MCQ Questions for Class 6 Maths Practical Geometry Quiz 3

Number of set squares in the geometry box is

0%
0
0%
1
0%
2
0%
3

The instrument to draw a circle is

0%
Ruler
0%
Protractor
0%
Driver
0%
Compass

Explanation

Option D is right

The

$\mathrm{compass}$ is used to draw a circle.

State whether the statement are true (T) or false (F).
Two perpendiculars can be drawn to a given line from a point not lying on it.

0%
True
0%
False

State whether the statement are true (T) or false (F).
Infinitely many perpendicular bisectors can be drawn to a given ray.

0%
True
0%
False

State whether the given statement is true (T) or false (F).

With a given center and a given radius, only one circle can be drawn.

0%
True
0%
False

Explanation

With a given center and given radius, only one circle can be drawn.

For example:

We can draw only one circle with center as

$O$ and radius

$r.$

1646638_1790063_ans_ed2bb12f969a487f9f8fcc7602502136.png

Which of the following angles cannot be constructed using a ruler and compass?

0%
$15^\circ$
0%
$45^\circ$
0%
$75^\circ$
0%
$85^\circ$

Write True or False and justify your answer in each of the following :
A circle of radius

$3\,cm$ can be drawn through two points A , B such that AB =

$6\,cm$ .

0%
True
0%
False

Explanation

Radius of circle

$= 3\,cm$

$\therefore \,$ Diameter of circle

$= 2 \times r$

$= 2 \times 3 \,cm$

$= 6\,cm$

Now , AB

$= 6\,cm$ , so the given statement is true because AB will be the diameter.

1660281_1795079_ans_d71819c0e5df490295eb9fed66699620.jpg

The instrument in a geometry box having the shape of a semicircle is called

0%
ruler
0%
divider
0%
compass
0%
protractor

Explanation

Protractor

It is used to draw or measure angles.

Hence, the correct answer is

$Op-D$

2004796_1800650_ans_44dd3e7bf21a4ab7975cc326bb5bc6a1.png

The number of perpendicular bisectors that can be drawn to a given line segment is:

0%
$0$
0%
$1$
0%
$2$
0%
Infinitely Many

Explanation

We know that perpendicular bisector divides the line segment into two equal parts. So, for a given line segment only one perpendicular bisector can exist.

Hence, option

$B$ is correct.

The number of perpendiculars that can be drawn to a line from a point, not on it is

0%
$1$
0%
$2$
0%
$4$
0%
infinitely many

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

$B$ .

Suppose you are given a circle. Give the construction to find its centre.

0%
The center lies on the perpendicular bisectors of any two chords of the circle
0%
The center lies on the mid point of any chord of the circle
0%
The center lies on the bisector of the angle betwee any two chords of the circle
0%
Both options B & C.

Explanation

$Given-\quad \\ A\quad circle.\\ To\quad find\quad out-\\ its\quad cenre\quad by\quad construction.$

$Construction-$

$(I)\quad Any\quad three\quad points\quad A,\quad B\quad \& \quad C\quad are\quad taken\quad on\quad the\quad circumference.$

$(II)\quad AB\quad \& \quad BC\quad are\quad joined.$

$(III)\quad The\quad right\quad bisectors\quad OM\quad \& \quad ON\quad of\quad AB\quad \& \quad BC\quad are\quad drawn.$

$OM\quad \& \quad ON\quad intersect\quad at\quad O.$

$So\quad O\quad is\quad the\quad centre\quad of\quad the\quad circle.$

$Justification-\\ OA,\quad OB\quad \& \quad OC\quad are\quad joined.$

$OA=OB\quad since\quad \quad OM\quad is\quad the\quad right\quad bisector\quad of\quad AB\quad and$

$OB=OC\quad since\quad \quad ON\quad is\quad the\quad right\quad bisector\quad of\quad AB.$

$\therefore \quad OA=OB=OC.$

$i.e\quad A,\quad B\quad \& \quad C\quad are\quad equdistant\quad from\quad O\quad since\quad the\quad centre\quad of\quad a\quad circle\\is\quad eqidistant\quad from\quad all\quad the\quad points\quad on\quad the\quad circumference.$

$\therefore \quad O\quad is\quad the\quad centre\quad of\quad the\quad circle.$

$Ans-\quad Option\quad A.\\ \\$

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

Bisecting means dividing into two ......... parts.

0%
Unequal
0%
Equal
0%
Trianguiar
0%
None of these

An angle

$\angle$ XYZ =

$75^\circ$ is bisected by an angular bisector YU, then the measure of

$\angle$ UYZ is .........

0%
$37^\circ$
0%
$37.5^\circ$
0%
$47.5^\circ$
0%
$47^\circ$

With ruler and compass, we can bisect any given line segment.

0%
True
0%
False

The angles is to be bisected to obtain an angle of 90

$^\circ$ is

0%
$60^\circ$ and $45^\circ$
0%
$60^\circ$ and $120^\circ$
0%
$120^\circ$ and $180^\circ$
0%
$30^\circ$ and $60^\circ$

Explanation

Given: angles to be bisected to make

$90^o$ by using compass and ruler first we will draw

$60^o$ angle and then making another

$60^o$ angle from previous angle made to get

$120^o$ . Now if we bisect

$60^o$ and

$120^o$ we will get

$90^o$ .

