NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1

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

## 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.

Thus, AOF = 90°
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°
3BOC = 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.

Thus, BOG = 45° or AOG = 45°
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°

Thus, BOD = 30° or AOD = 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

Thus, AOD = 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°.

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:
(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°

Thus, AOQ = 90° + 15° = 105°

(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.

Thus, ∆OBC is the required equilateral triangle.

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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