Q1 )
Construct an angle of

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then,

5. Draw the ray OF passing through D. Then,

6. Next, taking C and D as centres and with the radius more then (

7. Draw the ray OG. This ray OG is the bisector of the

Thus,

Justification:

i)Join BC.

Then, OC = OB = BC ...(By construction)

Thus,

Also,

ii)Join CD.

Then, OD = OC = CD ...(By construction)

Thus,

Also,

iii)Now, join CG and DG.

In

OD = OC ...((Radii of the same arc)

DC = CG .

(Arcs of equal radii)

and OG = OG ...(Given)

By SSS criterion, we get,

Hence,

So,

Thus,

Q2 )
Construct an angle of

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then,

5. Draw the ray OF passing through D. Then,

6. Next, taking C and D as centres and with the radius more then (

7. Draw the ray OG. This ray OG is the bisector of the

8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I, respectively.

9. Next, taking H and I as centres and with the radius more than (

10. Draw the raw OJ. This ray OJ is the required bisector of the

Thus,

Justification:

i)Join BC.

Then, OC = OB = BC ...(By construction)

Thus,

Also,

ii)Join CD.

Then, OD = OC = CD ...(By construction)

Therefore,

Thus,

Also,

iii)Now, join CG and DG.

In

OD = OC ...((Radii of the same arc)

DC = CG ...(Arcs of equal radii)

and OG = OG ...(Given)

By SSS criterion, we get,

Hence,

So,

Thus,

i.e.,

iv) Now, join HJ and IJ.

In

OI = OH ...((Radii of the same arc)

IJ = HJ ...(Arcs of equal radii)

and OJ = OJ ...(Given)

By SSS criterion, we get,

Hence,

So,

Q3 )
Construct the angles of the following measurements:

I)

II)

III)

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

I) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Draw the ray OE passing through C. Then,

4. Taking B and C as centres and with the radius more than (

5. Draw the ray OD, this ray OD is the bisector of the

II) Steps of construction:

1.Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, drawn an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then,

5. Draw the ray OF passing through D. Then,

6. Next, taking C and D as centres and with the radius more then (

7. Draw the ray OG. This ray OG is the bisector of the

Thus,

8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I, respectively.

9. Next, taking H and I as centres and with the radius more than

10. Draw the raw OJ. This ray OJ is the required bisector of the

Thus,

11. Now, taking O as centre and any radius, drawn an arc to intersect the rays OA and OJ, say at K and L, respectively.

12. Next, taking K and L as centres and with the radius more than

13. Draw the ray OM. This ray OM is the bisector of the

III) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an intersecting the previously drawn arc, say at a point c.

3. Draw the ray OE passing through C. Then,

4. Next, taking B and C as centres and with the radius more then

5. Draw the ray OD intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the the

6. Now, taking B and F as centers and with the radius more then

7. Draw the ray OG. This ray OG is the bisector of the

Q4 )
Construct the following angles and verify by measuring then by a protractor:

I)

II)

III)

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

I) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then,

5. Draw the ray OF passing through D. Then,

6. Next, taking C and D as centres and with the radius more then

7. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the

8. Next, taking C and H as centres and with the radius more than

9. Draw the ray OI. This ray OI is the bisector of the

Thus,

On measuring the IOA by protractor, we find that

Thus, the construction is verified.

II) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then,

5. Draw the ray OF passing through D. Then,

6. Next, taking C and D as centres and with the radius more then

7. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the

i.e.,

Thus,

8. Next, taking H and D as centres and with the radius more then (

9. Draw the ray OI. This ray OI is the bisector of the

Thus,

On measuring the IOA by protractor, we find that

Thus, the construction is verified.

III) Steps of construction:

1. Produce AO to A' to form ray OA'.

2. Taking O as centre and some radius, draw an arc of a circle which intersects OA at a point B and OA' at a point B'.

3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.

4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at O.

5. Draw the ray OE passing through C. Then,

6. Draw the ray OF passing through D. Then,

7. Next, taking C and D as centres and with the radius more then

8. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the

i.e.,

Thus,

9. Next, taking B' and H as centres and with the radius more then

10. Draw the ray OI. This ray OI is the bisector of the

Thus,

On measuring the IOA by protractor, we find that

Thus, the construction is verified.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Steps of construction:

1. Take a ray AX with initial point A. From AX, cut off AB = 4cm.

2. Taking A as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point B.

3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

4. Draw the ray AE passing through C.

5. Next, taking B as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point A.

6. Taking A as centre and with the same radius as in step 5, draw an arc intersecting the previously drawn arc, say at a point C.

7. Draw the ray BF passing through C. Then, ABC is the required triangle with gives side 4 cm.

Justification:

AB = BC ...(By construction)

AB = AC ...(By construction)

Thus, AB = BC = CA

Therefore,

Hence, The construction is justified.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Given: In

Steps of construction:

1. Draw the base BC = 7 cm.

2. At the point B make an

3. Cut a line segment BD equal to AB + AC = 13 cm from the ray BX.

4. Join DC.

5. Make

6. Let CY intersect BX at A.

Hence, ABC is the required triangle.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Given: In

Steps of construction:

1. Draw the base BC = 8cm

2. At the point B make an

3. Cut the line segment BD equal to AB - AC = 3.5 cm from the ray BX.

4. Join DC.

5. Draw the perpendicular bisector, say PQ of DC.

6. Let it intersect BX at a point A.

7. Join AC.

Hence, ABC is the required triangle.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Given: In

Steps of construction:

1. Draw the base QR = 6 cm.

2. At the point Q make an

3. Cut line segment QS = PR - PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.

4. Let LM intersect QX at P.

5. Join PR.

Hence, ABC is the required triangle.

Q4 )
Construct a

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Given:

In

Steps of construction:

1. Draw a line segment BC = XY + YZ + ZX = 11cm.

2. Make

3. Bisect

4. Draw perpendicular bisectors DE of XB and FG of XC.

5. Let DE intersect BC at Y and FC intersect BC at Z.

6. Join XY and XZ.

Then,

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

Answer :

Given: In

Steps of construction:

1. Draw the base BC =12cm.

2. At the point B, make an

3. Cut a line segment BD = AB + AC = 18 cm from the ray BX.

4. Join DC.

5. Draw the perpendicular bisector PQ of CD to intersect BD at a point A.

6. Join AC.

Then,

There are total 10 questions present in ncert solutions for class 9 maths chapter 11 constructions

There are total 4 long question/answers in ncert solutions for class 9 maths chapter 11 constructions

There are total 2 exercise present in ncert solutions for class 9 maths chapter 11 constructions