NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-4-ex-4-5/

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.3
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5

Question 1.
Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonal in the interior or exterior of the quadrilateral?
Solution :

The meeting point O of the diagonals PR and QS of the quadrilateral PQRS is in the interior of the quadrilateral PQRS.

Question 2.
Draw a rough sketch of a quadrilateral KLMN. State :
(a) two pairs of opposite sides.
(b) two pairs of opposite angles.
(c) two pairs of adjacent sides.
(d) two pairs of adjacent angles.
Solution :
(a) \(\bar { KL }\), \(\bar { NM }\) and \(\bar { KN }\), \(\bar { ML }\)
(b) ∠K, ∠M and ∠N, ∠L
(c) \(\bar { KL }\) , \(\bar { KN }\) and \(\bar { NM }\), \(\bar { ML }\) or \(\bar { KL }\) , \(\bar { LM }\) and \(\bar { NM }\), \(\bar { ML }\)

(d) ∠K, ∠L, and ∠M. ∠N or ∠K. ∠L and ∠L, ∠M etc.

Question 3.
Investigate:
Use strips and fasteners to make a triangle and a quadrilateral. Try to push inward at any one vertex of the triangle. Do the same to the quadrilateral. Is the triangle distorted? Is the quadrilateral distorted? Is the triangle rigid? Why is it that structures like electric towers make use of triangular shapes and not quadrilaterals?
Solution :
On pushing inward at any one vertex of the triangle, the triangle is not distorted, However, the quadrilateral is distorted. Hence, a triangle is a rigid figure. This is why structures like electric towers make use of triangular shapes and not quadrilaterals.

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-4-ex-4-3/

• Basic Geometrical Ideas Class 6 Ex 4.1
• Basic Geometrical Ideas Class 6 Ex 4.2
• Basic Geometrical Ideas Class 6 Ex 4.4
• Basic Geometrical Ideas Class 6 Ex 4.5
• Basic Geometrical Ideas Class 6 Ex 4.6

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3

Question 1.
Name the angles in the given figure.

Solution :
∠A or ∠DAB; ∠B or ∠ABC; ∠C or ∠BCD; ∠Dor ∠CDA.

Question 2.
In the given diagram, name the point (s)
(a) in the interior of ∠DOE
(b) in the exterior of∠EOF
(c) on ∠EOF.

Solution :
(a) A
(b) C, A, D
(c) E, B,0, F.

Question 3.
Draw rough diagrams of two angles such that they have
(a) One point in common
(b) Two points in common
(c) Three points in common
(d) Four points in common
(e) One ray in common.
Solution :
(a)

∠AOB and ∠BOC have one point O in common.
(b) ∠AOB and ∠OBC have two points O and B in common.

(c) Not possible
(d) Not possible
(e)

∠AOB and ∠BOC have one ray \(\overrightarrow { OB }\) in common

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6.

• 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 14.5

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

Question 1.
Draw ∠POQ of measure 75° and find its line of symmetry.
Solution :

Step 1. Draw a ray \(\overline { OQ }\).
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark point P at 75°
Step 4. Join \(\overline { OP }\). Then, ∠POQ = 75°.
Step 5. With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.
Step 6. With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.
Step 7. With the same radius and with F as center, draw another arc in the interior of ∠POQ. Let
the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ which is also the line of symmetry of ∠POQ as ∠POR = ∠ROQ.

Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution :

Step 1. Draw \(\overline { OQ }\) of any length.
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark a point P at 147°.
Step 4. Join OP. Then, ∠POQ = 147°.
Step 5. With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.
Step 6. With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.
Step 7. With the same radius and with F as center, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

Question 3.
Draw a right angle and construct its bisector.
Solution :
Step 1. Draw a ray OQ.
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark point P at 90°.
Step 4. Join \(\overline { OP }\). Then, ∠POQ = 90°.
Step 5. With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.
Step 6. With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half th length Q’F.
Step 7. With the same radius and with P center, draw another arc in the interior of ∠POQ the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution :
Step 1. Draw a ray \(\overline { OQ }\).
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark a point P at 153°.
Step 4. Join OP. Then, ∠POQ = 153°.
Step 5. With O as the center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q.
Step 6. With Q’ as the center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.
Step 7. With the same radius and with F as a center, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

