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

*According to the latest update on the CBSE Syllabus 2023-24, this chapter has been removed.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.5 have been created by the faculty at BYJUâ€™S to help the students prepare for their exams efficiently. The perpendicular bisector of a line segment and the steps to be followed in their construction using a compass and ruler are explained in a simple way under Exercise 14.5. The steps in the construction of a line segment are discussed in an interactive manner in these NCERT Solutions, with the aim of making it easily understandable for the students.

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

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

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

Solutions:

The following steps are followed to construct of length 7.3 cm and to find its axis of symmetry:

(1) Draw a line segmentÂ  of 7.3 cm.

(2) Take A as the centre and draw a circle by using compasses. The radius of the circle should be more than half the length of
.

(3) Now, take B as the centre and draw another circle using compasses with the same radius as before. Let it cut the previous circle at points C and D.

(4) Join CD. Now is the axis of symmetry.

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

Solutions:

The following steps are observed to construct a line segment of length 9.5 cm and to construct its perpendicular bisector:

(1) Draw a line segmentÂ of 9.5 cm.

(2) Take point P as the centre and draw a circle by using compasses. The radius of the circle should be more than half the length of .

(3) Taking the centre at point Q, again draw another circle using compasses with the same radius as before. Let it cut the previous circle at R and S, respectively.

(4) Join RS. Now is the axis of symmetry, i.e., the perpendicular bisector of the line .

Â

3. Draw the perpendicular bisector of 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 , what can you say about the lengths MX and XY?

Solutions:

(1) Draw a line segment Â of 10.3 cm.

(2) Take point X as the centre and draw a circle by using compasses. The radius of the circle should be more than half the length of .

(3) Now, taking Y as the centre, draw another circle using compasses with the same radius as before. Let it cut at the previous circle at points A and B.

(4) Join AB. HereÂ is the axis of symmetry.

Â

(a) Take any point P onÂ . We may observe that the measure of lengths of PX and PY are the same,

Â being the axis of symmetry, any point lying onÂ will be at the same distance from both ends of .

(b) M is the midpoint of . Perpendicular bisectorÂ will be passing through point M. Hence, the length of
is double ofÂ or 2MX = XY.

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

Solutions:

(1) Draw a line segment Â of 12.8 cm.

(2) By taking point X as the centre and radius more than half of XY, draw a circle.

(3) Again, with the same radius and centre as Y, draw two arcs to cut the circle at points A and B. Join AB, which intersects Â at point M.

(4) By taking X and Y as centres, draw two circles with a radius more than half of .

(5) Taking M as the centre and with the same radius, draw two arcs to intersect these circles at P, Q and R, S.

(6) Join PQ and RS. These intersect Â at points T and U.

(7) The 4 equal parts of Â areÂ =Â =Â =Â .

By measuring these line segments with the help of a ruler, we may observe that each is 3.2 cm.

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

Solution:

(1) Draw a line segment Â of 6.1 cm.

(2) Take point P as the centre and radius more than half of , draw a circle.

(3) Again, with the same radius and Q as the centre, draw two arcs intersecting the circle at points R and S.

(4) Join RS, which intersects Â at T.

(5) Taking the centre as T and radius TP, draw a circle which passes through Q. Now, this is the required circle.

6. Draw a circle with centre C and a radius 3.4 cm. Draw any chord . Construct the perpendicular bisector of and examine if it passes through C.

Solutions:

(1) Mark any point C on the sheet.

(2) Adjust the compass up to 3.4 cm, and by putting the pointer of the compass at point C, turn the compass slowly to draw the circle. This is the required circle with 3.4 cm radius.

(3) Mark any chord Â in the circle.

(4) Now, taking A and B as centres, draw arcs on both sides of . Let these intersect each other at points D and E.

(5) Join DE. Now DE is the perpendicular bisector of AB.

IfÂ is extended, it will pass through point C.

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

Solutions:

(1) Mark any point C on the sheet.

(2) Adjust the compass up to 3.4 cm, and by putting the pointer of the compass at point C, turn the compass slowly to draw the circle. This is the required circle with 3.4 cm.

(3) Now mark any diameter Â in the circle.

(4) Now, taking A and B as centres, draw arcs on both sides of Â with a radius more than . Let these intersect each other at points D and E.

(5) Join DE, which is the perpendicular bisector of AB.

Now, we may observe thatÂ is passing through the centre C of the circle.

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?

Solutions:

(1) Mark any point O on the sheet. Now adjust the compass up to 4 cm, and by placing the pointer of the compass at point O, turn the compass slowly to draw the circle. This is the required circle with 4 cm radius.

(2) Take any two chords Â andÂ in the circle.

(3) By taking A and B as centres and a radius more than half of , draw arcs on both sides of AB. The arcs intersect each other at points E and F. Join EF, which is the perpendicular bisector of AB.

(4) Again, take C and D as centres and a radius more than half of , draw arcs on both sides of CD, such that they intersect each other at points G, H. Join GH, which is the perpendicular bisector of CD.

We may observe that when EF and GH are extended, they meet at point O, which is the centre of the circle.

9. Draw any angle with vertex O. Take point A on one of its arms and B on another, such that OA = OB. Draw the perpendicular bisectors of and . Let them meet at P. Is PA = PB?

Solutions:

(1) Draw any angle with vertex as O.

(2) By taking O as the centre and with a convenient radius, draw arcs on both rays of this angle. Let these points be A and B.

(3) Now take O and A as centres, and with a radius of more than half of OA, draw arcs on both sides of OA. Let these intersect at points C and D, respectively. Join CD.

(4) Similarly, we may find which is the perpendicular bisector of . These perpendicular bisectors, andÂ , intersect each other at point P. Now, measure PA and PB. They are equal in length.

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