NCERT Solutions Class 6th Maths Chapter -14 Practical Geometry Exercise – 14.5

Question 1. Draw AB of length 7.3 cm and find its axis of symmetry.

Solution:
Step I: Draw AB¯ = 7.3 cm

Step II: Taking A and B as centre and radius more than half of AB¯, draw two arcs which intersect each other at C and D.
Step III: Join C and D to intersect AB¯ at E. Thus, CD is the perpendicular bisector or axis of symmetry of AB¯.

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

Solution:
Step I: Draw a line segment PQ¯=9.5 cm

Step II: With centres P and Q and radius more than half of PQ, draw two arcs which meet each other at R and S.
Step III: Join R and S to meet PQ¯ at T.
Thus, RS is the perpendicular bisector of PQ.

Question 3. Draw the perpendicular bisector of 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 XY¯.What can you say about the length of MX and MY?

Solution:
Step I: Draw a line segment XY¯ = 10.3 cm.

Step II : With centre X and Y and radius more than half of XY, draw two arcs which meet each other at U and V.
Step III: Join U and V which meets XY¯ at M.
Step IV: Take a point P on UV¯ .
(a) On measuring, PX = PY = 5.6 cm.
(b) On measuring, MX¯ = MY¯ = 12 XY = 5.15 cm.

Question 4. Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.

Solution: Step I: Draw a line segment AB¯ = 12.8 cm.

Step II : With centre A and B and radius more than half of AB, draw two arcs which meet each other at D and E.
Step III : Join D and E which meets AB¯ at C which is the midpoint of AB¯.
Step IV : With centre A and C and radius more than half of AC, draw two arcs which meet each other at F and G.
Step V: Join F and G which meets AC¯ at H which is the midpoint of AC¯ .
Step VI : With centre C and B and radius more than half of CB, draw two arcs which meet each other at J and K.
Step VII : Join J and K which meets CB¯ at L which is the midpoint of CB¯.
Thus, on measuring, we find.
AH¯ = HC¯ = CL¯ = LB¯ = 3.2 cm.

Question 5. With PQ¯ of length 6.1 cm as diameter, draw a circle.

Solution:
Step I: Draw PQ¯ = 6.1 cm.
Step II: Draw a perpendicular bisector of PQ¯ which meets PQ¯ at R i.e. R is the midpoint of PQ¯.

Step III : With centre R and radius equal to RP¯ , draw a circle passing through P and Q.
Thus, the circle with diameter PQ¯ = 6.1 cm is the required circle.

Solution:
Step I: Draw a circle with centre C and radius 3.4 cm.
Step II: Draw any chord AB¯.
Step III : Draw the perpendicular bisector of AB¯ which passes through the centre C.

Question 7. Repeat Question number 6, if AB¯ happens to be a diameter.

Solution:
Step I: Draw a circle with centre C and radius 3.4 cm.
Step II : Draw a diameter AB of the circle.

Step III : Draw a perpendicular bisector of AB which passes through the centre C and on measuring, we find that C is the midpoint of AB¯ .

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?

Solution:
Step I: Draw a circle with centre 0 and radius 4 cm.

Step II: Draw any two chords AB¯ and CD¯ of the circle.
Step III : Draw the perpendicular bisectors of AB¯ and CD¯ i.e. I and m.
Step IV : On producing the two perpendicular bisectors meet each other at the centre O of the circle.

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 OA¯ and OB¯ . Let them meet at P. Is PA = PB?

Solution:
Step I: Draw an angle XOY with O as its vertex.
Step II : Take any point A on OY and B on OX, such that OA + OB.

Step III : Draw the perpendicular bisectors of OA and OB which meet each other at a point P.
Step IV : Measure the lengths of PA¯ and PB¯. Yes, PA¯ = PB¯.