NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1.

## NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

Ex 11.1 Class 9 Maths Question 1.

Construct an angle of 90° at the initial point of a given ray and justify the construction.

Solution:

Steprf of Construction:

Step I : Draw \(\overline { AB }\).

Step II : Taking O as centre and having a suitable radius, draw a semicircle, which cuts \(\overline { OA }\) at B.

Step III : Keeping the radius same, divide the semicircle into three equal parts such that \(\breve { BC } =\quad \breve { CD } \quad =\quad \breve { DE }\) .

Step IV : Draw \(\overline { OC }\) and \(\overline { OD }\).

Step V : Draw \(\overline { OF }\), the bisector of ∠COD.

Thus, ∠AOF = 90°

Justification:

∵ O is the centre of the semicircle and it is divided into 3 equal parts.

∴ \(\breve { BC } =\quad \breve { CD } \quad =\quad \breve { DE }\)

⇒ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]

And, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC + ∠BOC + ∠BOC = 180°

⇒ 3∠BOC = 180°

⇒ ∠BOC = 60°

Similarly, ∠COD = 60° and ∠DOE = 60°

∵ \(\overline { OF }\) is the bisector of ∠COD

∴ ∠COF = \(\frac { 1 }{ 2 }\) ∠COD = \(\frac { 1 }{ 2 }\) (60°) = 30°

Now, ∠BOC + ∠COF = 60° + 30°

⇒ ∠BOF = 90° or ∠AOF = 90°

Ex 11.1 Class 9 Maths Question 2.

Construct an angle of 45° at the initial point of a given ray and justify the construction.

Solution:

Steps of Construction:

Stept I : Draw \(\overline { OA }\).

Step II : Taking O as centre and with a suitable radius, draw a semicircle such that it intersects \(\overline { OA }\). at B.

Step III : Taking B as centre and keeping the same radius, cut the semicircle at C. Now, taking C as centre and keeping the same radius, cut the semicircle at D and similarly, cut at E, such that \(\breve { BC } =\quad \breve { CD } \quad =\quad \breve { DE }\)

Step IV : Draw \(\overline { OC }\) and \(\overline { OD }\).

Step V : Draw \(\overline { OF }\), the angle bisector of ∠BOC.

Step VI : Draw \(\overline { OG }\), the ajngle bisector of ∠FOC.

Thus, ∠BOG = 45° or ∠AOG = 45°

Justification:

∵ \(\breve { BC } =\quad \breve { CD } \quad =\quad \breve { DE }\)

∴ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]

Since, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC = 60°

∵ \(\overline { OF }\) is the bisector of ∠BOC.

∴ ∠COF = \(\frac { 1 }{ 2 }\) ∠BOC = \(\frac { 1 }{ 2 }\)(60°) = 30° …(1)

Also, \(\overline { OG }\) is the bisector of ∠COF.

∠FOG = \(\frac { 1 }{ 2 }\)∠COF = \(\frac { 1 }{ 2 }\)(30°) = 15° …(2)

Adding (1) and (2), we get

∠COF + ∠FOG = 30° + 15° = 45°

⇒ ∠BOF + ∠FOG = 45° [∵ ∠COF = ∠BOF]

⇒ ∠BOG = 45°

Ex 11.1 Class 9 Maths Question 3.

Construct the angles of the following measurements

(i) 30°

(ii) 22 \(\frac { 1 }{ 2 } \circ \)

(iii) 15°

Solution:

(i) Angle of 30°

Steps of Construction:

Step I : Draw \(\overline { OA }\).

Step II : With O as centre and having a suitable radius, draw an arc cutting \(\overline { OA }\) at B.

Step III : With centre at B and keeping the same radius as above, draw an arc to cut the previous arc at C.

Step IV : Join \(\overline { OC }\) which gives ∠BOC = 60°.

Step V : Draw \(\overline { OD }\), bisector of ∠BOC, such that ∠BOD = \(\frac { 1 }{ 2 }\)∠BOC = \(\frac { 1 }{ 2 }\)(60°) = 30°

Thus, ∠BOD = 30° or ∠AOD = 30°

(ii) Angle of 22 \(\frac { 1 }{ 2 } \circ \)

Steps of Construction:

Step I : Draw \(\overline { OA }\).