$\therefore$ option B is correct

2105356_418976_ans_136070e07aeb4a848b00f7a35f944eb0.png

An angle of

$15^\circ$ is drawn using a pair of compass and ruler by bisecting .........

0%
$60^\circ$
0%
$60^\circ$ and $120^\circ$
0%
$60^\circ$ and then bisecting it again
0%
$60^\circ$ and $180^\circ$

Explanation

To draw an angle of

$15^o$ by using compass and ruler, first, we will draw the angle of

$60^o$ , then bisecting it to get the angle of

$30^o$ and then bisecting it again to get the required angle of

$15^o$ .

$\therefore$ option C is correct.

2105349_418940_ans_68463420237c4d969e42fc237ad97237.png

A perpendicular is drawn to a line segment

$\overline{IP}$ at N using protractor and point M is marked on perpendicular then

0%
$\overline{MP}\, \perp \, \overline{NP}$
0%
$\overline{MN}\, \perp\, P$
0%
$M\, \perp\, P$
0%
$\overline{MN}\, \perp \, \overline{NP}$

Explanation

A perpendicular is drawn to a line segment

$\overline {IP}$ at

$N$ using protractor and point

$M$ is marked on perpendicular then

$\overline {MN} \bot \overline {NP}$

If AB is the perpendicular bisector of the line segment PQ, then

$\Delta APQ$ and

$\Delta BPQ$ are _____.

0%
isosceles
0%
scalene
0%
acute
0%
obtuse

Explanation

$AB$ is the perpendicular bisector of

$PQ$

i.e.

$PA=AQ$ and

$PB=BQ$

$\Rightarrow APQ , BPQ$ are isosceles triangles.

Each of the four angles at

$J$ are

$90^\circ$ and

$PJ=JQ$ . Therefore

$AB$ is _____ to

$PQ$ .

0%
parallel
0%
sector
0%
perpendicular
0%
diagonal

Explanation

Each of the four angles at

$J$ are

$90^\circ$ . Therefore

$AB$ is perpendicular to

$PQ$ .

$PQ$ is the perpendicular bisector of

$AB$ , then

$PQ$ divides

$AB$ in the ratio:

0%
$1:2$
0%
$1:3$
0%
$2:3$
0%
$1:1$

Explanation

Perpendicular bisector always divides a segment into two equal parts.
Therefore

$PQ$ divides

$AB$ into

$1:1$ .

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%
$3$

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 perpendicular is to be drawn to a line segment of length

$5$ cm at a distance of

$3$ cm from the left end. In what ratio, does it divide the given line?

0%
$3:2$
0%
$2:3$
0%
$1:2$
0%
$1:1$

The third step in the process will be:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Correct sequence of steps is :

Step 1: Draw segment

$AB$ and take a point

$P$ on it.

Step 2: From point

$P$ , mark two equidistant points from

$P$ on line

$AB$ , and name them

$C$ and

$D$

Step 3 : From points

$C$ and

$D$ mark two intersecting arcs on either side of the line

$AB$ .Name the intersection point as

$E$

Step 4: Join

$E$ and

$P$ .

$EP$ is the required perpendicular.

So the third step is

$!$

So option

$A$ is correct.

The third step in the process is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Correct sequence is :

Step 1 . Draw a line $PQ$ and take a point $A$ anywhere outside the line.

Step 2. Place the pointed end of the compass on $A$ and with an arbitrary radius, mark two points $D$ and $E$ on line $PQ$ with the same radius.

Step 3. From points $D$ and $E$ , mark two intersecting arcs on either side of $PQ$ and name them $R$ and $S.$

Step 4. Join $R−S$ passing through $A$ .

So the third step is $3$ .

The steps for constructing a perpendicular from point

$A$ to line

$PQ$ is given in jumbled order as follows:

$(A$ does not lie on

$PQ)$

Join

$R-S$ passing through

$A$ .

Place the pointed end of the compass on

$A$ and with an arbitrary radius, mark two points

$D$ and

$E$ on line

$PQ$ with the same radius.

From points

$D$ and

$E$ , mark two intersecting arcs on either side of

$PQ$ and name them

$R$ and

$S$ .

Draw a line

$PQ$ and take a point

$A$ anywhere outside the line.

The second step in the process is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Correct sequence is :

Step 1 . Draw a line $PQ$ and take a point $A$ anywhere outside the line.

Step 2. Place the pointed end of the compass on $A$ and with an arbitrary radius, mark two points $D$ and $E$ on line $PQ$ with the same radius.

Step 3. From points $D$ and $E$ , mark two intersecting arcs on either side of $PQ$ and name them $R$ and $S.$

Step 4. Join $R−S$ passing through $A$ .

So the second step is $2$ .

Option $B$ is correct.

CBSE Questions for Class 6 Maths Practical Geometry Quiz 3 - MCQExams.com

0:0:2

Practice Class 6 Maths Quiz Questions and Answers

CBSE Questions for Class 6 Maths Practical Geometry Quiz 3 - MCQExams.com

0:0:2

Practice Class 6 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.