Practical Geometry Class 6 Extra Questions Maths Chapter 14

Extra Questions for Class 6 Maths Chapter 14 Practical Geometry

Practical Geometry Class 6 Extra Questions Very Short Answer Type

Question 1.
If AB = 3.6 and CD = 1.6 cm, construct a line segment equal to Image may be NSFW.
Clik here to view. $\overline { AB }$ + Image may be NSFW.
Clik here to view. $\overline { CD }$ and measure the total length.
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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.
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Clik here to view.
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 Image may be NSFW.
Clik here to view. $\overleftrightarrow { PQ }$ and take any point A on it.
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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 Image may be NSFW.
Clik here to view. $\overleftrightarrow { PQ }$ .

Question 3.
Construct an angle of 60° and bisect it.
Solution:
Step I: Draw a line segment Image may be NSFW.
Clik here to view. $\overline { AB }$ .
Image may be NSFW.
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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 Image may be NSFW.
Clik here to view. $\overline { OA }$ .
Image may be NSFW.
Clik here to view. Practical Geometry Class 6 Extra Questions Maths Chapter 14
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 Image may be NSFW.
Clik here to view. $\frac { { 1 }^{ \circ } }{ 2 }$ .
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
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Step IV : Draw the bisector OD of ∠COA.
Image may be NSFW.
Clik here to view.
Thus ∠DOA is the required angle of 37 Image may be NSFW.
Clik here to view. $\frac { { 1 }^{ \circ } }{ 2 }$ .

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.
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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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Practical Geometry Class 6 Extra Questions Maths Chapter 14