Step 8. With O as a center and using compasses, draw an arc that cuts both rays of ∠ROQ. Label the points of intersection as B and A.
Step 9. With A as a center, draw (in the interior of ∠ROQ) an arc whose radius is more than half the length AB.
Step 10. With the same radius and with B as a center, draw another arc in the interior of ∠ROQ. Let the two arcs intersect at S. Then, \(\overline { OS }\) is the bisector of ∠ROQ.
Step 11. With O as a center and using compasses, draw an arc that cuts both rays of ∠POR. Label the points of intersection as D and C.
Step 12. With C as a center, draw (in the interior of ∠POR) an arc whose radius is more than half the length CD.
Step 13. With the same radius and with D as centre, draw another arc in the interior of ∠POR. Let the two arcs intersect at T. Then, \(\overline { OT }\) is the bisector of ∠POR. Thus, \(\overline { OS }\), \(\overline { OR }\) and \(\overline { OT }\) divide ∠POQ = 153° into four equal parts.

Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°.
Solution :
(a) Construction of an angle of measure 60°

Step 1. Draw a line PQ and mark a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line PQ at a point say A.
Step 3. With the pointer at A (as center), now draw an arc that passes through O.
Step 4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.

(b) Construction of an angle of measure 30°

Step 1. Draw a line PQ and mark a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line PQ at a point say A.
Step 3. With the pointer at A (as center), now draw an arc that passes through O.
Step 4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
Step 5. With O as a center and using compasses, draw an arc that cuts both rays of ∠BOA. Label the points of intersection as D and C.
Step 6. With C as a center, draw (in the interior of ∠BOA) an arc whose radius is more than half the length CD.
Step 7. With the same radius and with D as a center, draw another arc in the interior of ∠BOA. Let the two arcs intersect at E. Then, \(\overline { OE }\) is the bisector of ∠BOA, i.e., ∠BOE = ∠EOA = 30°.

(c) Construction of an angle of measure 90°

Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc where the radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COL. Let
the two arcs intersect at F. Join \(\overline { OF }\). Then, \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠ FOQ = 90°.

(d) Construction of an angle of measure 120°

Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compass es at O and draw an arc of convenient radius which puts the line at A. • ‘
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the i alius > on the compasses and with B as a center, draw an arc which cuts the first arc at C. .
Step 5. Join OC. Then, ∠COA is the required angle whose measure is 120°.

(e) Construction of an angle of measure 45°

Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let the two arcs intersect at F. Join OF. Then, ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays to ∠FOQ. Label the points of the intersection as G and H.
Step 10. With H as a center, draw (in the interior of. ∠FOQ) an arc whose radius is more than half the length HG.
Step 11. With the same radius and with G as a center, draw another arc in the interior of ∠FOQ. Let the two arcs intersect at I. Join OI. Then, OI is the bisector of ∠FOH, i.e., ∠FOI = ∠IOH. Now,
∠FOI = ∠IOH = 45°.

(f) Construction of an angle of measure 135° it.

Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as the center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as the center, draw another arc in the interior of ∠COB. Let the two arcs intersect at F. Join \(\overline { OF }\). Then, \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays of ∠POF. Label the points of intersection as G and H.
Step 10. With H as center draw (in the interior of ∠POF) an arc whose radius is more than half the length HG.
Step 11. With the same radius and with G as a center, draw another arc in the interior of ∠POF. Let
the two arcs intersect at I. Join 01. Then, \(\overline { OI }\) is the bisector of ∠POF, i.e., ∠GOI = ∠IOF. Now, ∠IOQ = 135°.

Question 6.
Draw an angle of measure 45° and bisect it.
Solution :

Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A. ’
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let
the two arcs intersect at F. Join OF. Then ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays of ∠FOQ. Label the points of intersection on G and H.
Step 10. With G as a center, draw in the interior of ∠FOQ) an arc whose radius is more than half the length HG.
Step 11. With the same radius and with G as a center, draw another arc in the interior of ∠FOQ. Let the two arcs intersect at I. Join OI. Then OI is the bisector of ∠FOQ, i.e., ∠FOI = ∠IOH. Now, ∠FOI = ∠IOH = 45°.
Step 12. With O as a center and using compasses, draw an arc that cuts both rays of ∠IOH. Label the points of intersection as J and K.
Step 13. With K as a center, draw (in the interior of ∠IOH) an arc whose radius is more than half the length KJ.
Step 14. With the same radius and with J as a center, draw another arc in the interior of ∠IOH. Let the two arcs intersect at L. Join OL. Then OL is the bisector of ∠IOH, i.e., ∠IOL = ∠LOK \(22\frac { 1^{ \circ } }{ 2 }\) .

