1d4834712.png

The diagram represents the construction of triangle ABC with which of the following dimensions?

0%
$\displaystyle BC=4.8$ $cm$ , $\displaystyle \angle B=60^{\circ}$ and $\displaystyle AB-AC=12\ cm$ .
0%
$\displaystyle BC=4.8$ $cm$ , $\displaystyle \angle B=30^{\circ}$ and $\displaystyle AB-AC=15\ cm$ .
0%
$\displaystyle BC=4.8$ $cm$ , $\displaystyle \angle B=30^{\circ}$ and $\displaystyle AB-AC=12\ cm$ .
0%
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:

0%
$4$
0%
$7$
0%
$5$
0%
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

0%
$3.18$
0%
$4.32$
0%
$5.12$
0%
None of these

Explanation

Given :

$AC=5cm$

$\angle BAC=60°$

$BC-AB=1.2cm$

$BC=1.2+AB$

$\Longrightarrow \cos 60°=\cfrac{{AC}^{2}+{AB}^{2}-{BC}^{2}}{2AC\times AB}$

$\Longrightarrow \cfrac{1}{2}=\cfrac{25+{x}^{2}-{(1.2+x)}^{2}}{2\times 5\times x}$

$\Longrightarrow 5x=25+{x}^{2}-1.44-{x}^{2}-2.4x$

$\Longrightarrow 7.4x=23.56$

$\Longrightarrow x=3.18$

$\therefore AB=3.18, BC=4.38$

1016149_599639_ans_e0267217fe244a57a0f44828e2d57f87.png

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 ?

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

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,

$BD=14cm$ .

1372789_1042034_ans_86aa874d78294bbb940213112bcddcd4.PNG

Write True or False in each of the following . Give reason for your answer:
An angle of

$42.5^{\circ}$ can be constructed .

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

0%
True
0%
False

CBSE Questions for Class 9 Maths Constructions Quiz 2 - MCQExams.com

0:0:3

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Constructions Quiz 2 - MCQExams.com

0:0:3

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.