Class IX Math NCERT Solution for Constructions 11.1
NCERT TEXTBOOK QUESTIONS SOLVED
EXERCISE: 11.1 (Page 191)
Q1. Construct an angle of 90° at the initial point of a given ray and juste the construction.
Sol. Steps of construction:
I. Draw a ray OA.
II. Taking 0 as centre and suitable radius, draw a semicircle, which cuts OA at B.
III. Keeping the radius same, divide the semicircle into three equal parts such that 
Thus, ∠AOF = 90°.
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 = 180° ⇒ ∠BOC + ∠BOC + ∠BOC = 180°
⇒3∠BOC = 180°
∴ ∠BOC = 60°
Similarly, ∠COD = 60° and ∠DOE = 60°
∵OF is the bisector of ∠COD.
Q2. Construct an angle of 45� at the initial point of a given ray and juste the construction.
Sol.
Steps of construction:
I. Draw a ray 
II. Taking O as centre and with a suitable radius, draw a semicircle such that it intersects at B.
at B.
III. Taking Bas 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.
Join OC and produce.
IV. Divide B�lC into two equal parts, such that 
V. Draw OG, the angle bisector of ∠FOC.
Thus, ∠BOG = 45°
or ∠AOG = 45°
Justification:
Q3. Construct the angles of the following measurements:
Solution: (i) Angle of 30°
Steps of construction:
I. Draw a ray OA.
II. With 0 as centre and a suitable radius, draw an arc, cutting 
at B.
at B.
III. With centre at B and the same radius as above, draw an arc to cut the previous arc at C.
IV. Join 
and produce, such that ∠BOC = 60°.
and produce, such that ∠BOC = 60°.
V. Draw 
bisector of ∠BOC, such that
bisector of ∠BOC, such that
Q4. Construct the following angles and verb by measuring them by a protractor:
(i) 75°
(ii) 105°
(iii) 135°
Sol. (i) Angle of 75°:
Hint: 75° = 60° + 15°
Steps of construction:
I. Draw .
.
II. With O as centre and having a suitable radius, draw an arc which meets at B.
at B.
III. With centre B and keeping the radius same, mark a point C on the previous arc.'
IV. With centre C and the same radius, mark another point D on the arc of step II.
VI. Draw 
, the bisector of ∠COP, such that ∠COQ = 15°
, the bisector of ∠COP, such that ∠COQ = 15°
Thus, ∠BOQ = 60° + 15° = 75°
or ∠AOQ = 75°.
(ii) Angle of 105°:
Hint: 105° = 90° + 15°
Steps of construction:
I . Draw .
.
II. With centre 0 and having a suitable radius, draw an arc which meet OA at B.
III. With centre B and keeping the same radius, mark a point C on the arc of step II.
VI. With centre C and keeping the same radius, mark another point D on the arc of step II.
V. Draw OP, the bisector of 
.
.
VI. Draw OQ, the bisector of 
.
.
Thus, ∠AOQ = 105°
(iii) Angle of 135°:
Hint: 120° + 15° = 135°
Steps of construction:
I. Draw a ray 
.
.
II. With centre O and having a suitable radius draw an arc to meet OP at A.
III. Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II.
VI. Draw 
, the bisector of 
.
, the bisector of .
V. Draw OM, the bisector of 
.
.
Thus, ∠POM = 135°.
Q5. Construct an equilateral triangle, given its side and justify the construction.
Sol. Let us construct an equilateral triangle, each of whose side = PQ
Steps of construction:
I. Draw a ray 
II. Taking O as centre and radius equal to PQ, draw an arc to cut OA at B such that OB = PQ
III. Taking B as centre and radius = OB, draw an arc, to intersect the previous arc at C.
IV. Join OC and OB.
Thus, ΔOBC is the required equilateral triangle.
Justification:
SOME CONSTRUCTIONS OF TRIANGLES
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