NCERT MCQ Questions for Class 6 Maths Chapter 14 Practical Geometry with Answers. MCQ Questions for Class 6 Maths with Answers were prepared based on the latest exam pattern. We have provided Knowing Our Numbers Class 6 Maths MCQs Questions with Answers to help students understand the concept very well
Table of Contents
Chapter- 14 Practical Geometry Class 6th Maths Important Questions & MCQS
MCQ
Question 1.
A line segment TP−→− is bisected at I. What is the measure of TI−→?
(a) 12TP−→−
(b) IP−→
(c) TP−→−
(d) 13 TP−→−
Answer
Answer: (b) IP−→
Hint:
TI−→ = 13 TP−→− = IP
Question 2.
Which of the following can be drawn on a piece of paper?
(a) A line
(b) A line segment
(c) A ray
(d) A plane
Answer
Answer: (b) A line segment
Question 3.
l || m. P and Q are points on land m respectively such that PQ ⊥ lR is a point on a line n in the same plane such that PQ¯¯¯¯¯¯¯¯ = QR¯¯¯¯¯¯¯¯. Which of the following is true?
(a) l || n
(b) m || n
(c) Both [a] and [b]
(d) Neither [a] nor [b]
Answer
Answer: (c) Both [a] and [b]
Hint:
Clearly, from the given data and the figure, l || n and m || n.
Question 4.
At 7 a.m. the angle between the Sun’s ray and the ground at a point is 43°. What would be the angle at 10 a.m.?
(a) 40°
(b) 90°
(c) Between 43° and 90°
(d) Greater than 90°
Answer
Answer: (c) Between 43° and 90°
Hint:
Let QP be the sun’s ray and RP be the ground. The angle between QP and PR at P is 43° at 7 a.m. At 10 a.m., the sun’s ray is Q’P. We know that at 12 noon the sun is exactly above our head. So, the sun’s ray will be perpendicular to the ground. So, clearly at 10 am, the required angle will be between 43° and 90°.
Question 5.
XY−→− divides ∠MXN = 72° in the ratio 1 : 2. What is the measure of ∠YXN?
(a) 48°
(b) 24°
(c) 72°
(d) 96°
Answer
Answer: (a) 48°
Hint:
Given ∠MXN = 72∘ and XY−→− divides ∠MXN in the ratio 1 : 2.
∠YXN = 23 ∠MXN = 23 × 72° = 48°
Question 6.
MN−→− is the perpendicular bisector of AB←→. Which of the given statements is correct?
(i) ∠ANM + ∠MNB = 90∘
(ii) AN¯¯¯¯¯¯¯¯ = NB¯¯¯¯¯¯¯¯
(iii) AN¯¯¯¯¯¯¯¯ = 2 NB¯¯¯¯¯¯¯¯
(iv) ∠MNB = 12 ∠ANM
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i) and (ii) only
(d) (ii) and (iii) only
Answer
Answer: (c) (i) and (ii) only
Hint:
NM−→− ⊥ AB←→ and NM−→− divides AB←→ into two congruent parts.
Clearly ∠ANM = ∠MNB = 90∘ is true. AN¯¯¯¯¯¯¯¯ = NB¯¯¯¯¯¯¯¯ is true since NM−→− ⊥ AB←→, AN¯¯¯¯¯¯¯¯ = 2NB¯¯¯¯¯¯¯¯ is false, and ∠MNB = 12 ∠ANM is false. Thus, only (i) and (ii) are correct.
Question 7
Identify the uses of a ruler.
(a) To draw a line segment of a given length
(b) To draw a copy of a given segment.
(c) To draw a diameter of a circle.
(d) All the above.
Answer
Answer: (d) All the above.
Hint:
A ruler is used to draw a line segment of a given length, to draw the copy of a given segment, and to draw a diameter of a circle. Thus, all the given options are correct.
Question 8.
P is the midpoint of AB¯¯¯¯¯¯¯¯. M and N are midpoints of AP¯¯¯¯¯¯¯¯ respectively. What is the measure of MN¯¯¯¯¯¯¯¯¯¯?
(a) 13 AB¯¯¯¯¯¯¯¯
(b) 12 AB¯¯¯¯¯¯¯¯
(c) 12 AP¯¯¯¯¯¯¯¯
(d) 32 AB¯¯¯¯¯¯¯¯
Answer
Answer: (b) 12 AB¯¯¯¯¯¯¯¯
Hint:
P is the midpoint of AB¯¯¯¯¯¯¯¯ ⇒ AP = PB M and N are midpoints of AP¯¯¯¯¯¯¯¯ and PB¯¯¯¯¯¯¯¯ ⇒ AM¯¯¯¯¯¯¯¯¯ = MP¯¯¯¯¯¯¯¯¯ and PN¯¯¯¯¯¯¯¯ = NB¯¯¯¯¯¯¯¯
∴ MN¯¯¯¯¯¯¯¯¯¯ = MP¯¯¯¯¯¯¯¯¯ + PN¯¯¯¯¯¯¯¯ = 12 AP¯¯¯¯¯¯¯¯ + 12 PB¯¯¯¯¯¯¯¯ = 12 (AP¯¯¯¯¯¯¯¯ + PB¯¯¯¯¯¯¯¯ ) = 12 (AB¯¯¯¯¯¯¯¯)
Question 9.
