### 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.