Ncert Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 Practical Geometry Exercise 14.5 Ncert Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 Chapter 14 Practical Geometry Class 6 Chapter 14 Maths Chapter 14 Class 6 Maths Maths Class 6 Chapter 14 Maths Chapter 14 Class 6 Ncert Maths Class 6 Chapter 14 Ncert Solutions For Class 6 Maths Chapter 14 Practical Geometry Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 Class 6 Maths Ncert Solutions

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

Question 1.

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

Solution :

**Step 1.** Draw a line segment AB¯¯¯¯¯¯¯¯ of length 7.3 cm.

**Step 2.** With A ascentere, using compasses, drawthe circle. The radius of this circle should be more than half of the length of AB¯¯¯¯¯¯¯¯.

**Step 3.** With the same radius and with B as a centre, draw another circle using compasses. Let it cut the previous circle at C and D.

**Step 4.** Join CD. Then, CD¯¯¯¯¯¯¯¯ is the axis of symmetry of AB¯¯¯¯¯¯¯¯.

Question 2.

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

Solution :

**Step 1.** Draw a line segment AB¯¯¯¯¯¯¯¯ of length 9.5 cm.

**Step 2.** With A as a centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of AB¯¯¯¯¯¯¯¯.

**Step 3.** With the same radius and with B as a centre, draw another circle using compasses. Let it cut the previous circle at C and D.

**Step 4.** Join CD. Then CD¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment AB¯¯¯¯¯¯¯¯.

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 midpoint of XY¯¯¯¯¯¯¯¯, what can you say about the lengths MX and XY ?

Solution :

**Step 1.** Draw a line segment XY¯¯¯¯¯¯¯¯ of length 10.3 cm.

**Step 2.** With X as a centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of XY¯¯¯¯¯¯¯¯.

**Step 3.** With the same radius and with Y as a centre, draw another circle using compasses. Let it cuts the previous circle at A and B.

**Step 4.** Join AB. Then AB¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment XY¯¯¯¯¯¯¯¯.

**(a)** On examination, we find that PX = PY.

**(b)** We can say that the lengths of MX is half of the length of XY.

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 1.** Draw a line segment AB¯¯¯¯¯¯¯¯ of length 12.8 cm.

**Step 2.** With A as centre, using compasses, draw two arcs on either side of AB. The radius of this arc should be more than fialf of the length of AB¯¯¯¯¯¯¯¯.

**Step 3.** With the same radius and with B as centre, draw another arcs using compasses. Let it cut the previous arcs at C and D.

**Step 4.** Join CD¯¯¯¯¯¯¯¯. It cuts AB¯¯¯¯¯¯¯¯ at E. Then CD¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment AB¯¯¯¯¯¯¯¯.

**Step 5.** With A as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of AE.

**Step 6.** With the same radius and with E as centre, draw another circle using compasses; Let it cut the previous circle at F and G.

**Step 7.** Join FG¯¯¯¯¯¯¯¯. It cuts AE¯¯¯¯¯¯¯¯ at H. Then FG¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment AE¯¯¯¯¯¯¯¯.

**Step 8.** With E as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of EB.

**Step 9.** With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at I and J.

**Step 10.** Join IJ¯¯¯¯¯¯. It cuts EB¯¯¯¯¯¯¯¯ at K. Then IJ¯¯¯¯¯¯ is the perpendicular bisector of the line segment EB¯¯¯¯¯¯¯¯. Now, the points H, E and K divide AB into four equal parts, i.e., AH¯¯¯¯¯¯¯¯ = HE¯¯¯¯¯¯¯¯ = EK¯¯¯¯¯¯¯¯ = KB¯¯¯¯¯¯¯¯ By measurement, AH¯¯¯¯¯¯¯¯ = HE¯¯¯¯¯¯¯¯ = EK¯¯¯¯¯¯¯¯ = KB¯¯¯¯¯¯¯¯ = 3.2 cm.

Question 5.

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

Solution :

**Step 1.** Draw a line segment PQ¯¯¯¯¯¯¯¯ of length 6.1 cm.

**Step 2.** With P as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of PQ¯¯¯¯¯¯¯¯.

**Step 3.** With the same radius and with Q as centre, draw another circle using compasses. Let it cut the previous circle at A and B.

**Step 4.** Join AB¯¯¯¯¯¯¯¯. It cuts PQ¯¯¯¯¯¯¯¯ at C. Then AB¯¯¯¯¯¯¯¯ is

the perpendicular bisector of the line segment PQ .

