NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1
NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for 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:
Steps of Construction:
Step I : Draw a ray OB.
Step II : Taking O as centre and having a
suitable radius, draw a semicircle, which cuts OA at B.
Step III : Keeping the radius same and centre as
B, draw an arc at C and again keeping the same radius and centre as C, draw an
arc at D.
Step IV : Draw the rays OC and OD.
Step V : Draw OF, the bisector of ∠COD.
Justification:
∵ O is the centre of the semicircle and it is divided into 3 equal parts.
∴ BC = CD = 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°
∵ OF is the
bisector of ∠COD.
∴ ∠COF = ½ ∠COD = ½ (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:
Step I : Draw a ray OA.
Step II : Taking O as centre and with a suitable radius, draw a semicircle such
that it intersects 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 BC = CD = DE
Step IV : Draw OC and OD.
Step V : Draw OF, the angle
bisector of ∠BOC.
Step
VI : Draw OG, the angle
bisector of ∠FOC.
Justification:
∵ BC = CD = DE
∴ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]
Since, ∠BOC + ∠COD + ∠DOE = 180°
⇒ ∠BOC = 60°
∵ OF is the bisector of ∠BOC.
∴ ∠BOF = ∠COF = ½ ∠BOC = ½ (60°) = 30° …..…(1)
Also, OG is the bisector of ∠COF.
∠FOG = ½ ∠COF = ½ (30°) = 15° …..…(2)
Adding (1) and (2), we get
∠BOF + ∠FOG = 30° + 15° = 45°
⇒ ∠BOG = 45°
Ex 11.1 Class 9 Maths Question 3.
Construct the angles of the
following measurements.
(i) 30°
(ii) 22½ ∘
(iii) 15°
Solution:
(i) Angle of 30°
Steps of Construction:
Step I : Draw a ray OA.
Step II : With O as centre and having a suitable
radius, draw an arc cutting 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 OC which gives ∠BOC = 60°.
Step V : Draw OD, bisector of ∠BOC, such that ∠BOD = ½ ∠BOC = ½ (60°) = 30°
(ii)
Angle of 22½ ∘
Steps of Construction:
Step I : Draw a ray OA.
Step II : Construct ∠AOB = 90°
Step III : Draw OC, the bisector of ∠AOB, such that ∠AOC = ½ ∠AOB = ½ (90°) = 45°
Step IV : Now, draw OD, the bisector of ∠AOC, such that ∠AOD = ½ ∠AOC = ½ (45°) = 22½ ∘
(iii)
Angle of 15°
Steps of Construction:
Step I : Draw a ray OA.
Step II : Construct ∠AOB = 60°.
Step III : Draw OC, the bisector of ∠AOB, such that ∠AOC = ½ ∠AOB = ½ (60°) = 30°. Step IV : Draw OD, the bisector
of ∠AOC such that ∠AOD = ½ ∠AOC = ½ (30°) = 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:
(i) Steps of Construction:
Step
I : Draw a ray OA.
Step II : With O as centre and having a suitable
radius, draw an arc which cuts 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 OC and OD, which gives ∠COD = 60° = ∠BOC.
Step VI : Draw OP, the bisector of ∠COD, such that ∠COP = ½ ∠COD = ½ (60°) = 30°.
Step VII: Draw OQ, the bisector of ∠COP, such that ∠COQ = ½ ∠COP = ½ (30°) = 15°.
Thus,
∠BOQ = 60° + 15° = 75°or ∠AOQ = 75°
(ii) Steps of Construction:
Step I : Draw a ray OA.
Step II : With centre O and having a suitable
radius, draw an arc which cuts 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 OQ, the bisector of ∠EOD such that ∠POQ = 15°
(iii)
Steps of Construction:
Step I : Draw a ray OP.
Step II : With centre O and having a suitable radius, draw an arc which cuts 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 AQ = QR = RS.
Step IV : Draw OL, the bisector
of RS which cuts the arc RS at T.
Step V : Draw OM, the bisector
of RT.
Thus, ∠POM = 135°
Ex
11.1 Class 9 Maths Question 5.
Construct an equilateral triangle, given its side
and justify the construction.
Solution:
Let us construct an equilateral triangle, each of
whose side = 3 cm (say).
Steps of Construction:
Step I : Draw a ray OA.
Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut 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.
Justification:
∵ The arcs OC and BC are drawn with the same radius.
⇒ OC = BC [Chords corresponding to equal arcs
are equal]
∵ OC = OB = BC
∴ OBC is an equilateral triangle.
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