Step II : Construct ∠AOB = 90°

Step III : Draw \(\overline { OC }\), the bisector of ∠AOB, such that

∠AOC = \(\frac { 1 }{ 2 }\)∠AOB = \(\frac { 1 }{ 2 }\)(90°) = 45°

Step IV : Now, draw OD, the bisector of ∠AOC, such that

∠AOD = \(\frac { 1 }{ 2 }\)∠AOC = \(\frac { 1 }{ 2 }\)(45°) = 22 \(\frac { 1 }{ 2 } \circ \)

Thus, ∠AOD = 22 \(\frac { 1 }{ 2 } \circ \)

(iii) Angle of 15°

Steps of Construction:

Step I : Draw \(\overline { OA }\).

Step II : Construct ∠AOB = 60°.

Step III : Draw OC, the bisector of ∠AOB, such that

∠AOC = \(\frac { 1 }{ 2 }\)∠AOB = \(\frac { 1 }{ 2 }\)(60°) = 30°

i.e., ∠AOC = 30°

Step IV : Draw OD, the bisector of ∠AOC such that

∠AOD = \(\frac { 1 }{ 2 }\)∠AOC = \(\frac { 1 }{ 2 }\)(30°) = 15°

Thus, ∠AOD = 15°

Ex 11.1 Class 9 Maths Question 4.

Construct the following angles and verify by measuring them by a protractor

(i) 75°

(ii) 105°

(iii) 135°

Solution:

Step I : Draw \(\overline { OA }\).

Step II : With O as centre and having a suitable radius, draw an arc which cuts \(\overline { OA }\) at B.

Step III : With centre B and keeping the same radius, mark a point C on the previous arc.

Step IV : With centre C and having the same radius, mark another point D on the arc of step II.

Step V : Join \(\overline { OC }\) and \(\overline { OD }\), which gives ∠COD = 60° = ∠BOC.

Step VI : Draw \(\overline { OP }\), the bisector of ∠COD, such that

∠COP = \(\frac { 1 }{ 2 }\)∠COD = \(\frac { 1 }{ 2 }\)(60°) = 30°.

Step VII: Draw \(\overline { OQ }\), the bisector of ∠COP, such that

∠COQ = \(\frac { 1 }{ 2 }\)∠COP = \(\frac { 1 }{ 2 }\)(30°) = 15°.

Thus, ∠BOQ = 60° + 15° = 75°∠AOQ = 75°

(ii) Steps of Construction:

Step I : Draw \(\overline { OA }\).

Step II : With centre O and having a suitable radius, draw an arc which cuts \(\overline { OA }\) at B.

Step III : With centre B and keeping the same radius, mark a point C on the previous arc.

Step IV : With centre C and having the same radius, mark another point D on the arc drawn in step II.

Step V : Draw OP, the bisector of CD which cuts CD at E such that ∠BOP = 90°.

Step VI : Draw \(\overline { OQ }\), the bisector of \(\breve { BC }\) such that ∠POQ = 15°

Thus, ∠AOQ = 90° + 15° = 105°

(iii) Steps of Construction:

Step I : Draw \(\overline { OP }\).

Step II : With centre O and having a suitable radius, draw an arc which cuts \(\overline { OP }\) at A

Step III : Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that \(\breve { AQ } =\quad \breve { QR } \quad =\quad \breve { RS }\) .

StepIV :Draw \(\overline { OL }\), thebisector of \(\breve { RS }\) which cuts the arc \(\breve { RS }\) at T.

Step V : Draw \(\overline { OM }\), the bisector of \(\breve { RT }\).

Thus, ∠POQ = 135°

Ex 11.1 Class 9 Maths Question 5.

Construct an equilateral triangle, given its side and justify the construction.

Solution:

pt us construct an equilateral triangle, each of whose side = 3 cm(say).

Steps of Construction:

Step I : Draw \(\overline { OA }\).

Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut \(\overline { OA }\) at B such that OB = 3 cm

Step III : Taking B as centre and radius equal to OB, draw an arc to intersect the previous arc at C.