XY−→− bisects∠AXB. If ∠YXB = 37.5∘, what is the measure of ∠AXB?
(a) 37.5°
(b) 74°
(c) 64°
(d) 75°
Answer
Answer: (d) 75°
Question 10.
X and Y are two distinct points in a plane. How many lines can be drawn passing through both X and Y?
(a) 0
(b) 1
(c) Only 2
(d) Infinitely many
Answer
Answer: (b) 1
Question 11.
Lines a, b, p, q, m, n and x have a point P common to all of them. What is the name of P?
(a) Point of concurrence
(b) Point of intersection
(c) Common point
(d) Distinct point
Answer
Answer: (a) Point of concurrence
Hint:
A point common to multiple lines is called a point of concurrence as the lines are concurrent lines.
Question 12.
QZ−→− is the bisector of ∠PQZ = ∠PQR. Which of the following is true?
(a) ∠PQZ = ∠PQR
(b) ∠PQZ = ∠ZQR
(c) ∠PQZ = 12 ∠ZQR
(d) Both [b] and [c]
Answer
Answer: (d) Both [b] and [c]
Hint:
QZ−→− bisects ∠PQR (Given)
Thus. ∠PQZ = ∠ZQR = 12 ∠PQR
Question 13.
Identify the pair of parallel lines.
(i) Lines m and n have two points in common.
(ii) Lines p and q do not have any point in common
(iii) Lines p and q have a point X in common.
(a) (i) and (ii) only
(b) (ii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Answer
Answer: (b) (ii) only
Hint:
Parallel lines do not have any point in common.
Question 14.
Identify the one with no definite length.
(a) AB←→
(b) PQ¯¯¯¯¯¯¯¯
(c) -XYZ
(d) MN¯¯¯¯¯¯¯¯¯¯
Answer
Answer: (a) AB←→
Hint:
AB←→ has no definite length.
Question 15.
If two lines have only one point in common, what are they called?
(a) Parallel lines
(b) Intersecting lines
(c) Perpendicular lines
(d) Transversal
Answer
Answer: (b) Intersecting lines
Hint:
Intersecting lines have only one point in common.
Question 16.
Two lines are said to be perpendicular to each other when they meet at ____angle.
(a) 180°
(b) 90°
(c) 60°
(d) 360°
Answer
Answer: (b) 90°
Important Questions
Question 1.
If AB = 3.6 and CD = 1.6 cm, construct a line segment equal to AB¯¯¯¯¯¯¯¯ + CD¯¯¯¯¯¯¯¯ and measure the total length.
Solution:
Step I: Draw a ray OX.
Step II : With centre 0 and radius equal to the length of AB (3.6 cm) mark a point P on the ray.
Step III: With centre P and radius equal to the length of CD (1.6 cm) mark another point Q on the ray.
Thus OQ is the required segment such that OQ = 3.6 cm + 1.6 cm = 5.2 cm.
Question 2.
Construct a perpendicular to a given line segment at point on it.
Solution:
Step IDraw a line PQ←→ and take any point A on it.
Step II : With centre A draw an arc which meets PQ at C and D.
Step III : Join AB and produce.
Step IV: With centres C and D and radius equal to half of the length of the previous arc, draw two arcs which meets each other at B.
Thus AB is the required perpendicular to PQ←→.
Question 3.
Construct an angle of 60° and bisect it.
Solution:
Step I: Draw a line segment AB¯¯¯¯¯¯¯¯.
Step II: With centre B and proper radius, draw an arc which meets AB at C.
Step III : With C as centre and the same radius as in step II, draw an arc cutting the previous arc at D.
Step IV : Join B to D and produce.
Step V : Draw the bisector BE of ∠ABD.
Thus BE is the required bisector of ∠ABD.
Question 4.
Draw an angle of 120° and hence construct an angle of 105°.
Solution:
Step I : Draw a line segment OA¯¯¯¯¯¯¯¯.
Step II : With centre O and proper radius, draw an arc which meets OA at C.
Step III : With centre C and radius same, mark D and E on the previous arc.
Step IV : Join O to E and produce.
Step V : ∠EOA is the required angle of 120°.
Step VI : Construct an angle of 90° which meets the previous arc at F.
Step VII : With centre E and F and proper radius, draw two arcs which meet each other at G.
Step VIII : Join OG and produce.
Thus ∠GOA is the required angle of 105°.
Question 5.
Using compasses and ruler, draw an angle of
75° and hence construct an angle of 37 1∘2.
Solution:
Step I: Draw a line segment OA.
Step II : Construct ∠BOA = 90° and ∠EOA = 60°
Step III : Draw OC as the bisector of ∠BOE , which equal to
Step IV : Draw the bisector OD of ∠COA.
Thus ∠DOA is the required angle of 37 1∘2 .
Question 6.
Draw ∆ABC. Draw perpendiculars from A, B and C respectively on the sides BC, CA and AB. Are there perpendicular concurrent? (passing through the same points).
Solution:
Step I: Draw any ∆ABC.
Step II : Draw the perpendicular AD from A to BC.
Step III : Draw the perpendicular BE from B to AC.
Step IV : Draw the perpendicular CF from C to AB.
We observe that the perpendiculars AD, BE and CF intersect each other at P.
Thus, P is the point of intersection of the three perpendiculars.
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