NCERT Solutions for Class 9 Maths Chapter 11 - Exercise 11.1 Constructions

Basic Constructions

A geometrical construction means to draw geometrical figures, such as an angle, a circle, a triangle, a quadrilateral, and a polygon, etc.
We normally use all or some of the following instruments for drawing geometrical figures:

• A protractor
• A pair of compasses
• A pair of set squares
• A pair of dividers
• A graduated scale

Exercise 11.1

Q1. Construct an angle of 90º at the initial point of a given ray and justify the construction.
Ans: Steps of construction:

• Draw a ray OA.
• Taking O as the centre and suitable radius, draw a semicircle, which cuts OA at B.
• Keeping the radius the same, divide the semicircle into three equal parts such that
  
• Draw
  
• Draw, the bisector of ∠COD.

Thus, ∠AOF = 90º.

Justification:

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

∴
∠BOC = ∠COD = ∠DOE
∵ Equal chords subtend equal angles at the centre
∴ ∠BOC + ∠COD + ∠DOE = 180º
∠BOC + ∠BOC + ∠BOC = 180º
3∠BOC = 180°
∴ ∠BOC = 60º
Similarly, ∠COD = 60º and ∠DOE = 60º
∵ OF is the bisector of ∠COD.
∴ = 30º
Now, ∠BOC + ∠COF = 60º + 30º
∠BOF = 90º or ∠AOF = 90º

Q2. Construct an angle of 45º at the initial point of a given ray and justify the construction.
Ans: Steps of construction:

• Draw a ray.
• Taking O as the centre and with a suitable radius, draw a semicircle such that it intersects  at B.
  
• Taking B as centre and keeping the same radius, cut the semicircle at C. Similarly, cut the semicircle at D and E, such that . Join OC and produce.
• Divideinto two equal parts, such that
  
• Draw OG, the angle bisector of ∠FOC.

Thus, ∠BOG = 45º or ∠AOG = 45º
Justification:
∵
∴ ∠BOC = ∠COD = ∠DOE
∵ Equal chords subtend equal angles at the centre
∴ ∠BOC + ∠COD + ∠DOE = 180º
∠BOC = 60º
∵  is the bisector of  ∠BOC/
∴    .. (1)
Also,  is the bisector of  ∠COF.
∴    ... (2)
Adding (1) and (2), we get∠COF + ∠FOG = 30º + 15º = 45º
∠BOF + ∠FOG = 45º   [∵ ∠COF = ∠BOF]
∠BOG = 45º

Q3. Construct the angles of the following measurements:
(a) 30º
(b)
(c) 15º
Ans: 

(a) Angle of 30º
Steps of construction:

• Draw a ray OA.
• With O as the centre and a suitable radius, draw an arc, cutting  at B.
• With centre at B and the same radius as above, draw an arc to cut the previous arc at C.
• Join  and produce, such that ∠BOC = 60º.
•   bisector of ∠BOC, such that
  

Thus, ∠BOD = 30º
(b) Angle of
Steps of construction:

• Draw a ray
  
• Draw an angle ∠AOB = 90º
• Draw OC, the bisector of ∠AOB, such that
  
• Now, draw OD, the bisector of ∠AOC, such that

Thus,

(c) Angle of 15º
Steps of construction:

• Draw a ray.
• Construct ∠AOB = 60º.
• Draw the bisector of ∠AOB, such thati.e. ∠AOC = 30º
• Draw  the angle bisector of ∠AOC such that

Thus, ∠AOD = 15º

Q4. Construct the following angles and verify by measuring them by a protractor:
(a) 75º
(b) 105º
(c) 135º
Ans: 

(a) Angle of 75º (Hint: 75º = 60º + 15º)
Steps of construction:

• Draw .
• With O as centre and having a suitable radius, draw an arc which meets
  
• With centre B and keeping the radius same, mark a point C on the previous arc.
• With centre C and the same radius, mark another point D on the arc of step 2.
• Draw the bisector of  such that ∠COP
  
• Draw  , the bisector of ∠COP, such that ∠COQ = 15º

Thus, ∠BOQ = 60º + 15º = 75ºor ∠AOQ = 75º.

(b) Angle of 105º (Hint: 105º = 90º + 15º)

Steps of construction:

• Draw
  
• With centre O and having a suitable radius, draw an arc that meets OA at B.
• With centre B and keeping the same radius, mark a point C on the arc of step 2.
• With centre C and keeping the same radius, mark another point D on the arc of step 2.
•  Draw OP, the bisector of
  
• Draw OQ, the bisector of.

Thus, ∠AOQ = 105º

(c) Angle of 135º (Hint: 120º + 15º = 135º)
Steps of construction:

• Draw a ray
  
• With centre O and having a suitable radius draw an arc to meet OP at A.
• Keeping the same radius and starting from A, mark points Q, R and S on the arc of step 2.
• Draw the bisector ofV. Draw the bisector of

Thus, ∠POM = 135º.

Q5. Construct an equilateral triangle, given its side and justify the construction.
Ans: Let us construct an equilateral triangle, each of whose side = PQ
Steps of construction:

• Draw a ray.
  
• Taking O as centre and radius equal to PQ, draw an arc to cut OA at B such that OB = PQ
• Taking B as centre and radius = OB, draw an arc, to intersect the previous arc at C.
• Join OC and OB.

Thus, ΔOBC is the required equilateral triangle.
Justification:
∵ The are drawn with the same radius.
∴ ⇒  
∵ Chords corresponding to equal arcs are equal.
∵ OC = OB = BC
∴ ΔOBC is an equilateral triangle.