**Step 5.** Place the pointer of the compasses at C and open the pencil upto P.

**Step 6.** Turn the compasses slowly to draw the circle.

Question 6.

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

Solution :

**Step 1.** Draw a point with a sharp pencil and mark it as C.

**Step 2.** Open the compasses for the required radius 3.4 cm, by putting the pointer on 0 and opening the pencil upto 3.4 cm.

**Step 3.** Place the pointer of the compasses at C. Step 4. Turn the compasses slowly to draw the

circle.

**Step 5.** Draw any chord AB¯¯¯¯¯¯¯¯ of this circle.

**Step 6.** With A as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of AB¯¯¯¯¯¯¯¯.

**Step 7.** With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at D and E.

**Step 8.** Join DE¯¯¯¯¯¯¯¯ . Then DE¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment AB¯¯¯¯¯¯¯¯. On examination, we find that it passes through C.

Question 7.

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

Solution :

**Step 1.** Draw a point with a sharp pencil and mark it as C.

**Step 2.** Open the compasses for the required radius 3.4 cm, by putting the pdinter of compasses on 0 of the scale and opening the pencil upto 3.4 cm.

**Step 3.** Place the pointer of the compasses at C.

**Step 4.** Turn the compasses slowly to draw the circle.

**Step 5.** Draw any diameter AB¯¯¯¯¯¯¯¯.

**Step 6.** With A as centre, using compasses, draw arcs on either side. The radius of this arc should be more than half of the length of AB¯¯¯¯¯¯¯¯.

**Step 7.** With the same radius and with B as centre, draw another arcs using compasses. Let it cut the previous arcs at D and E.

**Step 8.** Join DE¯¯¯¯¯¯¯¯ . Then DE¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment AB¯¯¯¯¯¯¯¯. On examination, we find that it passes through C.

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 1.** Draw a point with a sharp pencil and mark it as O.

**Step 2.** Open the compasses for the required radius of 4 cm. by putting the pointer on 0 and opening the pencil upto 4 cm.

**Step 3.** Place the pointer of the compasses at O.

**Step 4.** Turn the compasses slowly to draw the circle.

**Step 5.** Draw any two chords AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ of this circle.

**Step 6.** With A as centre, using compasses, draw two arcs on either side of AB. The radius of this arc should be more than half of the length of AB¯¯¯¯¯¯¯¯.

**Step 7.** With the same radius and with B as centre, draw another two arcs using compasses. Let it cut the previous circle at E and F.

**Step 8.** Join EF¯¯¯¯¯¯¯¯. Then EF¯¯¯¯¯¯¯¯ is the perpendicular bisector of the chord AB¯¯¯¯¯¯¯¯.

**Step 9.** With C a< centre, using compasses, draw two arcs on either side of CD. The radius of this arc should be more than half of the length of CD¯¯¯¯¯¯¯¯.

**Step 10.** With the same radius and with D as centre, draw another two arcs using compasses. Let it cut the previous circle at G and H.

**Step 11.** Join GH¯¯¯¯¯¯¯¯. Then GH¯¯¯¯¯¯¯¯ is the perpendi¬cular bisector of the chord CD¯¯¯¯¯¯¯¯. We find that the perpendicular bisectors EF¯¯¯¯¯¯¯¯ and GH¯¯¯¯¯¯¯¯ meet at O, the centre 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 1.** Draw any angle POQ with vertex O.

**Step 2.** Take a point A on the arm OQ and another point B on the arm OP such that OA¯¯¯¯¯¯¯¯ = OB¯¯¯¯¯¯¯¯.

**Step 3.** With O as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of OA¯¯¯¯¯¯¯¯.

**Step 4.** With the same radius and with A as centre, draw another circle using compasses. Let it cut the previous circle at C and D.

**Step 5.** Join CD¯¯¯¯¯¯¯¯. Then CD¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment OA¯¯¯¯¯¯¯¯.

**Step 6.** With O as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of OB¯¯¯¯¯¯¯¯.

**Step 7.** With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at E and F.

**Step 8.** Join EF¯¯¯¯¯¯¯¯. Then EF¯¯¯¯¯¯¯¯ is the perpendicular bisector of the line segment OB. The two perpendicu¬lar bisectors meet at P.

**Step 9.** Join PA¯¯¯¯¯¯¯¯ and PB¯¯¯¯¯¯¯¯. We find that PA¯¯¯¯¯¯¯¯ = PB¯¯¯¯¯¯¯¯.