Solutions For All Chapters Maths Class 6
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.5
Ex 14.5 Class 6 Maths Question 1.
Draw AB of length 7.3 cm and find its axis of symmetry.
Solution:
Step I: Draw A̅B̅ = 7.3 cm
Step II: Taking A and B as centre and radius more than half of A̅B̅, draw two arcs which intersect each other at C and D.
Step III: Join C and D to intersect A̅B̅ at E. Thus, CD is the perpendicular bisector or axis of symmetry of A̅B̅.
Ex 14.5 Class 6 Maths Question 2.
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Solution:
Step I: Draw a line segment P̅Q̅ =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 P̅Q̅ at T.
Thus, RS is the perpendicular bisector of PQ.
Ex 14.5 Class 6 Maths Question 3.
Draw the perpendicular bisector of X̅Y̅ whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the midpoint of X̅Y̅ . What can you say about the length of MX and MY?
Solution:
Step I: Draw a line segment X̅Y̅ = 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 X̅Y̅ at M.
Step IV: Take a point P on U̅V̅ .
(a) On measuring, PX = PY = 5.6 cm.
(b) On measuring, M̅X̅ = M̅Y̅ = 1/2 XY = 5.15 cm.
Ex 14.5 Class 6 Maths 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 A̅B̅ = 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 A̅B̅ at C which is the midpoint of A̅B̅.
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 A̅C̅ at H which is the midpoint of A̅C̅ .
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 C̅B̅ at L which is the midpoint of C̅B̅ .
Thus, on measuring, we find
A̅H̅ = H̅C̅ = C̅L̅ = L̅B̅ = 3.2 cm.
Ex 14.5 Class 6 Maths Question 5.
With P̅Q̅ of length 6.1 cm as diameter, draw a circle.
Solution:
Step I: Draw P̅Q̅ = 6.1 cm
Step II: Draw a perpendicular bisector of P̅Q̅ which meets P̅Q̅ at R i.e. R is the midpoint of P̅Q̅.
Step III : With centre R and radius equal to R̅P̅ , draw a circle passing through P and Q.
Thus, the circle with diameter P̅Q̅ = 6.1 cm is the required circle.
Ex 14.5 Class 6 Maths Question 6.
Draw a circle with centre C and radius 3.4 cm. Draw any chord A̅B̅ . Construct the perpendicular bisector of A̅B̅ and examine if it passes through C.
Solution:
Step I: Draw a circle with centre C and radius 3.4 cm.
Step II: Draw any chord A̅B̅.
Step III : Draw the perpendicular bisector of A̅B̅ which passes through the centre C.
Ex 14.5 Class 6 Maths Question 7.
Repeat Question number 6, if A̅B̅ 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 A̅B̅ .
Ex 14.5 Class 6 Maths 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 A̅B̅ and C̅D̅ of the circle.
Step III : Draw the perpendicular bisectors of A̅B̅ and C̅D̅ i.e. I and m.
Step IV : On producing the two perpendicular bisectors meet each other at the centre O of the circle.
Ex 14.5 Class 6 Maths 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 O̅A̅ and O̅B̅ . 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 P̅A̅ and P̅B̅. Yes, P̅A̅ = P̅B̅.