NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5.

- Practical Geometry Class 6 Ex 14.1
- Practical Geometry Class 6 Ex 14.2
- Practical Geometry Class 6 Ex 14.3
- Practical Geometry Class 6 Ex 14.4
- Practical Geometry Class 6 Ex 14.6

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Practical Geometry |

Exercise |
Ex 14.5 |

Number of Questions Solved |
9 |

Category |
NCERT Solutions |

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

Question 1.

Draw \(\overline { AB }\) of length 7.3 cm and find its axis of symmetry.

Solution :

**Step 1.** Draw a line segment \(\overline { AB }\) 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 \(\overline { AB }\).

**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, \(\overline { CD }\) is the axis of symmetry of \(\overline { AB }\).

Question 2.

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

Solution :

**Step 1.** Draw a line segment \(\overline { AB }\) 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 \(\overline { AB }\).

**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 \(\overline { CD }\) is the perpendicular bisector of the line segment \(\overline { AB }\).

Question 3.

Draw the perpendicular bisector of \(\overline { XY }\) 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 \(\overline { XY }\), what can you say about the lengths MX and XY ?

Solution :

**Step 1.** Draw a line segment \(\overline { XY }\) 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 \(\overline { XY }\).

**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 \(\overline { AB }\) is the perpendicular bisector of the line segment \(\overline { XY }\).

**(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 \(\overline { AB }\) 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 \(\overline { AB }\).

**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 \(\overline { CD }\). It cuts \(\overline { AB }\) at E. Then \(\overline { CD }\) is the perpendicular bisector of the line segment \(\overline { AB }\).

**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 \(\overline { FG }\). It cuts \(\overline { AE }\) at H. Then \(\overline { FG }\) is the perpendicular bisector of the line segment \(\overline { AE }\).

**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 \(\overline { IJ }\). It cuts \(\overline { EB }\) at K. Then \(\overline { IJ }\) is the perpendicular bisector of the line segment \(\overline { EB }\). Now, the points H, E and K divide AB into four equal parts, i.e., \(\overline { AH }\) = \(\overline { HE }\) = \(\overline { EK }\) = \(\overline { KB }\) By measurement, \(\overline { AH }\) = \(\overline { HE }\) = \(\overline { EK }\) = \(\overline { KB }\) = 3.2 cm.

Question 5.

With \(\overline { PQ }\) of length 6.1 cm as diameter draw a circle.

Solution :

**Step 1.** Draw a line segment \(\overline { PQ }\) 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 \(\overline { PQ }\).

**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 \(\overline { AB }\). It cuts \(\overline { PQ }\) at C. Then \(\overline { AB }\) 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 \(\overline { AB }\). Construct the perpendicular bisector of \(\overline { AB }\) 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 \(\overline { AB }\) 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 \(\overline { AB }\).

**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 \(\overline { DE }\) . Then \(\overline { DE }\) is the perpendicular bisector of the line segment \(\overline { AB }\). On examination, we find that it passes through C.

Question 7.

Repeat Question 6, if \(\overline { AB }\) 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 \(\overline { AB }\).

**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 \(\overline { AB }\).

**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 \(\overline { DE }\) . Then \(\overline { DE }\) is the perpendicular bisector of the line segment \(\overline { AB }\). 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 \(\overline { AB }\) and \(\overline { CD }\) 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 \(\overline { AB }\).

**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 \(\overline { EF }\). Then \(\overline { EF }\) is the perpendicular bisector of the chord \(\overline { AB }\).

**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 \(\overline { CD }\).

**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 \(\overline { GH }\). Then \(\overline { GH }\) is the perpendi¬cular bisector of the chord \(\overline { CD }\). We find that the perpendicular bisectors \(\overline { EF }\) and \(\overline { GH }\) 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 \(\overline { OA }\) and \(\overline { OB }\). 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 \(\overline { OA }\) = \(\overline { OB }\).

**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 \(\overline { OA }\).

**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 \(\overline { CD }\). Then \(\overline { CD }\) is the perpendicular bisector of the line segment \(\overline { OA }\).

**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 \(\overline { OB }\).

**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 \(\overline { EF }\). Then \(\overline { EF }\) is the perpendicular bisector of the line segment OB. The two perpendicu¬lar bisectors meet at P.

**Step 9.** Join \(\overline { PA }\) and \(\overline { PB }\). We find that \(\overline { PA }\) = \(\overline { PB }\).

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