Question 7.
Draw an angle of measure 135° and bisect it.
Solution :

Step 1. Draw any line PQ and take a point O on.
Step 2. Place the pointer of the compasses at’ and draw an arc of convenient radius which cul line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let
the two arcs intersect at F. Join \(\overline { OF }\). Then \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays of ∠POF. Label the points of intersection as G and H.
Step 10. With G as a center, draw in the interior of ∠POF an arc whose radius is more than half the length HG.
Step 11. With the same radius and with H as a centre, draw another arc in the interior of ∠POF. Let
the two arcs intersect at I. Join \(\overline { OI }\). Then \(\overline { OI }\) is the bisector of ∠POF, i.e., ∠POI = ∠IOF. Now, ∠IOQ = 135°. .
Step 12. With O as centre and using compasses, draw an arc that cuts both rays of ∠IOQ. Label the points of intersection as J and K.
Step 13. With K as centre, draw (in the interior of ∠IOQ) an arc whose radius is more than half the length KJ.
Step 14. With the same radius and with J as centre, draw another arc in the interior of ∠IOQ. Let the two arcs intersect at L. Join \(\overline { OL }\). Then \(\overline { OL }\) is the bisector of ∠IOQ, i.e., ∠IOL = ∠LOQ.

Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Solution :
Steps of construction
1. Construct an angle ABC = 70°.
2. Take a line z and mark a point D on it.
3. Fix the compasses pointer on B and draw an arc which cuts the sides of ∠ABC at D and E.
4. Without changing the compasses setting, place the pointer on P and draw an arc which cuts ∠ at Q.
5. Open the compasses equal to length DE.


6. Without disturbing the radius on compasses, place its pointer at Q and draw an arc which cuts the previous arc at R.
7. Join PR and draw ray PR. It gives ∠RPQ which is the required angle whose measure is equal to the measure of ∠ABC.

Question 9.
Draw an angle of 40°. Copy its supplementary angle.
Solution :
Steps of construction


1. Draw ∠CAB = 40°.
2. Draw a line I and mark a point P on it.
3. Place the pointer of the compasses on A and draw an arc which cuts extended BA at E and AC at F.
4. Without changing the radius on compasses, place its pointer at P and draw an arc which cuts l at Q.
5. Open the length of compasses equal to EF.
6. Without disturbing the radius on compasses, place its pointer at Q and draw an arc which cuts the previous arc at R.
7. Join QR and draw ray QR. It gives ∠RQS which is the required angle whose measure

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6, drop a comment below and we will get back to you at the earliest.

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

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 \(\overline { PA }\) = \(\overline { PB }\).

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4.

• 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 14.5
• Practical Geometry Class 6 Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4

Question 1.
Draw any line segment \(\overline { AB }\) . Mark any point M on it. Through M draw a perpendicular to \(\overline { AB }\). (use ruler and compasses).
Solution :
Step 1. Given a point M on any line \(\overline { AB }\).

Step 2. With M as centre and a convenient radius, construct a part circle (arc) intersecting the line segment \(\overline { AB }\) at two points C and D.
Step 3. With C and D as centres and a radius greater than CM, construct two arcs which cut each other at N.
Step 4. Join \(\overline { MN }\). Then \(\overline { MN }\) is perpendicular to \(\overline { AB }\) at M, i.e., \(\overline { MN }\) \(\overline { AB }\).

Question 2.
Draw any’line segment \(\overline { PQ }\). Take any point R not on it. Through R draw a perpendicular to \(\overline { PQ }\). (use ruler and set-square).
Solution :
Step 1. Let \(\overline { PQ }\) be the given line segment and R be a point not on it.

Step 2. Place a set-square on \(\overline { PQ }\) such that one arm of the right angle aligns along \(\overline { PQ }\).
Step 3. Place a ruler along the edge opposite of the right angle.
Step 4. Hold the ruler fixed. Slide the set-square along the ruler all the point R touches the arm of the set-square.
Step 5. Join RS along the edge through R, meeting \(\overline { PQ }\) at S. Now \(\overline { RS }\) \(\overline { PQ }\).

Question 3.
Draw a line l and a point X on it. Through X, draw a line segment \(\overline { XY }\) perpendicular to l. Now draw a perpendicular to XY at Y. (use ruler and compasses)
Solution :
Step 1. Given a point X on a line l.
Step 2. With X as centre and a convenient radius, construct a part circle (arc) intersecting the line l at two points A and B.
Step 3. With A and B as centres and a radius greater than AX, construct two arcs which cut each other at Y.

Step 4. Join \(\overline { XY }\). Then \(\overline { XY }\) is perpendicular to l atX, i.e., \(\overline { XY }\) l.

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4, drop a comment below and we will get back to you at the earliest.