Step IV : Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

Justification:

∵ The arcs \(\breve { OC }\) and \(\breve { BC }\) are drawn with the same radius.

∴ \(\breve { OC }\) = \(\breve { BC }\)

⇒ OC = BC [Chords corresponding to equal arcs are equal]

∵ OC = OB = BC

∴ OBC is an equilateral triangle.

### NCERT Solutions for Class 9 Maths Chapter 11 Constructions (Hindi Medium) Ex 11.1

### NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

Ex 11.2 Class 9 Maths Question 1.

Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

Solution:

Steps of Construction:

Step I : Draw \(\overline { BX }\).

Step II : Along \(\overline { BX }\), cut off a line segment BC = 7 cm.

Step III : At B, construct ∠CBY = 75°

Step IV : From \(\overline { BY }\), cut off BD = 13 cm (= AB + AC)

Step V : Join DC.

Step VI : Draw a perpendicular bisector of CD which meets BD at A.

Step VII: Join AC.

Thus, ∆ABC is the required triangle.

Ex 11.2 Class 9 Maths Question 2.

Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.

Solution:

Steps of Construction:

Step I : Draw \(\overline { BX }\).

Step II : Along \(\overline { BX }\), cut off a line segment BC = 8 cm.

Step III : At B, construct ∠CBY = 45°

Step IV : From \(\overline { BX }\), cut off BD = 3.5 cm (= AB – AC)

Step V : Join DC.

Step VI : Draw PQ, perpendicular bisector of DC, which intersects \(\overline { BY }\) at A.

Step VII: Join AC.

Thus, ∆ABC is the required triangle.

Ex 11.2 Class 9 Maths Question 3.

Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.

Solution:

Steps of Construction:

Step I : Draw \(\overline { QX }\).

Step II : Along \(\overline { QX }\), cut off a line segment QR = 6 cm.

Step III : Construct a line YQY’ such that ∠RQY = 60°.

Step IV : Cut off QS = 2 cm (= PR – PQ) on QY’.

Step V : Join SR.

Step VI : Draw MN, perpendicular bisector of SR, which intersects QY at P.

Step VII: Join PR.

Thus, ∆PQR is the required triangle.

Ex 11.2 Class 9 Maths Question 4.

Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.

Solution:

Steps of Construction:

Step I : Draw a line segment AB = 11 cm = (XY+YZ + ZX)

Step II : Construct ∠BAP = 30°

Step III : Construct ∠ABQ = 90°

Step IV : Draw AR, the bisector of ∠BAP.

Step V : Draw BS, the bisector of ∠ABQ. Let AR and BS intersect at X.

Step VI : Draw perpendicular bisector of \(\overline { AX }\), which intersects AB at Y.

Step VII: Draw perpendicular bisector of \(\overline { XB }\), which intersects AB at Z.

Step VIII: Join XY and XZ.

Thus, ∆XYZ is the required triangle.

Ex 11.2 Class 9 Maths Question 5.

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Solution:

Steps of Construction:

Step I : Draw BC = 12 cm.

Step II : At B, construct ∠CBY = 90°.

Step III : Along \(\overline { BY }\), cut off a line segment BX = 18 cm.

Step IV : Join CX.

Step V : Draw PQ, perpendicular bisector of CX, which meets BX at A.

Step VI : Join AC.

Thus, ∆ABC is the required triangle.

## NCERT Solutions for Class 9 Maths

- Chapter 1 Number systems
- Chapter 2 Polynomials
- Chapter 3 Coordinate Geometry
- Chapter 4 Linear Equations in Two Variables
- Chapter 5 Introduction to Euclid Geometry
- Chapter 6 Lines and Angles
- Chapter 7 Triangles
- Chapter 8 Quadrilaterals
- Chapter 9 Areas of Parallelograms and Triangles
- Chapter 10 Circles
- Chapter 11 Constructions
- Chapter 12 Heron’s Formula
- Chapter 13 Surface Areas and Volumes
- Chapter 14 Statistics
- Chapter 15 Probability
- Class 9 Maths (Download PDF)

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