NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.5

Question 1.
Draw \(\overline{\mathrm{AB}}\) of length 7.3 cm and find its axis of symmetry.
Answer:
Axis of symmetry of line segment \(\overline{\mathrm{AB}}\) will be the perpendicular bisector of \(\overline{\mathrm{AB}}\). So, draw the perpendicular bisector of \(\overline{\mathrm{AB}}\).
Steps of construction:

  • Draw a line segment \(\overline{\mathrm{AB}}\) = 7.3 cm
  • Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
  • Join CD. Then CD is the axis of symmetry of the line segment AB.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 1

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

Question 2.
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Answer:
Steps of construction:

  • Draw a line segment \(\overline{\mathrm{AB}}\) = 9.5 cm
  • Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
  • Join CD. Then CD is the perpendicular bisector of \(\overline{\mathrm{AB}}\) .

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 2

Question 3.
Draw the perpendicular bisector of \(\overline{\mathrm{XY}}\) whose length is 10.3 cm.
(a) Take any point P on the bisector drawn.
Examine whether PX = PY.
(b) If M is the mid-point of \(\overline{\mathrm{XY}}\) , what can you say about the lengths MX and XY?
Answer:
Steps of construction:

  • Draw a line segment \(\overline{\mathrm{XY}}\) = 10.3 cm
  • Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
  • Join CD. Then CD is the required perpendicular bisector of \(\overline{\mathrm{XY}}\) .

Now:
(a) Take any point P on the bisector drawn. With the help of divider we can check that \(\overline{\mathrm{PX}}\) = \(\overline{\mathrm{PY}}\)
(b) If M is the mid-point of \(\overline{\mathrm{PX}}\) and \(\overline{\mathrm{MX}}\) – 1/2\(\overline{\mathrm{XY}}\)
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 3

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

Question 4.
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 4

  • Draw a line segment AB = 12.8 cm
  • Draw the perpendicular bisector of \(\overline{\mathrm{AB}}\) which cuts it at C. Thus, C is the midpoint of \(\overline{\mathrm{AB}}\).
  • Draw the perpendicular bisector of \(\overline{\mathrm{AC}}\) which cuts it at D. Thus D is the midpoint of.
  • Again, draw the perpendicular bisector of \(\overline{\mathrm{CB}}\) which cuts it at E. Thus, E is the mid-point of \(\overline{\mathrm{CB}}\).
  • Now, point C, D and E divide the line segment \(\overline{\mathrm{AB}}\) in the four equal parts.
  • By actual measurement, we find that \(\overline{\mathrm{AD}}=\overline{\mathrm{DC}}=\overline{\mathrm{CE}}=\overline{\mathrm{EB}}\) = 3.2 cm

Question 5.
With \(\overline{\mathrm{PQ}}\) of length 6.1 cm as diameter,
draw a circle.
Answer:
Steps of construction:

  • Draw a line segment \(\overline{\mathrm{PQ}}\) = 6.1 cm.
  • Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point of \(\overline{\mathrm{PQ}}\).
  • Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment \(\overline{\mathrm{PQ}}\).

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 5

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

Question 6.
Draw a circle with centre C and radius 3.4 cm. Draw any chord \(\overline{\mathrm{AB}}\). Construct the perpendicular bisector \(\overline{\mathrm{AB}}\) and examine if it passes through C.
Answer:
Steps of construction:

  • Draw a circle with centre C and radius 3.4 cm.
  • Draw any chord \(\overline{\mathrm{AB}}\).
  • Taking A and B as centres and radius more than half of \(\overline{\mathrm{AB}}\), draw two arcs which cut each other at P and Q.
  • Join PQ. Then PQ is the perpendicular bisector of \(\overline{\mathrm{AB}}\).
  • This perpendicular bisector of \(\overline{\mathrm{AB}}\) passes through the centre C of the circle.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 6

Question 7.
Repeat Q6, if \(\overline{\mathrm{AB}}\) happens to be a diameter.
Answer:
Steps of construction:

  • Draw a circle with centre C and radius 3.4 cm.
  • Draw its diameter \(\overline{\mathrm{AB}}\).
  • Taking A and B as centres and radius more than half of it, draw two arcs which intersect each other at P and Q.
  • Join PQ. Then PQ is the perpendicular bisector of \(\overline{\mathrm{AB}}\) .
  • We observe that this perpendicular bisector of \(\overline{\mathrm{AB}}\) passes through the centre C of the circle.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 7

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

Question 8.
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords.
Where do they meet?
Answer:
Steps of construction:

  • Draw the circle with O and radius 4 cm.
  • Draw any two chords \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) in this circle.
  • Taking A and B as centres and radius more than half AB, draw two arcs which intersect each other at E and F.
  • Join EF. Thus EF is the perpendicular bisector of chord \(\overline{\mathrm{CD}}\).
  • Similarly draw GH the perpendicular bisector of chord \(\overline{\mathrm{CD}}\).
  • These two perpendicular bisectors meet at O, the centre of the circle.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 8

Question 9.
Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\). Let them meet at P. Is PA = PB?
Answer:
Steps of construction:

  • Draw any angle with vertex O.
  • Take a point A on one of its arms and B on another such that OA = OB.
  • Draw perpendicular bisector of \(\overline{\mathrm{OA}}\) and OB.
  • Let them meet at P. Join PA and PB.
  • With the help of divider, we check that \(\overline{\mathrm{PA}}=\mathrm{PB}\)

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 9

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